problem solving involving polynomial function with solutions

Polynomial Questions and Problems with Solutions

Polynomial questions and problems related to graphs, x and y intercepts, coefficients, degree, leading coefficients, ... with detailed solutions are presented.

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Course: Algebra 2   >   Unit 5

• Graphs of polynomials

Graphs of polynomials: Challenge problems

• Polynomial graphs: FAQ

What you should be familiar with before taking this lesson

• End behavior of polynomials
• Zeros of polynomials and their graphs

What you will do in this lesson

• (Choice A)   A
• (Choice B)   B
• (Choice C)   C
• (Choice D)   D

As x → + ∞ ‍   , y → − ∞ ‍   and as x → − ∞ ‍   , y → + ∞ ‍   .

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Zeros of Polynomial Functions

Solve real-world applications of polynomial equations.

We have now introduced a variety of tools for solving polynomial equations. Let’s use these tools to solve the bakery problem from the beginning of the section.

Example 8: Solving Polynomial Equations

A new bakery offers decorated sheet cakes for children’s birthday parties and other special occasions. The bakery wants the volume of a small cake to be 351 cubic inches. The cake is in the shape of a rectangular solid. They want the length of the cake to be four inches longer than the width of the cake and the height of the cake to be one-third of the width. What should the dimensions of the cake pan be?

Begin by writing an equation for the volume of the cake. The volume of a rectangular solid is given by [latex]V=lwh[/latex]. We were given that the length must be four inches longer than the width, so we can express the length of the cake as [latex]l=w+4[/latex]. We were given that the height of the cake is one-third of the width, so we can express the height of the cake as [latex]h=\frac{1}{3}w[/latex]. Let’s write the volume of the cake in terms of width of the cake.

Substitute the given volume into this equation.

Descartes’ rule of signs tells us there is one positive solution. The Rational Zero Theorem tells us that the possible rational zeros are [latex]\pm 3,\pm 9,\pm 13,\pm 27,\pm 39,\pm 81,\pm 117,\pm 351[/latex], and [latex]\pm 1053[/latex]. We can use synthetic division to test these possible zeros. Only positive numbers make sense as dimensions for a cake, so we need not test any negative values. Let’s begin by testing values that make the most sense as dimensions for a small sheet cake. Use synthetic division to check [latex]x=1[/latex].

Since 1 is not a solution, we will check [latex]x=3[/latex].

Since 3 is not a solution either, we will test [latex]x=9[/latex].

Synthetic division gives a remainder of 0, so 9 is a solution to the equation. We can use the relationships between the width and the other dimensions to determine the length and height of the sheet cake pan.

The sheet cake pan should have dimensions 13 inches by 9 inches by 3 inches.

A shipping container in the shape of a rectangular solid must have a volume of 84 cubic meters. The client tells the manufacturer that, because of the contents, the length of the container must be one meter longer than the width, and the height must be one meter greater than twice the width. What should the dimensions of the container be?

3 meters by 4 meters by 7 meters

• Precalculus. Authored by : Jay Abramson, et al.. Provided by : OpenStax. Located at : http://cnx.org/contents/[email protected] . License : CC BY: Attribution . License Terms : Download For Free at : http://cnx.org/contents/[email protected].

Polynomial Equation Word Problems (Mixed Operations)

These lessons help Algebra students learn how to write and solve polynomial equations for algebra word problems.

Related Pages Solving Challenging Word Problems Math Word Problems More Algebra Lessons

How To Solve Polynomial Equation Word Problem?

Example: A tree is supported by a wire anchored in the ground 5 feet from its base. The wire is 1 foot longer than the height that it reaches on the tree. Find the length of the wire.

Polynomial Equation Word Problem

Example: A gymnast dismounts the uneven parallel bars. Her height, h, depends on the time, t, that she is in the air as follows: h = -16t 2 + 8t + 8 a) How long will it take the gymnast to reach the ground? b) When will the gymnast be 8 feet above the ground?

How To Solve Word Problems With Polynomial Equations?

• The sum of a number and its square is 72. Find the number.
• The area of a triangle is 44m 2 . Find the lengths of the legs if one of the legs is 3m longer than the other leg.
• The top of a 15-foot ladder is 3 feet farther up a wall than the foo is from the bottom of the wall. How far is the ladder from the bottom of the wall?
• A projectile is launched upward from ground level with an initial speed of 98m/s. How high will it go? When will it return to the ground?

How To Write Polynomials For Word Problems?

Learn to write a polynomial for Word problems involving perimeter and area of rectangles and circles.

Learn How To Write And Solve Polynomial Equations

Learn to write and solve polynomial equations for special integers, consecutive integers.

Example 1: Find a number that is 56 less than its square. Let n be the number. Example 2: Find two consecutive odd integers whose sum is 130.

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o         Polynomial function

o         Subscript

o         Coefficient

o         Degree (of a polynomial)

o         Nonlinear function

o         Quadratic function

o         Parabola

o         Factor

o         Quadratic formula

o         Recognize a polynomial function

o         Know how many solutions a polynomial equation has

o         Learn how to factor quadratic expressions

o         Know how to use the quadratic formula

Polynomial Functions

Another type of function (which actually includes linear functions, as we will see) is the polynomial. A polynomial function is a function that is a sum of terms that each have the general form ax n , where a and n are constants and x is a variable. Thus, a polynomial function p ( x ) has the following general form:

Note that we use subscripts with the a factors (also referred to as coefficients ) to make the representation of the function more uniform and lucid. When the function has a finite number of terms, the term with the largest value of n determines the degree of the polynomial: we say that the function is a polynomial of degree n (or an n th degree polynomial). Thus, a polynomial of degree n can be written as follows:

Notice, then, that a linear function is a first-degree polynomial:

Polynomials of a degree higher than one are nonlinear functions ; that is, they do not plot graphically as a straight line. Instead, polynomials can have any particular shape depending on the number of terms and the coefficients of those terms. Finding the zeros of a polynomial function (recall that a zero of a function f ( x ) is the solution to the equation f ( x ) = 0) can be significantly more complex than finding the zeros of a linear function. For simplicity, we will focus primarily on second-degree polynomials, which are also called quadratic functions.

Quadratic Functions

A quadratic function is a polynomial of degree two. Because it is common, we'll use the following notation when discussing quadratics:

f ( x ) = ax 2 + bx + c

Let's take a look at the shape of a quadratic function on a graph. We'll just graph f ( x ) = x 2 .

The graph shows that the function is obviously nonlinear; the shape of a quadratic is actually a parabola . The quadratic function can be oriented either up (when a > 0, as in the above graph) or down (when a < 0), and it can be translated to any position in the plane (through variation of b and c ). Note also that, depending on its location, the parabola can cross the x -axis in two places, in one place (as in the above graph), or nowhere at all. Generally, however, a quadratic equation has two solutions, which may or may not correspond to the same number. Furthermore, the solutions to a quadratic equation may be complex numbers. (In fact, it is generally the case that an equation consisting of a polynomial of degree n has n solutions.) There are two essential approaches to solving a quadratic equation: factoring and the quadratic formula.

Solving Quadratic Equations: Factoring

A solution to any equation f ( x ) = 0 is the value or values of x for which f ( x ) is zero (that is, for which it crosses the x -axis). Let's look at the example quadratic function above:

f ( x ) = x 2 = ( x )( x )

What we have done here is factor the original expression. We can now see that this quadratic function has two zeros, both of which are at x = 0. But what if the function is more complicated? Let's first consider the following general quadratic expression. We can expand the expression by carefully applying the rule of distributivity.

f ( x ) = ( px + q )( rx + s ) = px ( rx + s ) + q ( rx + s ) = prx 2 + pxs + qrx + qs

f ( x ) = prx 2 + ( ps + qr ) x + qs

Note that this is fundamentally the same form as f ( x ) = ax 2 + bx + c , but different names are used for the coefficients. The zeros of the function are the x values for which either factor is equal to zero-thus, we can see that there are generally two solutions to a quadratic. We can find these solutions by setting each factor equal to zero and solving for x . Of course, going from the factored form to the standard form is much more difficult than the reverse process, but in many cases the factored form can be found without too much difficulty. The process of factoring a quadratic function is usually a process of trial and error, but with practice, you can learn to spot how to factor some quadratics.

Practice Problem: Factor the expression x 2 - 4.

Solution : This is our first problem involving factoring; start with what you know. We can write a quadratic expression as follows:

( px + q )( rx + s ) = prx 2 + ( ps + qr ) x + qs

More simply, we can assume p = r = 1 and write

( x + q )( x + s ) = x 2 + ( s + q ) x + qs

Now, we see that s must be equal to – q and that the product of s and q must be –4.

s = – q

sq = –4

–s 2 = –4

Now, s can be either 2 or –2, but it can only be one or the other. Let's just pick one: s = 2. Then, q = –2. Let's write the new expression and check our result:

( x + 2)( x – 2) = x ( x – 2) + 2( x – 2) = x 2  – 2 x + 2 x  – 4 = x 2  – 4

This result checks out.

Practice Problem : Factor the expression x 2 + 3 x + 2.

Solution : We can follow the same approach here as in the previous problem: our solution should have the following form:

( x + a )( x + b ) = x 2 + ( a + b ) x + ab = x 2 + 3 x + 2

So, we know that ab = 2 and a + b = 3. Here, trial and error is the best route. By inspection, a = 1 and b = 2 satisfies these equations.

ab = (1)(2) = 2 

a + b = 1 + 2 = 3

The factored form is thus ( x + 1)( x + 2). (Note that the zeros of the function are then x = –1 and x = –2.)

Practice Problem : Factor the expression x 2 + 1.

Solution : Here we run into a slight problem. Let's take a look.

( x + q )( x + s ) = x 2 + ( s + q ) x + qs = x 2 + 1

We see that q = – s , but also qs = 1.

qs = – s 2 = 1

s 2 = –1

Thus, we see that complex numbers can be involved in factors and zeros of quadratic functions. We can still handle this problem recalling what we already know about complex numbers.

q = – i

So, the factored expression is ( x + i )( x  – i ). Let's expand this just to check.

( x + i )( x – i ) = x 2 + ix  – ix  – i 2 = x 2  – (–1) = x 2 + 1

The zeros of this expression are then i and – i . (As it turns out, complex solutions exist when the graph of the quadratic function does not cross the x -axis anywhere--try graphing the expression in this problem to see.)

Solving Quadratic Functions: The Quadratic Formula

A more general and direct way to find the zeros of a quadratic function (and, also, to find the factors) is through the quadratic formula . We will not derive the quadratic formula here, but suffice it to say you can derive it using algebra. Given a quadratic function ax 2 + bx + c , the zeros of the function are at

Practice Problem : Find the solutions to the equation x 2 – 4 = 0.

Solution : We can use the factoring approach, as we did in a previous practice problem, or we can use the quadratic formula with a = 1, b = 0, and c = –4. Let's try this latter approach to compare.

The solutions to the equation are then x = 2 and x = –2. In some cases, the use of the quadratic equation is faster, even though factoring of the quadratic expression is still an option. 

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4.4 Solve Polynomial Equations by Factoring

Learning objectives.

• Review general strategies for factoring.
• Solve polynomial equations by factoring.
• Find roots of a polynomial function.
• Find polynomial equations given the solutions.

Reviewing General Factoring Strategies

We have learned various techniques for factoring polynomials with up to four terms. The challenge is to identify the type of polynomial and then decide which method to apply. The following outlines a general guideline for factoring polynomials:

• Check for common factors. If the terms have common factors, then factor out the greatest common factor (GCF).

Determine the number of terms in the polynomial.

• Factor four-term polynomials by grouping.
• Factor trinomials (3 terms) using “trial and error” or the AC method.

Factor binomials (2 terms) using the following special products:

• Look for factors that can be factored further.
• Check by multiplying.

Note : If a binomial is both a difference of squares and a difference cubes, then first factor it as difference of squares. This will result in a more complete factorization. In addition, not all polynomials with integer coefficients factor. When this is the case, we say that the polynomial is prime.

If an expression has a GCF, then factor this out first. Doing so is often overlooked and typically results in factors that are easier to work with. Furthermore, look for the resulting factors to factor further; many factoring problems require more than one step. A polynomial is completely factored when none of the factors can be factored further.

Factor: 54 x 4 − 36 x 3 − 24 x 2 + 16 x .

This four-term polynomial has a GCF of 2 x . Factor this out first.

54 x 4 − 36 x 3 − 24 x 2 + 16 x = 2 x ( 27 x 3 − 18 x 2 − 12 x + 8 )

Now factor the resulting four-term polynomial by grouping and look for resulting factors to factor further.

Answer: 2 x ( 3 x − 2 ) 2 ( 3 x + 2 ) . The check is left to the reader.

Factor: x 4 − 3 x 2 − 4 .

This trinomial does not have a GCF.

x 4 − 3 x 2 − 4 = ( x 2             ) ( x 2             ) = ( x 2 + 1 ) ( x 2 − 4 )          D i f f e r e n c e   o f   s q u a r e s = ( x 2 + 1 ) ( x + 2 ) ( x − 2 )

The factor ( x 2 + 1 ) is prime and the trinomial is completely factored.

Answer: ( x 2 + 1 ) ( x + 2 ) ( x − 2 )

Factor: x 6 + 6 x 3 − 16 .

Begin by factoring x 6 = x 3 ⋅ x 3 and look for the factors of 16 that add to 6.

The factor ( x 3 − 2 ) cannot be factored any further using integers and the factorization is complete.

Answer: ( x 3 − 2 ) ( x + 2 ) ( x 2 + 2 x + 4 )

Try this! Factor: 9 x 4 + 17 x 2 − 2

Answer: ( 3 x + 1 ) ( 3 x − 1 ) ( x 2 + 2 )

Solving Polynomial Equations by Factoring

In this section, we will review a technique that can be used to solve certain polynomial equations. We begin with the zero-product property A product is equal to zero if and only if at least one of the factors is zero. :

a ⋅ b = 0    if and only if    a = 0  or  b = 0

The zero-product property is true for any number of factors that make up an equation. In other words, if any product is equal to zero, then at least one of the variable factors must be equal to zero. If an expression is equal to zero and can be factored into linear factors, then we will be able to set each factor equal to zero and solve for each equation.

Solve: 2 x ( x − 4 ) ( 5 x + 3 ) = 0 .

Set each variable factor equal to zero and solve.

2 x = 0   or   x − 4 = 0   or   5 x + 3 = 0 2 x 2 = 0 2 x = 4 5 x 5 = − 3 5 x = 0 x = − 3 5

To check that these are solutions we can substitute back into the original equation to see if we obtain a true statement. Note that each solution produces a zero factor. This is left to the reader.

Answer: The solutions are 0, 4, and − 3 5 .

Of course, most equations will not be given in factored form.

Solve: 4 x 3 − x 2 − 100 x + 25 = 0 .

Begin by factoring the left side completely.

4 x 3 − x 2 − 100 x + 25 = 0 F a c t o r   b y   g r o u p i n g . x 2 ( 4 x − 1 ) − 25 ( 4 x − 1 ) = 0 ( 4 x − 1 ) ( x 2 − 25 ) = 0 F a c t o r   a s   a   d i f f e r e n c e   o f   s q u a r e s . ( 4 x − 1 ) ( x + 5 ) ( x − 5 ) = 0

Set each factor equal to zero and solve.

4 x − 1 = 0 or x + 5 = 0 or x − 5 = 0 4 x = 1 x = − 5 x = 5 x = 1 4

Answer: The solutions are 1 4 , −5, and 5.

Using the zero-product property after factoring an equation that is equal to zero is the key to this technique. However, the equation may not be given equal to zero, and so there may be some preliminary steps before factoring. The steps required to solve by factoring The process of solving an equation that is equal to zero by factoring it and then setting each variable factor equal to zero. are outlined in the following example.

Solve: 15 x 2 + 3 x − 8 = 5 x − 7 .

Step 1: Express the equation in standard form, equal to zero. In this example, subtract 5 x from and add 7 to both sides.

15 x 2 + 3 x − 8 = 5 x − 7 15 x 2 − 2 x − 1 = 0

Step 2: Factor the expression.

( 3 x − 1 ) ( 5 x + 1 ) = 0

Step 3: Apply the zero-product property and set each variable factor equal to zero.

3 x − 1 = 0         or         5 x + 1 = 0

Step 4: Solve the resulting linear equations.

3 x − 1 = 0 or 5 x + 1 = 0 3 x = 1 5 x = − 1 x = 1 3 x = − 1 5

Answer: The solutions are 1 3 and − 1 5 . The check is optional.

Solve: ( 3 x + 2 ) ( x + 1 ) = 4 .

This quadratic equation appears to be factored; hence it might be tempting to set each factor equal to 4. However, this would lead to incorrect results. We must rewrite the equation equal to zero, so that we can apply the zero-product property.

( 3 x + 2 ) ( x + 1 ) = 4 3 x 2 + 3 x + 2 x + 2 = 4 3 x 2 + 5 x + 2 = 4 3 x 2 + 5 x − 2 = 0

Once it is in standard form, we can factor and then set each factor equal to zero.

                ( 3 x − 1 ) ( x + 2 ) = 0 3 x − 1 = 0               or           x + 2 = 0               3 x = 1                                                 x = − 2                     x = 1 3

Answer: The solutions are 1 3 and −2.

Finding Roots of Functions

Recall that any polynomial with one variable is a function and can be written in the form,

f ( x ) = a n x n + a n − 1 x n − 1 + ⋯ + a 1 x + a 0

A root A value in the domain of a function that results in zero. of a function is a value in the domain that results in zero. In other words, the roots occur when the function is equal to zero, f ( x ) = 0 .

Find the roots: f ( x ) = ( x + 2 ) 2 − 4 .

To find roots we set the function equal to zero and solve.

f ( x ) = 0 ( x + 2 ) 2 − 4 = 0 x 2 + 4 x + 4 − 4 = 0 x 2 + 4 x = 0 x ( x + 4 ) = 0

Next, set each factor equal to zero and solve.

x = 0   or     x + 4 = 0 x = − 4

We can show that these x -values are roots by evaluating.

f ( 0 ) = ( 0 + 2 ) 2 − 4         f ( − 4 ) = ( − 4 + 2 ) 2 − 4 = 4 − 4   = ( − 2 ) 2 − 4 = 0         ✓ = 4 − 4 = 0         ✓

Answer: The roots are 0 and −4.

If we graph the function in the previous example we will see that the roots correspond to the x -intercepts of the function. Here the function f is a basic parabola shifted 2 units to the left and 4 units down.

Find the roots: f ( x ) = x 4 − 5 x 2 + 4 .

f ( x ) = 0 x 4 − 5 x 2 + 4 = 0 ( x 2 − 1 ) ( x 2 − 4 ) = 0 ( x + 1 ) ( x − 1 ) ( x + 2 ) ( x − 2 ) = 0

x + 1 = 0         or         x − 1 = 0         or         x + 2 = 0         or         x − 2 = 0 x = − 1 x = 1 x = − 2 x = 2

Answer: The roots are −1, 1, −2, and 2.

Graphing the previous function is not within the scope of this course. However, the graph is provided below:

Notice that the degree of the polynomial is 4 and we obtained four roots. In general, for any polynomial function with one variable of degree n , the fundamental theorem of algebra Guarantees that there will be as many (or fewer) roots to a polynomial function with one variable as its degree. guarantees n real roots or fewer. We have seen that many polynomials do not factor. This does not imply that functions involving these unfactorable polynomials do not have real roots. In fact, many polynomial functions that do not factor do have real solutions. We will learn how to find these types of roots as we continue in our study of algebra.

Find the roots: f ( x ) = − x 2 + 10 x − 25 .

f ( x ) = 0 − x 2 + 10 x − 25 = 0 − ( x 2 − 10 x + 25 ) = 0 − ( x − 5 ) ( x − 5 ) = 0

Next, set each variable factor equal to zero and solve.

x − 5 = 0 or x − 5 = 0 = 5 x = 5

A solution that is repeated twice is called a double root A root that is repeated twice. . In this case, there is only one solution.

Answer: The root is 5.

The previous example shows that a function of degree 2 can have one root. From the factoring step, we see that the function can be written

f ( x ) = − ( x − 5 ) 2

In this form, we can see a reflection about the x -axis and a shift to the right 5 units. The vertex is the x -intercept, illustrating the fact that there is only one root.

Try this! Find the roots of f ( x ) = x 3 + 3 x 2 − x − 3 .

Answer: ±1, −3

Assuming dry road conditions and average reaction times, the safe stopping distance in feet is given by d ( x ) = 1 20 x 2 + x , where x represents the speed of the car in miles per hour. Determine the safe speed of the car if you expect to stop in 40 feet.

We are asked to find the speed x where the safe stopping distance d ( x ) = 40 feet.

d ( x ) = 40 1 20 x 2 + x = 40

To solve for x , rewrite the resulting equation in standard form. In this case, we will first multiply both sides by 20 to clear the fraction.

20 ( 1 20 x 2 + x ) = 20 ( 40 ) x 2 + 20 x = 800 x 2 + 20 x − 800 = 0

Next factor and then set each factor equal to zero.

x 2 + 20 x − 800 = 0 ( x + 40 ) ( x − 20 ) = 0 x + 40 = 0 o r x − 20 = 0 x = − 40 x = 20

The negative answer does not make sense in the context of this problem. Consider x = 20 miles per hour to be the only solution.

Answer: 20 miles per hour

Finding Equations with Given Solutions

We can use the zero-product property to find equations, given the solutions. To do this, the steps for solving by factoring are performed in reverse.

Find a quadratic equation with integer coefficients, given solutions − 3 2 and 1 3 .

Given the solutions, we can determine two linear factors. To avoid fractional coefficients, we first clear the fractions by multiplying both sides by the denominator.

x = − 3 2 or x = 1 3 2 x = − 3 3 x = 1 2 x + 3 = 0 3 x − 1 = 0

The product of these linear factors is equal to zero when x = − 3 2 or x = 1 3 .

( 2 x + 3 ) ( 3 x − 1 ) = 0

Multiply the binomials and present the equation in standard form.

6 x 2 − 2 x + 9 x − 3 = 0 6 x 2 + 7 x − 3 = 0

We may check our equation by substituting the given answers to see if we obtain a true statement. Also, the equation found above is not unique and so the check becomes essential when our equation looks different from someone else’s. This is left as an exercise.

Answer: 6 x 2 + 7 x − 3 = 0

Find a polynomial function with real roots 1, −2, and 2.

Given solutions to f ( x ) = 0 we can find linear factors.

x = 1 or x = − 2 or x = 2 x − 1 = 0 x + 2 = 0 x − 2 = 0

Apply the zero-product property and multiply.

( x − 1 ) ( x + 2 ) ( x − 2 ) = 0 ( x − 1 ) ( x 2 − 4 ) = 0 x 3 − 4 x − x 2 + 4 = 0 x 3 − x 2 − 4 x + 4 = 0

Answer: f ( x ) = x 3 − x 2 − 4 x + 4

Try this! Find a polynomial equation with integer coefficients, given solutions 1 2 and − 3 4 .

Answer: 8 x 2 + 2 x − 3 = 0

Key Takeaways

• Factoring and the zero-product property allow us to solve equations.
• To solve a polynomial equation, first write it in standard form. Once it is equal to zero, factor it and then set each variable factor equal to zero. The solutions to the resulting equations are the solutions to the original.
• Not all polynomial equations can be solved by factoring. We will learn how to solve polynomial equations that do not factor later in the course.
• A polynomial function can have at most a number of real roots equal to its degree. To find roots of a function, set it equal to zero and solve.
• To find a polynomial equation with given solutions, perform the process of solving by factoring in reverse.

Topic Exercises

Part a: general factoring.

Factor completely.

50 x 2 − 18

12 x 3 − 3 x

10 x 3 + 65 x 2 − 35 x

15 x 4 + 7 x 3 − 4 x 2

6 a 4 b − 15 a 3 b 2 − 9 a 2 b 3

8 a 3 b − 44 a 2 b 2 + 20 a b 3

36 x 4 − 72 x 3 − 4 x 2 + 8 x

20 x 4 + 60 x 3 − 5 x 2 − 15 x

3 x 5 + 2 x 4 − 12 x 3 − 8 x 2

10 x 5 − 4 x 4 − 90 x 3 + 36 x 2

x 4 − 23 x 2 − 50

2 x 4 − 31 x 2 − 16

− 2 x 5 − 6 x 3 + 8 x

− 36 x 5 + 69 x 3 + 27 x

54 x 5 − 78 x 3 + 24 x

4 x 6 − 65 x 4 + 16 x 2

x 6 − 7 x 3 − 8

x 6 − 25 x 3 − 54

3 x 6 + 4 x 3 + 1

27 x 6 − 28 x 3 + 1

Part B: Solving Polynomial Equations by Factoring

( 6 x − 5 ) ( x + 7 ) = 0

( x + 9 ) ( 3 x − 8 ) = 0

5 x ( 2 x − 5 ) ( 3 x + 1 ) = 0

4 x ( 5 x − 1 ) ( 2 x + 3 ) = 0

( x − 1 ) ( 2 x + 1 ) ( 3 x − 5 ) = 0

( x + 6 ) ( 5 x − 2 ) ( 2 x + 9 ) = 0

( x + 4 ) ( x − 2 ) = 16

( x + 1 ) ( x − 7 ) = 9

( 6 x + 1 ) ( x + 1 ) = 6

( 2 x − 1 ) ( x − 4 ) = 39

x 2 − 15 x + 50 = 0

x 2 + 10 x − 24 = 0

3 x 2 + 2 x − 5 = 0

2 x 2 + 9 x + 7 = 0

1 10 x 2 − 7 15 x − 1 6 = 0

1 4 − 4 9 x 2 = 0

6 x 2 − 5 x − 2 = 30 x + 4

6 x 2 − 9 x + 15 = 20 x − 13

5 x 2 − 23 x + 12 = 4 ( 5 x − 3 )

4 x 2 + 5 x − 5 = 15 ( 3 − 2 x )

( x + 6 ) ( x − 10 ) = 4 ( x − 18 )

( x + 4 ) ( x − 6 ) = 2 ( x + 4 )

4 x 3 − 14 x 2 − 30 x = 0

9 x 3 + 48 x 2 − 36 x = 0

1 3 x 3 − 3 4 x = 0

1 2 x 3 − 1 50 x = 0

− 10 x 3 − 28 x 2 + 48 x = 0

− 2 x 3 + 15 x 2 + 50 x = 0

2 x 3 − x 2 − 72 x + 36 = 0

4 x 3 − 32 x 2 − 9 x + 72 = 0

45 x 3 − 9 x 2 − 5 x + 1 = 0

x 3 − 3 x 2 − x + 3 = 0

x 4 − 5 x 2 + 4 = 0

4 x 4 − 37 x 2 + 9 = 0

Find the roots of the given functions.

f ( x ) = x 2 + 10 x − 24

f ( x ) = x 2 − 14 x + 48

f ( x ) = − 2 x 2 + 7 x + 4

f ( x ) = − 3 x 2 + 14 x + 5

f ( x ) = 16 x 2 − 40 x + 25

f ( x ) = 9 x 2 − 12 x + 4

g ( x ) = 8 x 2 + 3 x

g ( x ) = 5 x 2 − 30 x

p ( x ) = 64 x 2 − 1

q ( x ) = 4 x 2 − 121

f ( x ) = 1 5 x 3 − 1 x 2 − 1 20 x + 1 4

f ( x ) = 1 3 x 3 + 1 2 x 2 − 4 3 x − 2

g ( x ) = x 4 − 13 x 2 + 36

g ( x ) = 4 x 4 − 13 x 2 + 9

f ( x ) = ( x + 5 ) 2 − 1

g ( x ) = − ( x + 5 ) 2 + 9

f ( x ) = − ( 3 x − 5 ) 2

g ( x ) = − ( x + 2 ) 2 + 4

Given the graph of a function, determine the real roots.

The sides of a square measure x − 2 units. If the area is 36 square units, then find x .

The sides of a right triangle have lengths that are consecutive even integers. Find the lengths of each side. (Hint: Apply the Pythagorean theorem)

The profit in dollars generated by producing and selling n bicycles per week is given by the formula P ( n ) = − 5 n 2 + 400 n − 6000 . How many bicycles must be produced and sold to break even?

The height in feet of an object dropped from the top of a 64-foot building is given by h ( t ) = − 16 t 2 + 64 where t represents the time in seconds after it is dropped. How long will it take to hit the ground?

A box can be made by cutting out the corners and folding up the edges of a square sheet of cardboard. A template for a cardboard box of height 2 inches is given.

What is the length of each side of the cardboard sheet if the volume of the box is to be 98 cubic inches?

The height of a triangle is 4 centimeters less than twice the length of its base. If the total area of the triangle is 48 square centimeters, then find the lengths of the base and height.

A uniform border is to be placed around an 8 × 10 inch picture.

If the total area including the border must be 168 square inches, then how wide should the border be?

The area of a picture frame including a 3-inch wide border is 120 square inches.

If the width of the inner area is 2 inches less than its length, then find the dimensions of the inner area.

Assuming dry road conditions and average reaction times, the safe stopping distance in feet is given by d ( x ) = 1 20 x 2 + x where x represents the speed of the car in miles per hour. Determine the safe speed of the car if you expect to stop in 75 feet.

A manufacturing company has determined that the daily revenue in thousands of dollars is given by the formula R ( n ) = 12 n − 0.6 n 2 where n represents the number of palettes of product sold ( 0 ≤ n < 20 ) . Determine the number of palettes sold in a day if the revenue was 45 thousand dollars.

Part C: Finding Equations with Given Solutions

Find a polynomial equation with the given solutions.

Find a function with the given roots.

2 5 , − 1 3

5 double root

−3 double root

Recall that if | X | = p , then X = − p or X = p . Use this to solve the following absolute value equations.

| x 2 − 8 | = 8

| 2 x 2 − 9 | = 9

| x 2 − 2 x − 1 | = 2

| x 2 − 8 x + 14 | = 2

| 2 x 2 − 4 x − 7 | = 9

| x 2 − 3 x − 9 | = 9

Part D: Discussion Board

Explain to a beginning algebra student the difference between an equation and an expression.

What is the difference between a root and an x -intercept? Explain.

Create a function with three real roots of your choosing. Graph it with a graphing utility and verify your results. Share your function on the discussion board.

Research and discuss the fundamental theorem of algebra.

2 ( 5 x + 3 ) ( 5 x − 3 )

5 x ( x + 7 ) ( 2 x − 1 )

3 a 2 b ( 2 a + b ) ( a − 3 b )

4 x ( x − 2 ) ( 3 x + 1 ) ( 3 x − 1 )

x 2 ( 3 x + 2 ) ( x + 2 ) ( x − 2 )

( x 2 + 2 ) ( x + 5 ) ( x − 5 )

• − 2 x ( x 2 + 4 ) ( x − 1 ) ( x + 1 )

6 x ( x + 1 ) ( x − 1 ) ( 3 x + 2 ) ( 3 x − 2 )

( x + 1 ) ( x 2 − x + 1 ) ( x − 2 ) ( x 2 + 2 x + 4 )

( 3 x 3 + 1 ) ( x + 1 ) ( x 2 − x + 1 )

0, 5 2 , − 1 3

− 1 2 , 1, 5 3

− 5 3 , 1 2

0, − 3 2 , 5

± 1 3 , 1 5

−3, −1, 0, 2

20 or 60 bicycles

30 miles per hour

x 2 − 2 x − 15 = 0

3 x 2 − 7 x + 2 = 0

x 2 + 4 x = 0

x 2 − 49 = 0

x 3 − x 2 − 9 x + 9 = 0

f ( x ) = 6 x 2 − 7 x + 2

f ( x ) = 16 x 2 − 9

f ( x ) = x 2 − 10 x + 25

f ( x ) = x 3 − 2 x 2 − 3 x

Answer may vary

PRACTICE PROBLEMS ON SOLVING POLYNOMIAL EQUATIONS

(1)  Solve the cubic equation : 2x 3 − x 2 −18x + 9 = 0, if sum of two of its roots vanishes         Solution

(2)   Solve the equation 9x 3 − 36x 2 + 44x −16 = 0 if the roots form an arithmetic progression.         Solution

(3)  Solve the equation 3x 3  − 26x 2  + 52x − 24 = 0 if its roots form a geometric progression.         Solution

(4)  Determine k and solve the equation 2x 3  − 6x 2  + 3x + k = 0 if one of its roots is twice the sum of the other two  roots.         Solution

(5)  Find all zeros of the polynomial x 6  − 3x 5  − 5x 4  + 22x 3  − 39x 2  − 39x + 135, if it is known that 1 + 2i an d  √ 3 are two of its zeros.              Solution

(6)   Solve the cubic equation

(i) 2x 3  − 9x 2  +10x = 3

(ii)  8x 3  − 2x 2  − 7x + 3 = 0.        Solution

(7)  Solve the equation x 4  −14x 2  + 45 = 0    Solution

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5.3: Applications of Polynomials

• Last updated
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• Page ID 19879

• David Arnold
• College of the Redwoods

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In this section we investigate real-world applications of polynomial functions.

Example \(\PageIndex{1}\)

The average price of a gallon of gas at the beginning of each month for the period starting in November 2010 and ending in May 2011 are given in the margin. The data is plotted in Figure \(\PageIndex{1}\) and fitted with the following third degree polynomial, where t is the number of months that have passed since October of 2010.

\[p(t)=-0.0080556 t^{3}+0.11881 t^{2}-0.30671 t+3.36 \label{Eq5.3.1} \]

Use the graph and then the polynomial to estimate the price of a gallon of gas in California in February 2011.

\[\begin{array}{c|c}\hline \text { Month } & {\text { Price }} \\ \hline \text { Nov. } & {3.14} \\ {\text { Dec. }} & {3.21} \\ {\text { Jan }} & {3.31} \\ {\text { Mar. }} & {3.87} \\ {\text { Apr. }} & {4.06} \\ {\text { May }} & {4.26} \\ \hline\end{array} \nonumber \]

Locate February (\(t = 4\)) on the horizontal axis. From there, draw a vertical arrow up to the graph, and from that point of intersection, a second horizontal arrow over to the vertical axis (see Figure \(\PageIndex{2}\)). It would appear that the price per gallon in February was approximately \(\$3.51\).

Next, we’ll use the fitted third degree polynomial to approximate the price per gallon for the month of February, 2011. Start with the function defined by Equation \ref{Eq5.3.1} and substitute \(4\) for \(t\).

\[\begin{aligned} p(t) &=-0.0080556 t^{3}+0.11881 t^{2}-0.30671 t+3.36 \\ p(4) &=-0.0080556(4)^{3}+0.11881(4)^{2}-0.30671(4)+3.36 \end{aligned} \nonumber \]

Use the calculator to evaluate \(p(4)\) (see Figure \(\PageIndex{3}\)). Rounding to the nearest

penny, the price in February was \(\$3.52\) per gallon.

Example \(\PageIndex{2}\)

If a projectile is fired into the air, its height above ground at any time is given by the formula

\[y=y_{0}+v_{0} t-\dfrac{1}{2} g t^{2} \label{Eq5.3.2} \]

\(y\) = height above ground at time \(t\)

\(y_0\) = initial height above ground at time \(t = 0\)

\(v_0\) = initial velocity at time \(t = 0\)

\(g\) = acceleration due to gravity

\(t\) = time passed since projectile’s firing

If a projectile is launched with an initial velocity of \(100\) meters per second (\(100\)m/s) from a rooftop \(8\) meters (\(8\)m) above ground level, at what time will the projectile first reach a height of \(400\) meters (\(400\)m)? Note: Near the earth’s surface, the acceleration due to gravity is approximately \(9.8\) meters per second per second (\(9.8\)(m/s)/s or \(9.8\)m/s 2 ).

We’re given the initial height is \(y_0 = 8\)m, the initial velocity is \(v_0 = 100\)m/s, and the acceleration due to gravity is \(g =9.8\)m/s 2 . Substitute these values in Equation \ref{Eq5.3.2}, then simplify to produce the following result:

\[\begin{array}{l}{y=y_{0}+v_{0} t-\dfrac{1}{2} g t^{2}} \\ {y=8+100 t-\dfrac{1}{2}(9.8) t^{2}} \\ {y=8+100 t-4.9 t^{2}}\end{array} \nonumber \]

Enter \(y=8+100 t-4.9 t^{2}\) as \(\mathbf{Y} \mathbf{1}=\mathbf{8}+100^{*} \mathbf{X}-4.9^{*} \mathbf{X} \wedge \mathbf{2}\) in the Y= menu (see the first image in Figure \(\PageIndex{4}\)). After some experimentation, we settled on the WINDOW parameters shown in the second image in Figure \(\PageIndex{4}\). Push the GRAPH button to produce the graph of \(y=8+100 t-4.9 t^{2}\) shown in the third image Figure \(\PageIndex{4}\).

In this example, the horizontal axis is actually the \(t\)-axis. So when we set \(\mathrm{Xmin}\) and \(\mathrm{Xmax}\), we’re actually setting bounds on the \(t\)-axis.

To find when the projectile reaches a height of \(400\) meters (\(400\)m), substitute \(400\) for \(y\) to obtain:

\[400=8+100 t-4.9 t^{2} \label{Eq5.3.3} \]

Enter the left-hand side of Equation \ref{Eq5.3.3} into \(\mathbf{Y} \boldsymbol{2}\) in the Y= menu , as shown in the first image in Figure \(\PageIndex{5}\). Push the GRAPH button to produce the result shown in the second image in Figure \(\PageIndex{5}\). Note that there are two points of intersection, which makes sense as the projectile hits \(400\) meters on the way up and \(400\) meters on the way down.

To find the first point of intersection, select 5:intersect from the CALC menu. Press ENTER in response to “First curve,” then press ENTER again in response to “Second curve.” For your guess, use the arrow keys to move the cursor closer to the first point of intersection than the second. At this point, press ENTER in response to “Guess.” The result is shown in the third image in Figure \(\PageIndex{5}\). The projectile first reaches a height of \(400\) meters at approximately \(5.2925359\) seconds after launch.

The parabola shown in Figure \(\PageIndex{6}\) is not the actual flight path of the projectile. The graph only predicts the height of the projectile as a function of time.

Reporting the solution on your homework: Duplicate the image in your calculator’s viewing window on your homework page. Use a ruler to draw all lines, but freehand any curves.

• Label the horizontal and vertical axes with \(t\) and \(y\), respectively (see Figure \(\PageIndex{6}\)). Include the units (seconds (s) and meters (m)).
• Place your WINDOW parameters at the end of each axis (see Figure \(\PageIndex{6}\)). Include the units (seconds (s) and meters (m)).
• Label each graph with its equation (see Figure \(\PageIndex{6}\)).
• Draw a dashed vertical line through the first point of intersection. Shade and label the point (with its \(t\)-value) where the dashed vertical line crosses the \(t\)-axis. This is the first solution of the equation \(400 = 8+100t−4.9t^2\) (see Figure \(\PageIndex{6}\)).

Rounding to the nearest tenth of a second, it takes the projectile approximately \(t ≈ 5.3\) seconds to first reach a height of \(400\) meters.

The phrase “shade and label the point” means fill in the point on the \(t\)-axis, then write the \(t\)-value of the point just below the shaded point.

Exercise \(\PageIndex{2}\)

If a projectile is launched with an initial velocity of \(60\) meters per second from a rooftop \(12\) meters above ground level, at what time will the projectile first reach a height of \(150\) meters?

\(\approx 3.0693987\) seconds

Zeros and \(x\)-intercepts of a Function

Recall that \(f(x)\) and \(y\) are interchangeable. Therefore, if we are asked to find where a function is equal to zero, then we need to find the points on the graph of the function that have a \(y\)-value equal to zero (see Figure \(\PageIndex{7}\)).

Zeros and \(x\)-intercepts

The points where the graph off crosses the \(x\)-axis are called the \(x\)-intercepts of the graph of \(f\). The \(x\)-value of each \(x\)-intercept is called a zero of the function \(f\).

The graph of \(f\) crosses the \(x\)-axis in Figure \(\PageIndex{7}\) at \((−3,0)\), \((−1,0)\), and \((3 ,0)\). Therefore:

• The \(x\)-intercepts of f are: \((−3,0)\), \((−1,0)\), and \((3,0)\)
• The zeros of \(f\) are: \(−3\), \(−1\), and \(3\)

A function is zero where its graph crosses the \(x\)-axis.

Example \(\PageIndex{3}\)

Find the zero(s) of the function \(f(x)=1 .5x +5 .25\).

Remember, \(f(x)=1 .5x +5 .25\) and \(y =1 .5x +5 .25\) are equivalent. We’re looking for the value of \(x\) that makes \(y = 0\) or \(f(x) = 0\). So, we’ll start with \(f(x) = 0\), then replace \(f(x)\) with \(1 .5x +5 .25\).

\[\begin{aligned} f(x) &= 0 \quad \color {Red} \text { We want the value of } x \text { that makes the function equal to zero. } \\ 1.5 x+5.25 &= 0 \quad \color {Red} \text { Replace } f(x) \text { with } 1.5 x+5.25 \end{aligned} \nonumber \]

Now we solve for \(x\).

\[\begin{aligned} 1.5 x &= -5.25 \quad \color {Red} \text { Subtract } 5.25 \text { from both sides. } \\ x &= \dfrac{-5.25}{1.5} \quad \color {Red} \text { Divide both sides by } 1.5 \\ x &= -3.5 \quad \color {Red} \text { Divide: }-5.25 / 1.5=-3.5 \end{aligned} \nonumber \]

Check: Substitute \(−3.5\) for \(x\) in the function \(f(x)=1 .5x +5 .25\).

\[\begin{aligned} f(x) &=1.5 x+5.25 \quad \color {Red} \text { The original function. } \\ f(-3.5) &=1.5(-3.5)+5.25 \quad \color {Red} \text { Substitute }-3.5 \text { for } x \\ f(-3.5) &=-5.25+5.25 \quad \color {Red} \text { Multiply: } 1.5(-3.5)=-5.25 \\ f(-3.5) &=0 \quad \color {Red} \text { Add. } \end{aligned} \nonumber \]

Note that \(−3.5\), when substituted for \(x\), makes the function \(f(x)=1 .5x+5 .25\) equal to zero. This is why \(−3.5\) is called a zero of the function.

Graphing calculator solution: We should be able to find the zero by sketching the graph of \(f\) and noting where it crosses the \(x\)-axis. Start by loading the function \(f(x)=1 .5x +5 .25\) into \(\mathbf{Y1} \) in the Y= menu (see the first image in Figure \(\PageIndex{8}\)).

Select 6:ZStandard from the ZOOM menu to produce the graph of \(f\) (see the second image in Figure \(\PageIndex{8}\)). Press 2ND CALC to open the the CALCULATE menu (see the third image in Figure \(\PageIndex{8}\)). To find the zero of the function \(f\):

• Select 2:zero from the CALCULATE menu. The calculator responds by asking for a “Left Bound?” (see the first image in Figure \(\PageIndex{9}\)). Use the left arrow button to move the cursor so that it lies to the left of the \(x\)-intercept of \(f\) and press ENTER .
• The calculator responds by asking for a “Right Bound?” (see the second image in Figure \(\PageIndex{9}\)). Use the right arrow button to move the cursor so that it lies to the right of the \(x\)-intercept of \(f\) and press ENTER .
• The calculator responds by asking for a “Guess?” (see the third image in Figure \(\PageIndex{9}\)). As long as your cursor lies between the left- and right-bound marks at the top of the screen (see the third image in Figure \(\PageIndex{9}\)), you have a valid guess. Since the cursor already lies between the left- and right-bounds, simply press ENTER to use the current position of the cursor as your guess.

The calculator responds by approximating the zero of the function as shown in Figure \(\PageIndex{10}\).

Note that the approximation found using the calculator agrees nicely with the zero found using the algebraic technique.

Exercise \(\PageIndex{3}\)

Find the zero(s) of the function \(f(x)=2 .6x−9.62\).

Example \(\PageIndex{4}\)

How long will it take the projectile in Example \(\PageIndex{2}\) to return to ground level?

In Example \(\PageIndex{2}\), the height of the projectile above the ground as a function of time is given by the equation \[y = 8 + 100t−4.9t^2 \nonumber \]When the projectile returns to the ground, its height above ground will be zero meters. To find the time that this happens, substitute \(y = 0\) in the last equation and solve for \(t\). \[0 = 8 + 100t−4.9t^2 \nonumber \]Enter the equation \(y = 8 + 100t − 4.9t^2\) into \(\mathbf{Y1} \) in the Y= menu of your calculator (see the first image in Figure \(\PageIndex{11}\)), then set the WINDOW parameters shown in the second image in Figure \(\PageIndex{11}\). Push the GRAPH button to produce the graph of the function shown in the third image in Figure \(\PageIndex{11}\).

In this example, the horizontal axis is actually the \(t\)-axis. So when we set \(\mathrm{Xmin}\) and \(\mathrm{Xmax}\) , we’re actually setting bounds on the \(t\)-axis.

To find the time when the projectile returns to ground level, we need to find where the graph of \(y = 8+100t−4.9t^2\) crosses the horizontal axis (in this case the t-axis) . Select 2:zero from the CALC menu. Use the arrow keys to move the cursor slightly to the left of the \(t\)-intercept, then press ENTER in response to “Left bound.” Move your cursor slightly to the right of the t-intercept, then press ENTER in response to “Right bound.” Leave your cursor where it is and press ENTER in response to “Guess.” The result is shown in Figure \(\PageIndex{11}\).

• Label the horizontal and vertical axes with \(t\) and \(y\), respectively (see Figure \(\PageIndex{12}\)). Include the units (seconds (s) and meters (m)).
• Place your WINDOW parameters at the end of each axis (see Figure \(\PageIndex{12}\)).
• Label the graph with its equation (see Figure \(\PageIndex{12}\)).
• Draw a dashed vertical line through the \(t\)-intercept. Shade and label the \(t\)-value of the point where the dashed vertical line crosses the \(t\)-axis. This is the solution of the equation \(0 = 8 + 100t−4.9t^2\) (see Figure \(\PageIndex{12}\)).

Rounding to the nearest tenth of a second, it takes the projectile approximately \(t ≈ 20.5\) seconds to hit the ground.

Exercise \(\PageIndex{4}\)

If a projectile is launched with an initial velocity of \(60\) meters per second from a rooftop \(12\) meters above ground level, at what time will the projectile return to ground level?

\(\approx 12.441734\) seconds

Grade 10 Mathematics Module: Solving Problems Involving Polynomial Functions

This Self-Learning Module (SLM) is prepared so that you, our dear learners, can continue your studies and learn while at home. Activities, questions, directions, exercises, and discussions are carefully stated for you to understand each lesson.

The SLM is composed of different parts. Each part shall guide you step-by-step as you discover and understand the lesson prepared for you.

Pretest is provided to measure your prior knowledge on lessons in each SLM. This will tell you if you need to proceed on completing this module or if you need to ask your facilitator or your teacher’s assistance for better understanding of the lesson. At the end of each module, you need to answer the posttest to self-check your learning. Answer Key is provided for each activity and test. We trust that you will be honest in using these.

Please use this module with care. Do not put unnecessary marks on any part of this SLM. Use a separate sheet of paper in answering the exercises and tests. And read the instructions carefully before performing each task.

If you have any questions in using this SLM or any difficulty in answering the tasks in this module, do not hesitate to consult your teacher or facilitator.

This module was designed and written with you in mind. It is here to help you solve problems involving polynomial functions applying the concepts learned in the previous modules. The scope of this module permits it to be used in many different learning situations. The language used recognizes the diverse vocabulary level of students. The lessons are arranged to follow the standard sequence of the course but the order in which you read and answer this module is dependent on your ability.

After going through this module, you are expected to solve problems involving polynomial functions.

Grade 10 Mathematics Quarter 2 Self-Learning Module: Solving Problems Involving Polynomial Functions

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