Mastering Solving Polynomial Equation Examples and Techniques

Linear and quadratic equations are useful for modeling certain real-life situations. However, there are also scenarios where an equation of a higher degree is needed. For this reason, it is important to learn the techniques crucial in solving equations containing polynomials.

Catch-Up and Review

Here are a few recommended readings before getting started with this lesson.

Factoring

Challenge

Finding All the Solutions

Tearrik is watching the clock on the wall just waiting for the school bell to ring so he can prepare for a fun weekend with friends and family. Just before the bell rang, his math teacher assigned the following challenge.

Homework for Monday: How many solutions does the equation x^3-125=0 have?

At first glance, he thought that the task was very simple since there is only one number whose cube is $125 .$ Is Tearrik right? Are there no more solutions to the equation? Find all the solutions to the equation.

Discussion

Equations Containing Polynomials

In Tearrik's courses, he previously learned that some real-life situations such as saving a constant amount of money weekly or shooting a basketball can be modeled by linear and quadratic equations, respectively.

Now, he wonders if there are situations involving other types of equations. Specifically, Tearrik wants to know if an equation could contain a polynomial. Luckily for Tearrik, his teacher is planning on introducing the topic next class.

Concept

Polynomial Equation

A polynomial equation is an equation that contains a polynomial expression on one or both sides of the equation. For example, consider the following equation.

polynomial x^{3} - x = polynomial 7 x^{2} + 3

By applying the Properties of Equality a polynomial equation can be rewritten in terms of a single polynomial. To do so, group all the terms on the same side of the equation.

x^{3} - x = 7 x^{2} + 3 ⇕ x^{3} - 7 x^{2} - x - 3 = 0

In order to solve polynomial equations, algebraic methods such as the Quadratic Formula or the Zero Product Property can be used. Alternatively, the equation can be solved graphically or using numerical methods.

Example

Roller Coasters and Polynomial Equations

On the weekend, Tearrik and his friends decided to go to the amusement park to have some fun.

a While lining up to get on a roller coaster, Tearrik saw a sign saying that the height, in meters, during the first few seconds of the ride is modeled by the polynomial

h (t) = \frac{1}{5} t^{3} - 2 t^{2} + \frac{2 1}{5} t + 7,

where

t

is measured in seconds.

$Graph of h(t)=\frac{1}{5}t^3-2t^2+\frac{21}{5}t+7$

It was then that he wondered how many seconds after the ride begins, will the car be

7

meters above the ground. Find those times, if any.

b On a second roller coaster, Tearrik found a similar sign, but this time the polynomial describing the height during the first few seconds of the ride is

h (t) = - \frac{1}{5} t^{3} + \frac{6}{5} t^{2} - \frac{1}{5} t + 10 .

Again, the height is measured in meters and the time in seconds.

$Graph of h(t)=-\frac{1}{5}t^3+\frac{6}{5}t^2-\frac{1}{5}t+10$

This time Tearrik wants to know how many seconds after the ride begins, will the car be

8.8

meters above the ground.

Hint

a Substitute

7

for

h (t)

and solve the resulting polynomial equation. Identify the GCF of the polynomial and factor it out. Then, use the Quadratic Formula. Note that Tearrik wants to know the moments after the ride begins.

b Substitute

8.8

for

h (t)

and solve the resulting polynomial equation. Factor the polynomial by grouping. Only real solutions make sense.

Solution

a Since

h (t)

models the height of the car during the first few seconds of the ride and Tearrik wants to know when the car will be

7

meters above the ground, substitute

7

for

h (t)

in the function rule.

7 = \frac{1}{5} t^{3} - 2 t^{2} + \frac{2 1}{5} t + 7

Next, group all the terms on the same side of the equation. This can be done by subtracting

7

from both sides.

7 = \frac{1}{5} t^{3} - 2 t^{2} + \frac{2 1}{5} t + 7

SubEqn

$LHS - 7 = RHS - 7$

0 = \frac{1}{5} t^{3} - 2 t^{2} + \frac{2 1}{5} t

RearrangeEqn

Rearrange equation

\frac{1}{5} t^{3} - 2 t^{2} + \frac{2 1}{5} t = 0

To get rid of the fractions, multiply both sides of the equation by

5 .

\frac{1}{5} t^{3} - 2 t^{2} + \frac{2 1}{5} t = 0

▼

Simplify

MultEqn

$LHS \cdot 5 = RHS \cdot 5$

(\frac{1}{5} t^{3} - 2 t^{2} + \frac{2 1}{5} t) 5 = 0 \cdot 5

ZeroPropMult

Zero Property of Multiplication

(\frac{1}{5} t^{3} - 2 t^{2} + \frac{2 1}{5} t) 5 = 0

Distr

Distribute $5$

\frac{5}{5} t^{3} - 10 t^{2} + \frac{1 0 5}{5} t = 0

CalcQuot

Calculate quotient

t^{3} - 10 t^{2} + 21 t = 0

Note that all the terms on the left-hand side have

t

as a common factor. Thus, it can be factored out.

t^{3} - 10 t^{2} + 21 t = 0 ⇓ t (t^{2} - 10 t + 21) = 0

By the Zero Product Property, either

t = 0

or the quadratic polynomial between the parentheses is equal to

0 .

t (t^{2} - 10 t + 21) = 0 ↙ ↘ t = 0 t^{2} - 10 t + 21 = 0

To find the solutions to the quadratic equation, the Quadratic Formula can be used.

t^{2} - 10 t + 21 = 0

▼

Use the Quadratic Formula:

a = 1, b = - 10, c = 21

SubstituteValues

Substitute values

t = \frac{- ( - 1 0 ) \pm ( - 1 0 ) ^{2} - 4 ( 1 ) ( 2 1 )}{2 ( 1 )}

CalcPowProd

Calculate power and product

t = \frac{- ( - 1 0 ) \pm 1 0 0 - 8 4}{2}

NegNeg

$- (- a) = a$

t = \frac{1 0 \pm 1 0 0 - 8 4}{2}

SubTerm

Subtract term

t = \frac{1 0 \pm 1 6}{2}

CalcRoot

Calculate root

t = \frac{1 0 \pm 4}{2}

FactorOut

Factor out $2$

t = \frac{2 ( 5 \pm 2 )}{2}

ReduceFrac

$\frac{a}{b} = \frac{a / 2}{b / 2}$

t = 5 \pm 2

StateSol

State solutions

t_{1} = 5 + 2 t_{2} = 5 - 2 (I) (II)

$(I), (II):$ Add and subtract terms

t_{1} = 7 t_{2} = 3

The solutions to the quadratic equation are

t = 7

and

t = 3 .

Since it was already stated that

t = 0

is a solution, and now two more solutions were found, the initial polynomial equation has three different solutions.

\frac{1}{5} t^{3} - 2 t^{2} + \frac{2 1}{5} t + 7 = 7 ⇓ t_{1} = 0 t_{2} = 3 t_{3} = 7

Notice that Tearrik is interested in the moments in which the car will be

7

meters above the ground after the ride starts. Thus,

t_{1} = 0

can be discarded. Consequently, based on the context, the car will be

7

meters above the ground twice — the first time will occur

3

seconds after the ride starts and the second time will occur

7

seconds after the ride starts.

$Graph of h(t)=\dfrac{1}{5}t^3-2t^2+\dfrac{21}{5}t+7 and y=7$

b As done in Part A, to determine when the car will be

8.8

meters above the ground, substitute

8.8

for

h (t)

in the function rule.

8.8 = - \frac{1}{5} t^{3} + \frac{6}{5} t^{2} - \frac{1}{5} t + 10

To get rid of the fractions and the decimal number, multiply both sides by

5 .

After that, group all the terms on the same side of the equation.

8.8 = - \frac{1}{5} t^{3} + \frac{6}{5} t^{2} - \frac{1}{5} t + 10

▼

Simplify

MultEqn

$LHS \cdot 5 = RHS \cdot 5$

8.8 \cdot 5 = (- \frac{1}{5} t^{3} + \frac{6}{5} t^{2} - \frac{1}{5} t + 10) 5

Multiply

44 = (- \frac{1}{5} t^{3} + \frac{6}{5} t^{2} - \frac{1}{5} t + 10) 5

Distr

Distribute $5$

44 = - \frac{5}{5} t^{3} + \frac{3 0}{5} t^{2} - \frac{5}{5} t + 50

CalcQuot

Calculate quotient

44 = - t^{3} + 6 t^{2} - t + 50

SubEqn

$LHS - 44 = RHS - 44$

0 = - t^{3} + 6 t^{2} - t + 6

RearrangeEqn

Rearrange equation

- t^{3} + 6 t^{2} - t + 6 = 0

MultEqn

$LHS \cdot (- 1) = RHS \cdot (- 1)$

(- t^{3} + 6 t^{2} - t + 6) (- 1) = 0 (- 1)

ZeroPropMult

Zero Property of Multiplication

(- t^{3} + 6 t^{2} - t + 6) (- 1) = 0

Distr

Distribute $(- 1)$

t^{3} - 6 t^{2} + t - 6 = 0

To solve the resulting polynomial equation, factor the polynomial on the left-hand side. This can be done by grouping.

t^{3} - 6 t^{2} + t - 6 = 0

AssociativePropAdd

Associative Property of Addition

(t^{3} - 6 t^{2}) + (t - 6) = 0

FactorOut

Factor out $t^{2}$

t^{2} (t - 6) + (t - 6) = 0

FactorOut

Factor out $(t - 6)$

(t - 6) (t^{2} + 1) = 0

According to the Zero Product Property, either the linear factor or the quadratic factor is equal to

0 .

(t - 6) (t^{2} + 1) = 0 ↙ ↘ t - 6 = 0 t^{2} + 1 = 0

Solving the left-hand side equation for

t

gives

6

as a solution. Because the quadratic equation is a sum of two squares, it can be factored using the following formula.

a^{2} + b^{2} = (a + b i) (a - b i)

Here,

i

is the imaginary unit. For that reason, the quadratic equation can be rewritten as follows.

t^{2} + 1 = (t + i) (t - i)

The solutions to the quadratic equation are

i

and

- i

— both imaginary numbers. While they are solutions to the polynomial equation, they do not make sense in the context of the ride. For this reason, during the first few seconds of the ride, the car will be

8.8

meters above the ground only once, and it will happen

6

seconds after the ride starts.

$Graph of h(t)=-\frac{1}{5}t^3+\frac{6}{5}t^2-\frac{1}{5}t+10 and y=8.8$

Example

Factoring Polynomial Equations

After arriving home and feeling excited about solving some polynomial equations at the amusement park, Tearrik sees a note written by his sister. Tearrik gets right to his homework so he can finish in time to watch a movie with his family!

Post it on the fridge door saying: Tonight, 20:30, movie

Tearrik's homework asks him to factor a pair of polynomial equations.

a Factor the equation

48 x^{9} - 108 x^{7} = 0

as a product of unfactorable factors with integer coefficients.

b Factor the equation

90 x^{6} + 75 x^{5} + 27 x^{7} = 0

as a product of unfactorable factors with integer coefficients.

Hint

a Start by factoring out the greatest common factor GCF of the terms. Notice that a difference of two squares is obtained.

b Factor out the GCF of the terms. The resulting polynomial is a perfect square trinomial. For that reason, it can be factored as the square of a binomial.

Solution

a To factor the given polynomial equation, identify the common factors between the terms.

48 x^{9} 108 x^{7} = \underline{2} \cdot \underline{2} \cdot 2 \cdot 2 \cdot \underline{3} \cdot \underline{x^{7}} \cdot x^{2} = \underline{2} \cdot \underline{2} \cdot \underline{3} \cdot 3 \cdot 3 \cdot \underline{x^{7}}

Comparing the factors of each term, the GCF of the two terms is

2 \cdot 2 \cdot 3 \cdot x^{7}

which, when the coefficients are multiplied, is the same as

12 x^{7} .

Next, rewrite each term of the polynomial in terms of the GCF.

48 x^{9} 108 x^{7} = 12 x^{7} \cdot 4 x^{2} = 12 x^{7} \cdot 9

Now, substitute these expressions into the polynomial equation and factor out the GCF.

48 x^{9} - 108 x^{7} = 0

SubstituteII

$48 x^{9} = 12 x^{7} \cdot 4 x^{2}$ , $108 x^{7} = 12 x^{7} \cdot 9$

12 x^{7} \cdot 4 x^{2} - 12 x^{7} \cdot 9 = 0

FactorOut

Factor out $12 x^{7}$

12 x^{7} (4 x^{2} - 9) = 0

The next task is factoring the quadratic expression. This can be done using the Quadratic Formula. However, notice that the terms in the parenthesis can be expressed differently. In fact,

4 x^{2}

can be expressed as

(2 x)^{2},

and

9

3^{2} .

Thus, the quadratic expression can be rewritten as a difference of two squares.

12 x^{7} (4 x^{2} - 9) = 0 ⇕ 12 x^{7} ((2 x)^{2} - 3^{2}) = 0

The difference of squares can be factored as the sum of the bases multiplied by the difference of the bases.

12 x^{7} ((2 x)^{2} - 3^{2}) = 0 ⇕ 12 x^{7} (2 x + 3) (2 x - 3) = 0

Notice that each of the factors cannot be factored further.

b As in Part A, start by identifying the common factors between the three terms forming the polynomial equation.

90 x^{6} 75 x^{5} 27 x^{7} = 2 \cdot \underline{3} \cdot 3 \cdot 5 \cdot \underline{x^{5}} \cdot x = \underline{3} \cdot 5 \cdot 5 \cdot \underline{x^{5}} = \underline{3} \cdot 3 \cdot 3 \cdot \underline{x^{5}} \cdot x^{2}

Taking the underlined factors into consideration, the GCF of the polynomial is

3 x^{5} .

Now, rewrite each term of the polynomial in terms of the GCF.

90 x^{6} 75 x^{5} 27 x^{7} = 3 x^{5} \cdot 30 x = 3 x^{5} \cdot 25 = 3 x^{5} \cdot 9 x^{2}

Next, substitute these expressions into the given polynomial and factor out the GCF.

90 x^{6} + 75 x^{5} + 27 x^{7} = 0

SubstituteExpressions

Substitute expressions

3 x^{5} \cdot 30 x + 3 x^{5} \cdot 25 + 3 x^{5} \cdot 9 x^{2} = 0

FactorOut

Factor out $3 x^{5}$

3 x^{5} (30 x + 25 + 9 x^{2}) = 0

To factor the quadratic expression between the parentheses, first rearrange its terms so that it is in standard form.

3 x^{5} (30 x + 25 + 9 x^{2}) = 0 ⇕ 3 x^{5} (9 x^{2} + 30 x + 25) = 0

Instead of using the Quadratic Formula, notice that

9 x^{2}

equals

(3 x)^{2},

30 x

equals

2 \cdot 3 x \cdot 5,

and

25

equals

5^{2} .

3 x^{5} (9 x^{2} + 30 x + 25) = 0 ⇕ 3 x^{5} ((3 x)^{2} + 2 \cdot 3 x \cdot 5 + 5^{2}) = 0

As can be seen, the quadratic expression has the form of a perfect square trinomial. Therefore, it can be rewritten as the square of the sum of

3 x

and

5 .

3 x^{5} ((3 x)^{2} + 2 \cdot 3 x \cdot 5 + 5^{2}) = 0 ⇕ 3 x^{5} (3 x + 5)^{2} = 0

The given polynomial has been factored as the product of unfactorable factors with integer coefficients.

Discussion

A New Factoring Formula

While checking the factorization methods he knows so far, Tearrik notices that he has a formula for factoring the sum and difference of two squares.

Sum of Squares	Difference of Squares
$a^{2} + b^{2} = (a + b i) (a - b i)$	$a^{2} - b^{2} = (a + b) (a - b)$

However, Tearrik wonders if a similar formula exists for the sum of two cubes. The good news is that such a formula does exist and, along with the Zero Product Property, is useful for solving polynomial equations.

Rule

Sum of Two Cubes

The sum of two cubes can be factored as the product of a binomial by a trinomial.

$a^{3} + b^{3} = (a + b) (a^{2} - a b + b^{2})$

Notice that the binomial is the sum of the bases $a$ and $b,$ and the trinomial is the sum of the bases squared minus the product of the bases.

Proof

Algebraic Proof

To show that the identity is true, the Distributive Property will be used to multiply the binomial by the trinomial. After simplifying, the left-hand side of the identity will be obtained.

(a + b) (a^{2} - a b + b^{2})

MultPar

Multiply parentheses

a \cdot a^{2} - a \cdot a b + a \cdot b^{2} + b \cdot a^{2} - b \cdot a b + b \cdot b^{2}

MultPow

$a^{m} \cdot a^{n} = a^{m + n}$

a^{3} - a^{2} b + a b^{2} + a^{2} b - a b^{2} + b^{3}

CommutativePropAdd

Commutative Property of Addition

a^{3} - a^{2} b + a^{2} b + a b^{2} - a b^{2} + b^{3}

AssociativePropAdd

Associative Property of Addition

a^{3} + (- a^{2} b + a^{2} b) + (a b^{2} - a b^{2}) + b^{3}

AddSubTerms

Add and subtract terms

a^{3} + b^{3}

After reading the formula and having in mind that a cube can also be thought of as a three-dimensional object, Tearrik wondered whether there is a geometric way of deducting the formula. Indeed, there is one way of visualizing the formula geometrically. First, consider two cubes, one of side $a$ and another of side $b .$

Two cubes, one of side a and another of side b.

Next, place the small cube on top of the bigger one and complete the missing parts to form a prism.

The sum of the volumes of the initial cubes,

a^{3} + b^{3},

is equal to the volume of the last prism minus the volumes of the auxiliary prisms.

Using the above expressions, an equation involving all the volumes can be written.

a^{3} + b^{3} = Prism Volume a^{2} (a + b) - Auxiliary Volumes b^{2} (a - b) - a b (a - b)

Finally, simplify the right-hand side expression to obtain the formula for factoring the sum of two cubes.

a^{3} + b^{3} = a^{2} (a + b) - b^{2} (a - b) - a b (a - b)

FactorOut

Factor out $- b (a - b)$

a^{3} + b^{3} = a^{2} (a + b) - b (a - b) (b + a)

FactorOut

Factor out $(a + b)$

a^{3} + b^{3} = (a + b) (a^{2} - b (a - b))

Distr

Distribute $b$

a^{3} + b^{3} = (a + b) (a^{2} - a b + b^{2}) ✓

Example

Relaxing and Factoring

In need of a break from such a fun weekend, Tearrik decided to just relax in his room. Looking around, he sees a die and a Rubik's cube that have been laying around forever. He wonders about the sum of their volumes. He knows that each side of the die measures $2$ centimeters but does not know the dimensions of the Rubik's cube.

a If each side of the Rubik's cube measures

y

centimeters, write an expression representing the sum of the volumes of the two cubes. Then, factor the expression so that each factor has integer coefficients.

b Additionally, Tearrik found his big brother's math notebook and opened it. There, he found an exercise asking for the sum of the solutions to the equation

8 x^{3} + 350 = 7

and decided to try to solve it. What result should Tearrik get?

Hint

a The sum of the volumes is

y^{3} + 2^{3},

which is a sum of two cubes.

b Group all non-zero numbers on the left-hand side. Notice that

8 x^{3}

is equal to

(2 x)^{3} .

Rewrite

343

as a perfect cube. The given equation is the sum of two cubes. Use the Quadratic Formula to solve the resulting quadratic equation.

Solution

a If each side of the cube is

y

centimeters long, its volume is

y^{3}

cubic centimeters. Similarly, the volume of the die is

2^{3}

cubic centimeters.

V_{Rubik} = y^{3} V_{Die} = 2^{3} ↘ ↙ V = y^{3} + 2^{3}

Notice that

V

has the form of the sum of two cubes. Therefore, it can be factored using the following formula.

a^{3} + b^{3} = (a + b) (a^{2} - a b + b^{2})

In this case,

a = y

and

b = 2 .

y^{3} + 2^{3} y^{3} + 2^{3} = (y + 2) (y^{2} - y \cdot 2 + 2^{2}) ⇓ = (y + 2) (y^{2} - 2 y + 4)

Now, before factoring the quadratic expression, its discriminant will be examined.

b^{2} - 4 a c

SubstituteValues

Substitute values

(- 2)^{2} - 4 (1) (4)

CalcPowProd

Calculate power and product

4 - 16

SubTerm

Subtract term

- 12

Since the discriminant is negative, the quadratic expression has no real solutions. Therefore, it cannot be factored using integer coefficients. Consequently, the factored form of

V

looks as follows.

V = (y + 2) (y^{2} - 2 y + 4)

b To solve the given equation, start by grouping all the non-zero numbers on the left-hand side. Thus, start by subtracting

7

from both sides.

8 x^{3} + 350 = 7

SubEqn

$LHS - 7 = RHS - 7$

8 x^{3} + 343 = 0

Using the fact that

8

is equal to

2^{3}

and

343

is equal to

7^{3},

the left-hand side expression can be rewritten.

8 x^{3} + 343 = 0

WritePow

Write as a power

2^{3} x^{3} + 7^{3} = 0

ProdPow

$a^{m} \cdot b^{m} = (a \cdot b)^{m}$

(2 x)^{3} + 7^{3} = 0

The left-hand side expression is a sum of cubes and can be factored using the same formula used in Part A. In this case,

a = 2 x

and

b = 7 .

(2 x)^{3} + 7^{3} = 0

$a^{3} + b^{3} = (a + b) (a^{2} - a b + b^{2})$

(2 x + 7) ((2 x)^{2} - 2 x \cdot 7 + 7^{2}) = 0

CalcPowProd

Calculate power and product

(2 x + 7) (4 x^{2} - 14 x + 49) = 0

▼

Solve using the Zero Product Property

ZeroPropMult

Zero Property of Multiplication

2 x + 7 = 0 4 x^{2} - 14 x + 49 = 0 (I) (II)

SubEqn

$(I):$ $LHS - 7 = RHS - 7$

2 x = - 7 4 x^{2} - 14 x + 49 = 0

DivEqn

$(I):$ $LHS / 2 = RHS / 2$

x = - \frac{7}{2} 4 x^{2} - 14 x + 49 = 0

So far, one solution to the equation has been found — namely,

x = - \frac{7}{2} .

Next, use the Quadratic Formula to solve the second equation.

4 x^{2} - 14 x + 49 = 0

▼

Solve using the quadratic formula

UseQuadForm

Use the Quadratic Formula: $a = 4, b = - 14, c = 49$

x = \frac{- ( - 1 4 ) \pm ( - 1 4 ) ^{2} - 4 ( 4 ) ( 4 9 )}{2 ( 4 )}

CalcPowProd

Calculate power and product

x = \frac{- ( - 1 4 ) \pm 1 9 6 - 7 8 4}{8}

NegNeg

$- (- a) = a$

x = \frac{1 4 \pm 1 9 6 - 7 8 4}{8}

SubTerm

Subtract term

x = \frac{1 4 \pm - 5 8 8}{8}

SqrtNegToSqrtI

$- a = a \cdot i$

x = \frac{1 4 \pm 5 8 8 i}{8}

SplitIntoFactors

Split into factors

x = \frac{1 4 \pm 1 9 6 \cdot 3 i}{8}

SqrtProd

$a \cdot b = a \cdot b$

x = \frac{1 4 \pm 1 9 6 \cdot 3 i}{8}

CalcRoot

Calculate root

x = \frac{1 4 \pm 1 4 3 i}{8}

FactorOut

Factor out $2$

x = \frac{2 ( 7 \pm 7 3 i )}{8}

ReduceFrac

$\frac{a}{b} = \frac{a / 2}{b / 2}$

x = \frac{7 \pm 7 3 i}{4}

WriteSumFrac

Write as a sum of fractions

x = \frac{7}{4} \pm \frac{7 3 i}{4}

Using the positive and negative signs, the two solutions to the quadratic equation are obtained. All the solution are written in the following table.

Equation	Solutions
$8 x^{3} + 350 = 7$	$x_{1} x_{2} x_{3} = - \frac{7}{2} = \frac{7}{4} + \frac{7 3}{4} i = \frac{7}{4} - \frac{7 3}{4} i$

Now that all the solutions are known, their sum can be calculated.

x_{1} + x_{2} + x_{3}

▼

Substitute values and simplify

SubstituteValues

Substitute values

- \frac{7}{2} + (\frac{7}{4} + \frac{7}{4}

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Catch-Up and Review

Hint

Solution

Hint

Solution

Proof

Hint

Solution