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 3}{4} i) + (\frac{7}{4} - \frac{7 3}{4} i)

CommutativePropAdd

Commutative Property of Addition

- \frac{7}{2} + \frac{7}{4} + \frac{7}{4} + \frac{7 3}{4} i - \frac{7 3}{4} i

SubTerm

Subtract term

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

ExpandFrac

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

- \frac{1 4}{4} + \frac{7}{4} + \frac{7}{4}

AddFrac

Add fractions

\frac{- 1 4 + 7 + 7}{4}

AddSubTerms

Add and subtract terms

\frac{0}{4}

$\frac{0}{a} = 0$

0

The sum of the solutions to the given equation is

0 .

Discussion

Removing a Cube From Another Cube

Tearrik liked the geometric way of deriving the sum of two cubes so much that he wants to investigate whether he can derive a formula for the difference of two cubes.

a^{3} - b^{3} = ?

To help himself with the geometric approach, he took his Rubik's cube and called

a

to the side lengths. Therefore, the volume of the cube is

a^{3} .

He then realizes that subtracting

b^{3}

from

a^{3}

is the same as removing a cube with volume

b^{3}

from the Rubik's cube. Thus, he calls

b

to the side length of each of the little cubes that make up the Rubik's cube. Then, he removes one of the little cubes that were placed in the corners.

Removing the corner of a Rubik's cube (a little cube).

Next, Tearrik breaks down the resulting solid into three different prisms and computes their volumes.

Disassembling the Rubik's cube into three prisms.

Finally, the sum of these three volumes equals

a^{3} - b^{3} .

a^{3} - b^{3} = V_{1} + V_{2} + V_{3}

SubstituteExpressions

Substitute expressions

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

FactorOut

Factor out $(a - b)$

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

CommutativePropAdd

Commutative Property of Addition

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

Just to be sure, Tearrik grabbed a book and confirmed the formula he got was correct.

Rule

Difference of Two Cubes

The difference 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 difference of the bases $a$ and $b,$ and the trinomial is the sum of the bases squared plus 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}

SubTerms

Subtract terms

a^{3} - b^{3} ✓

Example

Working With Wooden Cubes

On Sunday afternoon, Tearrik went to his tío Angelito's woodshop to lend a hand. There, tío Angelito asked for help in removing a cube with

3

inch sides from a larger cube so that the newly created shape has a volume of

37

cubic inches.

Removing a cube of side 3 from a bigger cube of side x.

However, tío Angelito has not been able to determine the side length of the larger cube. Excited, Tearrik told him that he knows how to calculate its length.

a If

x

represents the side length of the large cube, write an equation modeling the situation described by tío Angelito.

b Find all the solutions to the equation written in Part A.

c What side length should tío Angelito use?

Hint

a What is the volume of the large cube? What is the volume of the small cube? The difference between these volumes must be equal to

37

cubic inches.

b Group all the non-zero numbers on the left-hand side. Rewrite the expression as a difference of two cubes and factor it.

c In the context of the situation, only the real solutions make sense.

Solution

a The volume of a cube is equal to the cube of its side length. With this in mind, write the volume of both the large and small cubes.

V_{Large} V_{Small} = x^{3} = 3^{3} = 27

Since tío Angelito has to remove the small cube from the large one, the volume of the resulting object is equal to the volume of the large cube minus the volume of the small one.

V_{Object} V_{Object} = V_{Large} - V_{Small} ⇓ = x^{3} - 27

On the other hand, tío Angelito said that this object should have a volume of

37

cubic inches. Therefore, equate the previous expression to

37 .

x^{3} - 27 = 37

This equation models the situation described by tío Angelito.

b To solve the equation written in Part A, start by grouping all non-zero numbers on the same side of the equation. To do so, subtract

37

from both sides.

x^{3} - 27 = 37

SubEqn

$LHS - 37 = RHS - 37$

x^{3} - 64 = 0

Notice that

64

can be written as

4^{3} .

Doing this will allow seeing the equation as a difference of two cubes.

x^{3} - 64 = 0

Rewrite

Rewrite $64$ as $4^{3}$

x^{3} - 4^{3} = 0

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

(x - 4) (x^{2} + x \cdot 4 + 4^{2}) = 0

CalcPow

Calculate power

(x - 4) (x^{2} + 4 x + 16) = 0

By the Zero Product Property, either the linear factor or the quadratic factor is equal to zero.

(x - 4) (x^{2} + 4 x + 16) = 0 ↙ ↘ x - 4 = 0 x^{2} + 4 x + 16 = 0

Solving the left-hand side equation gives

x = 4

as a solution. Next, use the Quadratic Formula to solve the right-hand side equation.

x^{2} + 4 x + 16 = 0

▼

Solve using the quadratic formula

UseQuadForm

Use the Quadratic Formula: $a = 1, b = 4, c = 16$

x = \frac{- 4 \pm 4 ^{2} - 4 ( 1 ) ( 1 6 )}{2 ( 1 )}

CalcPowProd

Calculate power and product

x = \frac{- 4 \pm 1 6 - 6 4}{2}

SubTerm

Subtract term

x = \frac{- 4 \pm - 4 8}{2}

SqrtNegToSqrtI

$- a = a \cdot i$

x = \frac{- 4 \pm 4 8 \cdot i}{2}

SplitIntoFactors

Split into factors

x = \frac{- 4 \pm 1 6 \cdot 3 \cdot i}{2}

SqrtProd

$a \cdot b = a \cdot b$

x = \frac{- 4 \pm 1 6 \cdot 3 \cdot i}{2}

CalcRoot

Calculate root

x = \frac{- 4 \pm 4 3 \cdot i}{2}

FactorOut

Factor out $2$

x = \frac{2 ( - 2 \pm 2 3 \cdot i )}{2}

ReduceFrac

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

x = - 2 \pm 23 i

The solutions to the quadratic equation are

x = - 2 - 23 i

and

x = - 2 + 23 i .

Equation	Solutions
$x^{3} - 27 = 37$	$x_{1} = 4$ $x_{2} = - 2 - 23 i$ $x_{3} = - 2 + 23 i$

c From Part B, the equation modeling the situation described by tío Angelilto has three solutions — one real and two imaginary.

Real Solutions	Imaginary Solutions
$x = 4$	$x = - 2 - 23 i$
$x = 4$	$x = - 2 + 23 i$

However, in this context, the imaginary solutions do not make sense. For that reason, a real solution should be considered and, therefore, the large cube that tío Angelito needs to use should have a side length of

4

inches.

Discussion

Using the Graph to Solve Polynomial Equations

Since the graphic visualizations and examples facilitated the learning process so well, Tearrik wanted to continue using this approach. However, he noticed that the equations he solved previously involved cubic polynomials with only one variable term and one constant term.

x^{3} + C = 0

Now, Tearrik wants to solve equations involving polynomials with more terms whose degree is

3

or greater. In this case, the graphical approach can also be applied, but unfortunately there will be no three-dimensional solids that model the situation.

Method

Solving a Polynomial Equation Graphically

Polynomial equations could not be easy to solve algebraically, mostly because they are not always easy to factor. However, they can be solved graphically — although the solutions could be approximate. For example, consider the following equation.

P (x) 20 x^{3} - 200 x + 190 = Q (x) 44 x^{2} - 21 x + 50

To solve this equation graphically, the following five steps can be followed.

Group All the Terms on the Same Side

To facilitate the procedure, group all the terms on the same side of the equation. Notice that by doing this, instead of two functions, only one function will be graphed later.

20 x^{3} - 200 x + 190 = 44 x^{2} - 21 x + 50

SubEqn

$LHS - 44 x^{2} + 21 x - 50 = RHS - 44 x^{2} + 21 x - 50$

20 x^{3} - 44 x^{2} - 179 x + 140 = 0

Define a Polynomial Function

Consider the function defined by the polynomial obtained in the previous step.

f (x) = 20 x^{3} - 44 x^{2} - 179 x + 140

Graph the Function

Graph the polynomial function.

Graph of f(x) = 20x^3 - 44x^2 - 179x + 140

To make the graph, a graphing calculator could be used.

Identify the

x -

intercepts

Notice that the $y -$ coordinates of the $x -$ intercepts are equal to zero. That is, if the graph of $f$ cuts the $x -$ axis at $x_{0},$ then $f (x_{0}) = 0 .$ Consequently, the $x -$ intercepts are the solutions to the equation defined in the first step. Therefore, identify them.

There are three $x -$ intercepts, which means that there are three solutions. In this case, the two left-hand side solutions will be approximated.

Check the Solutions

Finally, check whether the values obtained in the previous step are indeed solutions. To do so, substitute them into the function. For example, start with

x = - 2.5 .

f (x) = 20 x^{3} - 44 x^{2} - 179 x + 140

▼

Substitute

- 2.5

for

x

and evaluate

Substitute

$x = - 2.5$

f (- 2.5) = 20 (- 2.5)^{3} - 44 (- 2.5)^{2} - 179 (- 2.5) + 140

CalcPowProd

Calculate power and product

f (- 2.5) = 20 (- 15.625) - 44 (6.25) + 447.5 + 140

Multiply

f (- 2.5) = - 312.5 - 275 + 447.5 + 140

AddSubTerms

Add and subtract terms

f (- 2.5) = 0

Since

f (- 2.5) = 0,

it can be concluded that

x = - 2.5

is a solution to the initial polynomial equation. The other two values can be checked similarly.

Value	Substitution	Solution?
$x \approx 0.7$	$f (0.7) = 20 (0.7)^{3} - 44 (0.7)^{2} - 179 (0.7) + 140 = 0$	$✓$
$x = 4$	$f (4) = 20 (4)^{3} - 44 (4)^{2} - 179 (4) + 140 = 0$	$✓$

As verified, the three values obtained in the fourth step are solutions to the polynomial equation. Note that these three values could also be obtained by graphing $y = P (x)$ and $y = Q (x)$ on the same coordinate plane and determining the $x -$ coordinates of their points of intersection.

Graph of P(x)=20x^3 - 200x + 190 and Q(x)=44x^2 - 21x + 50

Note that this method only gives the real solutions to a polynomial equation. If the equation has imaginary solutions, they have to be computed using another method.

Using this method, Tearrik noticed that a polynomial equation of the form $x^{3} + C = 0$ has only one $x -$ intercept. Therefore, it has only one real solution.

Graph of f(x)=x^3+C. There is only one x-intercept.

The previous conclusion makes perfect sense with the two factoring formulas Tearrik studied before.

Method	Formula
Sum of Two Cubes	$x^{3} + a^{3} = (x + a) (x^{2} - a x + a^{2})$
Difference of Two Cubes	$x^{3} - a^{3} = (x - a) (x^{2} + a x + a^{2})$

In both cases, the linear factor gives the real solution and the quadratic factor the two imaginary solutions.

Example

Diving Trajectory

Tearrik and tío Angelito are now taking a break from woodwork. Tío Angelito tells Tearrik to pull up a chair — he has a story to tell. Way back in the day, he used to go diving off the coast. Being that he loves math, when he dives he tries to follow the trajectory of a polynomial.

Polynomial D(t)=t^4-25t^3+210t^2-680t+614, with t between 1.44 and 11.4

By following the given polynomial, he was able to model the trajectory of one of his favorite dives. In this polynomial, $D (t)$ represents the depth, in meters, at which tío Angelito was $t$ minutes after he started diving. Negative values of $D (t)$ mean that he was underwater.

a Find the moments when tío Angelito was exactly

90

meters underwater during his dive.

b Did tío Angelito reach

200

meters underwater?

c How long was tío Angelito more than

150

meters below sea level?

Hint

a Since tío Angelito was underwater, the desired depth is negative. Given that fact, the function rule can be equated to

- 90 .

Next, solve the resulting equation.

b Substitute

- 200

for

D (t) .

Does the polynomial function intercepts the

x -

axis?

c Substitute

- 150

for

D (t) .

Approximate the distance between the solutions.

Solution

a It is desired to find the moments when tío Angelito was

90

meters underwater. Since tío Angelito was underwater, the depth is negative. This means that the function rule of the polynomial function should be set equal to

- 90 .

D (t) = t^{4} - 25 t^{3} + 210 t^{2} - 680 t + 614

Substitute

$D (t) = - 90$

- 90 = t^{4} - 25 t^{3} + 210 t^{2} - 680 t + 614

AddEqn

$LHS + 90 = RHS + 90$

0 = t^{4} - 25 t^{3} + 210 t^{2} - 680 t + 704

RearrangeEqn

Rearrange equation

t^{4} - 25 t^{3} + 210 t^{2} - 680 t + 704 = 0

The solutions to the previous equation represent the moments when tío Angelito was

90

meters underwater. These solutions can be found graphically. Begin by considering the function defined by the polynomial on the left.

f (t) = t^{4} - 25 t^{3} + 210 t^{2} - 680 t + 704

The next step is to graph

y = f (t)

which can be done using a table of values. Since

t

represents time, negative values do not make sense. Also, only values within the domain of

D (t)

will be considered.

$t$	$1.5$	$2$	$3$	$4$	$5$	$6$	$7$	$8$	$9$	$10$	$11$	$11.2$
$f (t)$	$\approx 77$	$0$	$- 40$	$0$	$54$	$80$	$60$	$0$	$- 70$	$- 96$	$0$	$\approx 42$

Use the values in the table to make a graph of

y = f (t) .

The

t -

intercepts represent the solutions to the equation

t^{4} -

25 t^{3} +

210 t^{2} -

680 t +

704 = 0 .

As the graph shows, there are four

t -

intercepts.

Given those four intercepts, it can be determined that there were four moments in which tío Angelito was exactly $90$ meters underwater — namely, at $2,$ $4,$ $8,$ and $11$ minutes after he began his dive.

b Following a similar process as done in Part A, equate the function rule function to

- 200

to determine whether tío Angelito reached

200

meter underwater.

D (t) = t^{4} - 25 t^{3} + 210 t^{2} - 680 t + 614 ⇓ - 200 = t^{4} - 25 t^{3} + 210 t^{2} - 680 t + 614

If this equation has at least one real solution, it means that tío Angelito reached

200

meters underwater. As in Part A, in this context, imaginary solutions do not make sense, the equation will be solved graphically, and solving for the solution can begin by grouping all the terms on the left-hand side.

- 200 = t^{4} - 25 t^{3} + 210 t^{2} - 680 t + 614

AddEqn

$LHS + 200 = RHS + 200$

0 = t^{4} - 25 t^{3} + 210 t^{2} - 680 t + 814

RearrangeEqn

Rearrange equation

t^{4} - 25 t^{3} + 210 t^{2} - 680 t + 814 = 0

Next, consider the function defined by the polynomial expression.

g (t) = t^{4} - 25 t^{3} + 210 t^{2} - 680 t + 814

To draw the graph of

y = g (t),

a table of values can be used. Again, only positive values of

t

within the domain of

D (t)

are going to be considered.

$t$	$1.5$	$2$	$3$	$4$	$5$	$6$	$7$	$8$	$9$	$10$	$11$	$11.2$
$g (t)$	$\approx 187$	$110$	$70$	$110$	$164$	$190$	$170$	$110$	$40$	$14$	$110$	$\approx 152$

At first glance, it can be seen that all the values are positive, which means that the graph does not intersect the

t -

axis. Nevertheless, it is best to graph the function to gain a better understanding.

As expected, the graph does not have

t -

intercepts. Therefore, the equation

t^{4} -

25 t^{3} +

210 t^{2} -

680 t +

814 = 0

has no real solutions. This implies that tío Angelito never reached

200

meters underwater.

c To calculate how long tío Angelito was more than

150

meters below sea level, first determine whether he reached that depth at all. To do that, substitute

150

for

D (t)

and solve the resulting equation.

D (t) = - 150 ⇓ - 150 = t^{4} - 25 t^{3} + 210 t^{2} - 680 t + 614

As in Part B, if the previous equation has at least one real solution, it means that tío Angelito, indeed, reached at least

150

meters below sea level. To determine this, begin by grouping all the terms on the left-hand side.

- 150 = t^{4} - 25 t^{3} + 210 t^{2} - 680 t + 614

AddEqn

$LHS + 150 = RHS + 150$

0 = t^{4} - 25 t^{3} + 210 t^{2} - 680 t + 764

RearrangeEqn

Rearrange equation

t^{4} - 25 t^{3} + 210 t^{2} - 680 t + 764 = 0

Consider the function defined by the polynomial on the left.

h (t) = t^{4} - 25 t^{3} + 210 t^{2} - 680 t + 764

The graph of

y = h (t)

can be made by making a table of values. Once more, only positive values of

t

within the domain of

D (t)

will be considered.

$t$	$1.5$	$2$	$3$	$4$	$5$	$6$	$7$	$8$	$9$	$10$	$11$	$11.2$
$h (t)$	$\approx 137$	$60$	$20$	$60$	$114$	$140$	$120$	$60$	$- 10$	$- 36$	$60$	$\approx 102$

By using the points in the table, graph

y = h (t) .

The graph of

y = h (t)

intersects the

t -

axis twice, which means that tío Angelito reached

150

meters below sea level twice — once going down and the other going up.

It can be seen that the

t -

intercepts occur around

9

and around

10.5 .

The difference between these approximations represents the amount of time tío Angelito was below the

150 -

meter mark.

10.5 - 9 \approx 1.5

Given this value, it can be determined that tío Angelito was under the

150 -

meter mark for between

1

and

2

minutes long.

Alternative Solution

Graphing

y = D (t)

Each of the questions can also be solved by analyzing the graph of $y = D (t) .$ As done in the previous parts, begin the process of graphing the equation by making a table of values.

$t$	$1.5$	$2$	$3$	$4$	$5$	$6$	$7$	$8$	$9$	$10$	$11$	$11.2$
$D (t)$	$\approx - 13$	$- 90$	$- 130$	$- 90$	$- 36$	$- 10$	$- 30$	$- 90$	$- 160$	$- 186$	$- 90$	$\approx - 48$

Use the values in the table to draw the graph of

y = D (t) .

Next, draw the lines

y = - 90,

y = - 200,

and

y = - 150

and mark the points of intersection.

Based on the drawn graph, some conclusions can be made.

The line $y = - 90$ intersects the graph of $y = D (t)$ four times. This means that tío Angelito was $90$ meters underwater at four different moments — at $2,$ $4,$ $8,$ and $11$ minutes after he started diving.
The line $y = - 200$ does not intersects the graph of $y = D (t) .$ Therefore, tío Angelito was never $200$ meters below sea level.
The line $y = - 150$ intersects the graph of $y = D (t)$ twice — one around $9$ and the other around $10.5 .$ The graph between these two values is below $- 150 .$ Given that fact, tío Angelito was more than $150$ meters below sea level around $1.5$ minutes, that is, between $1$ and $2$ minutes.

Closure

Graphing Polynomials

When dealing with polynomial equations, making a graph facilitates the solving process. The drawback of this method is that it requires knowing how to graph polynomials of different degrees by hand — which for polynomials of degree

3

or more is not easy.

0.05 x^{5} - 0.55 x^{4} - 0.940625 x^{3} + 24.4938 x^{2} - 80.0156 x + 77.9625 = 0

The most basic method for graphing functions is to make a table of values. However, a table of values reveals the behavior of the graph only in part of the domain, and sometimes the test values must be chosen carefully; otherwise, they might lead to wrong conclusions. For example, for the given polynomial, consider a table of values involving only integer values.

$x$	$- 4$	$- 3$	$- 2$	$- 1$	$0$	$1$	$2$	$3$	$4$	$5$	$6$	$7$
$y$	$\approx 658$	$\approx 507$	$\approx 333$	$\approx 183$	$\approx 78$	$\approx 21$	$\approx 1$	$\approx 0.5$	$\approx 0$	$\approx - 15$	$\approx - 47.5$	$\approx - 85$

According to the table, there is only one sign change, which means that between $- 4$ and $7$ there is only one $x -$ intercept — at $x = 4 .$ Therefore, the graph of the polynomial should look as follows.

Points from the table and a curve connecting them

Furthermore, from $- 3$ to the left the graph goes up, so it is expected to continue going up to the left of $x = - 4$ as well. Similarly, from $6$ onwards the graph decreases, so it is expected to continue descending to the right of $x = 7 .$

Extending the curve to the left and to the right

This is all that can be deduced from the table of values. However, some things are mistaken. According to a graphing calculator, the graph of the polynomial is the following.

Graph of y=0.05x^5 - 0.55x^4 - 0.940625x^3+24.4938x^2 - 80.0156x + 77.9625

As seen, the graph intersects the $x -$ axis twice between $2$ and $3 .$ Also, the behavior on both ends is contrary to the one previously found. Therefore, the table of values should include more values and also consider decimal numbers.

A table of values might not precisely reflect all the characteristics of a graph.

The good news for math lovers is that both the zeros of the polynomial and the degree also give information that allows for making a more precise graph. This concept will be explored further in another lesson.

Solving Polynomial Equations

Catch-Up and Review

Finding All the Solutions

Equations Containing Polynomials

Polynomial Equation

Roller Coasters and Polynomial Equations

Hint

Solution

Factoring Polynomial Equations

Hint

Solution

A New Factoring Formula

Sum of Two Cubes

Proof

Relaxing and Factoring

Hint

Solution

Removing a Cube From Another Cube

Difference of Two Cubes

Proof

Working With Wooden Cubes

Hint

Solution

Using the Graph to Solve Polynomial Equations

Solving a Polynomial Equation Graphically

Diving Trajectory

Hint

Solution

Alternative Solution

Graphing Polynomials

Solving Polynomial Equations

Recommended exercises

	12 Theory slides
	9 Exercises - Grade E - A
	Each lesson is meant to take 1-2 classroom sessions