Mastering Solving Polynomial Equation Examples and Techniques

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

Sum of Squares

Difference of Squares

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

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

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

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

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

Equation

Solutions

x^{3} - 27 = 37

x_{1} = 4

x_{2} = - 2 - 23 i

x_{3} = - 2 + 23 i

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

Real Solutions

Imaginary Solutions

x = 4

x = - 2 - 23 i

x = - 2 + 23 i

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

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

✓

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

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})

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

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

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

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

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

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

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

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

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

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

Find all the solutions to each of the following polynomial equations.

\frac{1}{3} x^{3} - \frac{5}{3} x^{2} - 8 (x - 3) - 4 = 20

x^{3} + 4 x^{2} - 8 x = 2 (2 x^{2} + 4 x)

To find the solutions of the given polynomial equation, let's start by distributing 8 to each of the terms inside the parentheses. 1/3x^3 - 5/3x^2 - 8(x-3) - 4 = 20 ⇓ 1/3x^3 - 5/3x^2 - 8x + 24 - 4 = 20 Now, let's group all the terms on the left-hand side of the equation.

To make computations easier, let's multiply the equation by 3 so that we get rid of the fractions.

Notice that all three terms on the left-hand side have x in common. Thus, let's factor it out. x^3 - 5 x^2 - 24 x = 0 ⇓ x(x^2 - 5x - 24) = 0 Next, we can apply the Zero Product Property and obtain two equations. x(x^2 - 5x - 24) = 0 ↙ ↘ x=0 x^2 - 5x - 24 = 0 From the above, x=0 is a solution. To find the remaining solutions, let's solve the quadratic equation by applying the Quadratic Formula.

The solutions to the quadratic equation are -3 and 8. Therefore, the solutions to the initial polynomial equation are -3, 0, and 8.

As in Part A, let's start by removing the parentheses and grouping all the terms on the left-hand side of the equation.

Since both terms on the left-hand side have x as a common factor, we can factor it out. x^3 -16x = 0 ⇓ x(x^2-16)=0 Once more, we can apply the Zero Product Property. x(x^2-16)=0 ↙ ↘ x = 0 x^2-16 = 0 The left-hand side equation tells us that x=0 is a solution to the initial equation. To find the solutions of the right-hand side equation, we can write 16 as 4^2. This way, the equation becomes a difference of two squares. x^2-16 = 0 ⇓ x^2 - 4^2 = 0 Remember that a difference of two squares can be factored using the following formula. a^2-b^2 = (a+b)(a-b) Therefore, we can rewrite the quadratic equation as follows. x^2 - 4^2 = 0 ⇓ (x+4)(x-4) = 0 Applying the Zero Product Property one more time gives us a pair of linear equations. (x+4)(x-4) = 0 ↙ ↘ x+4 = 0 x-4 = 0 If we solve each of the linear equations for x, we get x=-4 and x=4, respectively. Consequently, the solutions to the initial equation are -4, 0, and 4.

Find the sum of the solutions to the following polynomial equation.

4 x^{3} + 2 (4 x^{2} - 25 x) + x - 98 = 0

Let's start by distributing 2 to each of the terms inside the parentheses and combining like terms.

We can factor the resulting polynomial by grouping. Let's factor 4x^2 from the first two terms and -49 from the last two.

To factor the quadratic expression, we can rewrite 4x^2 as (2x)^2 and 49 as 7^2. (x+2)(4x^2-49)=0 ⇓ (x+2) ((2x)^2-7^2 )=0 Now, we notice that the second factor is a difference of squares. Thus, we can factor it as the sum of the bases multiplied by the difference of the bases. a^2-b^2 = (a+b)(a-b) Applying the previous formula we get the following equation. (x+2) ((2x)^2-7^2 )=0 ⇓ (x+2)(2x+7)(2x-7)=0 Next, let's apply the Zero Product Property to split the previous equation into three linear equations. Then, we will solve them one at a time.

Consequently, the solutions to the given polynomial equation are -2, - 72, and 72. Finally, let's calculate the sum of the three solutions. -2 + (-7/2) + 7/2 = -2 As seen, the sum of the solutions is equal to -2.

Find the product of the solutions to the following polynomial equation.

(x + 4)^{3} - 9 x - 36 = 0

Instead of expanding the cubic power, let's start by factoring out -9 from the last two terms of the given polynomial equation. (x+4)^3 - 9x - 36 = 0 ⇓ (x+4)^3 - 9(x + 4) = 0 Notice that both terms of the resulting equation have x+4 as a common factor. Therefore, let's factor it out. ( x+4)^3 - 9( x+4) = 0 ⇓ ( x+4)( (x+4)^2 - 9 ) = 0 If we rewrite 9 as 3^2, the second factor becomes a difference of squares. (x+4)( (x+4)^2 - 9 ) = 0 ⇓ (x+4)( (x+4)^2 - 3^2 ) = 0 Remember, we can rewrite a difference of squares as the sum of the bases multiplied by the difference of the bases.

Next, let's apply the Zero Product Property to split the previous equation into three linear equations. (x+4)(x+7)(x+1)=0 rcl ↙ & ↓ & ↘ x+4=0 & x+7=0 & x+1 = 0 ↓ & ↓ & ↓ x=-4 & x=-7 & x = -1 From the above, the solutions to the given polynomial equation are -4, -7, and -1. Finally, let's multiply the three solutions. (-4)(-7)(-1) = -28 Consequently, the product of the solutions is -28.

The given polynomial equation can also be solve by expanding the cubic power.

However, after expanding the power, it is not so clear how to factor the resulting polynomial. One option is rewriting 12x^2 as 11x^2+x^2 and 39x as 28x+11x.

To factor the quadratic polynomial, we can use the Quadratic Formula. Alternatively, we can rewrite 11x as 7x+4x and then factor out x from the first two terms and 4 from the last two.

To find the solutions to the last equation, we apply the Zero Product Property.

Finally, if we multiply the three solutions we get (-7)(-4)(-1)=-28.

The expression on the right-hand side represents the factorization of the polynomial on the left-hand side.

32 x^{11} + 32 x^{10} + 8 x^{9} = 2 x (2 x + 2 x)^{2}

Write the missing terms in the order they should be written.

To determine the terms of the right-hand side expression, let's factor the polynomial on the left-hand side. To do so, let's identify the common factors between the terms. 32x^(11) &= 2 * 2 * 2 * 2* 2 * x^9 * x^2 32x^(10) &= 2 * 2 * 2 * 2* 2 * x^9 * x 8x^9 &= 2 * 2 * 2 * x^9 If we compare the factors of each term, we can see that the greatest common factor (GCF) is 2* 2* 2* x^9. Performing the required multiplications we can conclude that the GCF is 8x^9. Next, let's rewrite each term of the polynomial in terms of the GCF. 32x^(11) &= 8x^9 * 4x^2 32x^(10) &= 8x^9 * 4x 8x^9 &= 8x^9 * 1 Now, we will substitute these expressions into the polynomial and factor out the GCF.

Notice that the quadratic expression seems like a perfect square trinomial. If that is the case, we can factor it using the following formula. a^2+2ab+b^2 = (a+b)^2 Thus, let's see if we can rewrite it as a perfect square trinomial. To do so, we can rewrite 4x^2 as ( 2x)^2 and 4x as 2* 2x* 1. 8x^9 ( 4x^2 + 4x + 1 ) = 8x^9 ( ( 2x)^2 + 2* 2x * 1 + 1 ) As suspected, the expression in parentheses is a perfect square trinomial with a= 2x and b= 1. Thus, we can rewrite the polynomial as follows. 8x^9 ( ( 2x)^2 + 2* 2x * 1 + 1 ) = 8x^9 ( 2x + 1)^2 Now that we factored the polynomial completely, we can see that the missing terms are 8x^9, 2x, and 1 in this particular order. 32x^(11) + 32x^(10) + 8x^9 = 8x^9 ( 2x+ 1 )^2

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

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