Precalculus Examples

x^{3} - 3 x^{2} - 4 x + 12

$x^{3} - 3 x^{2} - 4 x + 12$

Step 1

If a polynomial function has integer coefficients, then every rational zero will have the form

\frac{p}{q}

$\frac{p}{q}$ where

p

$p$ is a factor of the constant and

q

$q$ is a factor of the leading coefficient.

p = \pm 1, \pm 2, \pm 3, \pm 4, \pm 6, \pm 12

$p = \pm 1, \pm 2, \pm 3, \pm 4, \pm 6, \pm 12$

q = \pm 1

$q = \pm 1$

Step 2

Find every combination of

\pm \frac{p}{q}

$\pm \frac{p}{q}$ . These are the possible roots of the polynomial function.

\pm 1, \pm 2, \pm 3, \pm 4, \pm 6, \pm 12

$\pm 1, \pm 2, \pm 3, \pm 4, \pm 6, \pm 12$

Step 3

Substitute the possible roots one by one into the polynomial to find the actual roots. Simplify to check if the value is

0

$0$ , which means it is a root.

{(2)}^{3} - 3 {(2)}^{2} - 4 \cdot 2 + 12

${(2)}^{3} - 3 {(2)}^{2} - 4 \cdot 2 + 12$

Step 4

Simplify the expression. In this case, the expression is equal to

0

$0$ so

x = 2

$x = 2$ is a root of the polynomial.

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Step 4.1

Simplify each term.

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Step 4.1.1

Raise

2

$2$ to the power of

3

$3$ .

8 - 3 {(2)}^{2} - 4 \cdot 2 + 12

$8 - 3 {(2)}^{2} - 4 \cdot 2 + 12$

Step 4.1.2

Raise

2

$2$ to the power of

2

$2$ .

8 - 3 \cdot 4 - 4 \cdot 2 + 12

$8 - 3 \cdot 4 - 4 \cdot 2 + 12$

Step 4.1.3

Multiply

- 3

$- 3$ by

4

$4$ .

8 - 12 - 4 \cdot 2 + 12

$8 - 12 - 4 \cdot 2 + 12$

Step 4.1.4

Multiply

- 4

$- 4$ by

2

$2$ .

8 - 12 - 8 + 12

$8 - 12 - 8 + 12$

8 - 12 - 8 + 12

$8 - 12 - 8 + 12$

Step 4.2

Simplify by adding and subtracting.

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Step 4.2.1

Subtract

12

$12$ from

8

$8$ .

- 4 - 8 + 12

$- 4 - 8 + 12$

Step 4.2.2

Subtract

8

$8$ from

- 4

$- 4$ .

- 12 + 12

$- 12 + 12$

Step 4.2.3

Add

- 12

$- 12$ and

12

$12$ .

0

$0$

0

$0$

0

$0$

Step 5

Since

2

$2$ is a known root, divide the polynomial by

x - 2

$x - 2$ to find the quotient polynomial. This polynomial can then be used to find the remaining roots.

\frac{x^{3} - 3 x^{2} - 4 x + 12}{x - 2}

$\frac{x^{3} - 3 x^{2} - 4 x + 12}{x - 2}$

Step 6

Next, find the roots of the remaining polynomial. The order of the polynomial has been reduced by

1

$1$ .

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Step 6.1

Place the numbers representing the divisor and the dividend into a division-like configuration.

$2$ $2$	$1$ $1$	$- 3$ $- 3$	$- 4$ $- 4$	$12$ $12$

Step 6.2

The first number in the dividend

(1)

$(1)$ is put into the first position of the result area (below the horizontal line).

$2$ $2$	$1$ $1$	$- 3$ $- 3$	$- 4$ $- 4$	$12$ $12$

	$1$ $1$

Step 6.3

Multiply the newest entry in the result

(1)

$(1)$ by the divisor

(2)

$(2)$ and place the result of

(2)

$(2)$ under the next term in the dividend

(- 3)

$(- 3)$ .

$2$ $2$	$1$ $1$	$- 3$ $- 3$	$- 4$ $- 4$	$12$ $12$
		$2$ $2$
	$1$ $1$

Step 6.4

Add the product of the multiplication and the number from the dividend and put the result in the next position on the result line.

$2$ $2$	$1$ $1$	$- 3$ $- 3$	$- 4$ $- 4$	$12$ $12$
		$2$ $2$
	$1$ $1$	$- 1$ $- 1$

Step 6.5

Multiply the newest entry in the result

(- 1)

$(- 1)$ by the divisor

(2)

$(2)$ and place the result of

(- 2)

$(- 2)$ under the next term in the dividend

(- 4)

$(- 4)$ .

$2$ $2$	$1$ $1$	$- 3$ $- 3$	$- 4$ $- 4$	$12$ $12$
		$2$ $2$	$- 2$ $- 2$
	$1$ $1$	$- 1$ $- 1$

Step 6.6

Add the product of the multiplication and the number from the dividend and put the result in the next position on the result line.

$2$ $2$	$1$ $1$	$- 3$ $- 3$	$- 4$ $- 4$	$12$ $12$
		$2$ $2$	$- 2$ $- 2$
	$1$ $1$	$- 1$ $- 1$	$- 6$ $- 6$

Step 6.7

Multiply the newest entry in the result

(- 6)

$(- 6)$ by the divisor

(2)

$(2)$ and place the result of

(- 12)

$(- 12)$ under the next term in the dividend

(12)

$(12)$ .

$2$ $2$	$1$ $1$	$- 3$ $- 3$	$- 4$ $- 4$	$12$ $12$
		$2$ $2$	$- 2$ $- 2$	$- 12$ $- 12$
	$1$ $1$	$- 1$ $- 1$	$- 6$ $- 6$

Step 6.8

Add the product of the multiplication and the number from the dividend and put the result in the next position on the result line.

$2$ $2$	$1$ $1$	$- 3$ $- 3$	$- 4$ $- 4$	$12$ $12$
		$2$ $2$	$- 2$ $- 2$	$- 12$ $- 12$
	$1$ $1$	$- 1$ $- 1$	$- 6$ $- 6$	$0$ $0$

Step 6.9

All numbers except the last become the coefficients of the quotient polynomial. The last value in the result line is the remainder.

1 x^{2} + - 1 x - 6

$1 x^{2} + - 1 x - 6$

Step 6.10

Simplify the quotient polynomial.

x^{2} - x - 6

$x^{2} - x - 6$

x^{2} - x - 6

$x^{2} - x - 6$

Step 7

Factor

x^{2} - x - 6

$x^{2} - x - 6$ using the AC method.

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Step 7.1

Consider the form

x^{2} + b x + c

$x^{2} + b x + c$ . Find a pair of integers whose product is

c

$c$ and whose sum is

b

$b$ . In this case, whose product is

- 6

$- 6$ and whose sum is

- 1

$- 1$ .

- 3, 2

$- 3, 2$

Step 7.2

Write the factored form using these integers.

(x - 2) + (x - 3) (x + 2)

$(x - 2) + (x - 3) (x + 2)$

(x - 2) (x - 3) (x + 2)

$(x - 2) (x - 3) (x + 2)$

Step 8

Factor the left side of the equation.

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Step 8.1

Factor out the greatest common factor from each group.

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Step 8.1.1

Group the first two terms and the last two terms.

(x^{3} - 3 x^{2}) - 4 x + 12 = 0

$(x^{3} - 3 x^{2}) - 4 x + 12 = 0$

Step 8.1.2

Factor out the greatest common factor (GCF) from each group.

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

$x^{2} (x - 3) - 4 (x - 3) = 0$

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

$x^{2} (x - 3) - 4 (x - 3) = 0$

Step 8.2

Factor the polynomial by factoring out the greatest common factor,

x - 3

$x - 3$ .

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

$(x - 3) (x^{2} - 4) = 0$

Step 8.3

Rewrite

4

$4$ as

2^{2}

$2^{2}$ .

(x - 3) (x^{2} - 2^{2}) = 0

$(x - 3) (x^{2} - 2^{2}) = 0$

Step 8.4

Factor.

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Step 8.4.1

Since both terms are perfect squares, factor using the difference of squares formula,

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

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

a = x

$a = x$ and

b = 2

$b = 2$ .

(x - 3) ((x + 2) (x - 2)) = 0

$(x - 3) ((x + 2) (x - 2)) = 0$

Step 8.4.2

Remove unnecessary parentheses.

(x - 3) (x + 2) (x - 2) = 0

$(x - 3) (x + 2) (x - 2) = 0$

(x - 3) (x + 2) (x - 2) = 0

$(x - 3) (x + 2) (x - 2) = 0$

(x - 3) (x + 2) (x - 2) = 0

$(x - 3) (x + 2) (x - 2) = 0$

Step 9

If any individual factor on the left side of the equation is equal to

0

$0$ , the entire expression will be equal to

0

$0$ .

x - 3 = 0

$x - 3 = 0$

x + 2 = 0

$x + 2 = 0$

x - 2 = 0

$x - 2 = 0$

Step 10

Set

x - 3

$x - 3$ equal to

0

$0$ and solve for

x

$x$ .

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Step 10.1

Set

x - 3

$x - 3$ equal to

0

$0$ .

x - 3 = 0

$x - 3 = 0$

Step 10.2

Add

3

$3$ to both sides of the equation.

x = 3

$x = 3$

x = 3

$x = 3$

Step 11

Set

x + 2

$x + 2$ equal to

0

$0$ and solve for

x

$x$ .

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Step 11.1

Set

x + 2

$x + 2$ equal to

0

$0$ .

x + 2 = 0

$x + 2 = 0$

Step 11.2

Subtract

2

$2$ from both sides of the equation.

x = - 2

$x = - 2$

x = - 2

$x = - 2$

Step 12

Set

x - 2

$x - 2$ equal to

0

$0$ and solve for

x

$x$ .

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Step 12.1

Set

x - 2

$x - 2$ equal to

0

$0$ .

x - 2 = 0

$x - 2 = 0$

Step 12.2

Add

2

$2$ to both sides of the equation.

x = 2

$x = 2$

x = 2

$x = 2$

Step 13

The final solution is all the values that make

(x - 3) (x + 2) (x - 2) = 0

$(x - 3) (x + 2) (x - 2) = 0$ true.

x = 3, - 2, 2

$x = 3, - 2, 2$

Step 14