Precalculus Examples

$x^{4} + 4 x^{3} - 14 x^{2} - 36 x + 45$

Step 1

If a polynomial function has integer coefficients, then every rational zero will have the form $\frac{p}{q}$ where $p$ is a factor of the constant and $q$ is a factor of the leading coefficient.

$p = \pm 1, \pm 3, \pm 5, \pm 9, \pm 15, \pm 45$

$q = \pm 1$

Step 2

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

$\pm 1, \pm 3, \pm 5, \pm 9, \pm 15, \pm 45$

Step 3

Substitute the possible roots one by one into the polynomial to find the actual roots. Simplify to check if the value is $0$ , which means it is a root.

${(1)}^{4} + 4 {(1)}^{3} - 14 {(1)}^{2} - 36 \cdot 1 + 45$

Step 4

Simplify the expression. In this case, the expression is equal to $0$ so $x = 1$ is a root of the polynomial.

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Step 4.1

Simplify each term.

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Step 4.1.1

One to any power is one.

$1 + 4 {(1)}^{3} - 14 {(1)}^{2} - 36 \cdot 1 + 45$

Step 4.1.2

One to any power is one.

$1 + 4 \cdot 1 - 14 {(1)}^{2} - 36 \cdot 1 + 45$

Step 4.1.3

Multiply $4$ by $1$ .

$1 + 4 - 14 {(1)}^{2} - 36 \cdot 1 + 45$

Step 4.1.4

One to any power is one.

$1 + 4 - 14 \cdot 1 - 36 \cdot 1 + 45$

Step 4.1.5

Multiply $- 14$ by $1$ .

$1 + 4 - 14 - 36 \cdot 1 + 45$

Step 4.1.6

Multiply $- 36$ by $1$ .

$1 + 4 - 14 - 36 + 45$

Step 4.2

Simplify by adding and subtracting.

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Step 4.2.1

Add $1$ and $4$ .

$5 - 14 - 36 + 45$

Step 4.2.2

Subtract $14$ from $5$ .

$- 9 - 36 + 45$

Step 4.2.3

Subtract $36$ from $- 9$ .

$- 45 + 45$

Step 4.2.4

Add $- 45$ and $45$ .

$0$

Step 5

Since $1$ is a known root, divide the polynomial by $x - 1$ to find the quotient polynomial. This polynomial can then be used to find the remaining roots.

$\frac{x^{4} + 4 x^{3} - 14 x^{2} - 36 x + 45}{x - 1}$

Step 6

Next, find the roots of the remaining polynomial. The order of the polynomial has been reduced by $1$ .

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Step 6.1

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

$1$	$1$	$4$	$- 14$	$- 36$	$45$

Step 6.2

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

$1$	$1$	$4$	$- 14$	$- 36$	$45$

	$1$

Step 6.3

Multiply the newest entry in the result $(1)$ by the divisor $(1)$ and place the result of $(1)$ under the next term in the dividend $(4)$ .

$1$	$1$	$4$	$- 14$	$- 36$	$45$
		$1$
	$1$

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.

$1$	$1$	$4$	$- 14$	$- 36$	$45$
		$1$
	$1$	$5$

Step 6.5

Multiply the newest entry in the result $(5)$ by the divisor $(1)$ and place the result of $(5)$ under the next term in the dividend $(- 14)$ .

$1$	$1$	$4$	$- 14$	$- 36$	$45$
		$1$	$5$
	$1$	$5$

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.

$1$	$1$	$4$	$- 14$	$- 36$	$45$
		$1$	$5$
	$1$	$5$	$- 9$

Step 6.7

Multiply the newest entry in the result $(- 9)$ by the divisor $(1)$ and place the result of $(- 9)$ under the next term in the dividend $(- 36)$ .

$1$	$1$	$4$	$- 14$	$- 36$	$45$
		$1$	$5$	$- 9$
	$1$	$5$	$- 9$

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.

$1$	$1$	$4$	$- 14$	$- 36$	$45$
		$1$	$5$	$- 9$
	$1$	$5$	$- 9$	$- 45$

Step 6.9

Multiply the newest entry in the result $(- 45)$ by the divisor $(1)$ and place the result of $(- 45)$ under the next term in the dividend $(45)$ .

$1$	$1$	$4$	$- 14$	$- 36$	$45$
		$1$	$5$	$- 9$	$- 45$
	$1$	$5$	$- 9$	$- 45$

Step 6.10

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

$1$	$1$	$4$	$- 14$	$- 36$	$45$
		$1$	$5$	$- 9$	$- 45$
	$1$	$5$	$- 9$	$- 45$	$0$

Step 6.11

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

$1 x^{3} + 5 x^{2} + (- 9) x - 45$

Step 6.12

Simplify the quotient polynomial.

$x^{3} + 5 x^{2} - 9 x - 45$

Step 7

Factor out the greatest common factor from each group.

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Step 7.1

Group the first two terms and the last two terms.

$(x - 1) (x^{3} + 5 x^{2}) - 9 x - 45$

Step 7.2

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

$(x - 1) + x^{2} (x + 5) - 9 (x + 5)$

$(x - 1) x^{2} (x + 5) - 9 (x + 5)$

Step 8

Factor the polynomial by factoring out the greatest common factor, $x + 5$ .

$(x - 1) (x + 5) (x^{2} - 9)$

Step 9

Rewrite $9$ as $3^{2}$ .

$(x - 1) (x + 5) (x^{2} - 3^{2})$

Step 10

Factor.

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Step 10.1

Since both terms are perfect squares, factor using the difference of squares formula, $a^{2} - b^{2} = (a + b) (a - b)$ where $a = x$ and $b = 3$ .

$(x - 1) + (x + 5) ((x + 3) (x - 3))$

Step 10.2

Remove unnecessary parentheses.

$(x - 1) + (x + 5) (x + 3) (x - 3)$

$(x - 1) (x + 5) (x + 3) (x - 3)$

Step 11

Factor the left side of the equation.

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Step 11.1

Regroup terms.

$4 x^{3} - 36 x + x^{4} - 14 x^{2} + 45 = 0$

Step 11.2

Factor $4 x$ out of $4 x^{3} - 36 x$ .

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Step 11.2.1

Factor $4 x$ out of $4 x^{3}$ .

$4 x (x^{2}) - 36 x + x^{4} - 14 x^{2} + 45 = 0$

Step 11.2.2

Factor $4 x$ out of $- 36 x$ .

$4 x (x^{2}) + 4 x (- 9) + x^{4} - 14 x^{2} + 45 = 0$

Step 11.2.3

Factor $4 x$ out of $4 x (x^{2}) + 4 x (- 9)$ .

$4 x (x^{2} - 9) + x^{4} - 14 x^{2} + 45 = 0$

Step 11.3

Rewrite $9$ as $3^{2}$ .

$4 x (x^{2} - 3^{2}) + x^{4} - 14 x^{2} + 45 = 0$

Step 11.4

Factor.

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Step 11.4.1

Since both terms are perfect squares, factor using the difference of squares formula, $a^{2} - b^{2} = (a + b) (a - b)$ where $a = x$ and $b = 3$ .

$4 x ((x + 3) (x - 3)) + x^{4} - 14 x^{2} + 45 = 0$

Step 11.4.2

Remove unnecessary parentheses.

$4 x (x + 3) (x - 3) + x^{4} - 14 x^{2} + 45 = 0$

Step 11.5

Rewrite $x^{4}$ as ${(x^{2})}^{2}$ .

$4 x (x + 3) (x - 3) + {(x^{2})}^{2} - 14 x^{2} + 45 = 0$

Step 11.6

Let $u = x^{2}$ . Substitute $u$ for all occurrences of $x^{2}$ .

$4 x (x + 3) (x - 3) + u^{2} - 14 u + 45 = 0$

Step 11.7

Factor $u^{2} - 14 u + 45$ using the AC method.

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Step 11.7.1

Consider the form $x^{2} + b x + c$ . Find a pair of integers whose product is $c$ and whose sum is $b$ . In this case, whose product is $45$ and whose sum is $- 14$ .

$- 9, - 5$

Step 11.7.2

Write the factored form using these integers.

$4 x (x + 3) (x - 3) + (u - 9) (u - 5) = 0$

Step 11.8

Replace all occurrences of $u$ with $x^{2}$ .

$4 x (x + 3) (x - 3) + (x^{2} - 9) (x^{2} - 5) = 0$

Step 11.9

Rewrite $9$ as $3^{2}$ .

$4 x (x + 3) (x - 3) + (x^{2} - 3^{2}) (x^{2} - 5) = 0$

Step 11.10

Since both terms are perfect squares, factor using the difference of squares formula, $a^{2} - b^{2} = (a + b) (a - b)$ where $a = x$ and $b = 3$ .

$4 x (x + 3) (x - 3) + (x + 3) (x - 3) (x^{2} - 5) = 0$

Step 11.11

Factor $(x + 3) (x - 3)$ out of $4 x (x + 3) (x - 3) + (x + 3) (x - 3) (x^{2} - 5)$ .

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Step 11.11.1

Factor $(x + 3) (x - 3)$ out of $4 x (x + 3) (x - 3)$ .

$(x + 3) (x - 3) (4 x) + (x + 3) (x - 3) (x^{2} - 5) = 0$

Step 11.11.2

Factor $(x + 3) (x - 3)$ out of $(x + 3) (x - 3) (4 x) + (x + 3) (x - 3) (x^{2} - 5)$ .

$(x + 3) (x - 3) (4 x + x^{2} - 5) = 0$

Step 11.12

Let $u = x$ . Substitute $u$ for all occurrences of $x$ .

$(x + 3) (x - 3) (4 u + u^{2} - 5) = 0$

Step 11.13

Factor $4 u + u^{2} - 5$ using the AC method.

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Step 11.13.1

Consider the form $x^{2} + b x + c$ . Find a pair of integers whose product is $c$ and whose sum is $b$ . In this case, whose product is $- 5$ and whose sum is $4$ .

$- 1, 5$

Step 11.13.2

Write the factored form using these integers.

$(x + 3) (x - 3) ((u - 1) (u + 5)) = 0$

Step 11.14

Factor.

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Step 11.14.1

Replace all occurrences of $u$ with $x$ .

$(x + 3) (x - 3) ((x - 1) (x + 5)) = 0$

Step 11.14.2

Remove unnecessary parentheses.

$(x + 3) (x - 3) (x - 1) (x + 5) = 0$

Step 12

If any individual factor on the left side of the equation is equal to $0$ , the entire expression will be equal to $0$ .

$x + 3 = 0$

$x - 3 = 0$

$x - 1 = 0$

$x + 5 = 0$

Step 13

Set $x + 3$ equal to $0$ and solve for $x$ .

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Step 13.1

Set $x + 3$ equal to $0$ .

$x + 3 = 0$

Step 13.2

Subtract $3$ from both sides of the equation.

$x = - 3$

Step 14

Set $x - 3$ equal to $0$ and solve for $x$ .

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Step 14.1

Set $x - 3$ equal to $0$ .

$x - 3 = 0$

Step 14.2

Add $3$ to both sides of the equation.

$x = 3$

Step 15

Set $x - 1$ equal to $0$ and solve for $x$ .

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Step 15.1

Set $x - 1$ equal to $0$ .

$x - 1 = 0$

Step 15.2

Add $1$ to both sides of the equation.

$x = 1$

Step 16

Set $x + 5$ equal to $0$ and solve for $x$ .

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Step 16.1

Set $x + 5$ equal to $0$ .

$x + 5 = 0$

Step 16.2

Subtract $5$ from both sides of the equation.

$x = - 5$

Step 17

The final solution is all the values that make $(x + 3) (x - 3) (x - 1) (x + 5) = 0$ true.

$x = - 3, 3, 1, - 5$

Step 18