Precalculus Examples

$x^{4} - 8 x^{3} + 17 x^{2} - 8 x + 16$

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.

$p = \pm 1, \pm 2, \pm 4, \pm 8, \pm 16$

$q = \pm 1$

Step 2

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

$\pm 1, \pm 2, \pm 4, \pm 8, \pm 16$

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.

${(4)}^{4} - 8 {(4)}^{3} + 17 {(4)}^{2} - 8 \cdot 4 + 16$

Step 4

Simplify the expression. In this case, the expression is equal to $0$ so $x = 4$ 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 $4$ to the power of $4$ .

$256 - 8 {(4)}^{3} + 17 {(4)}^{2} - 8 \cdot 4 + 16$

Step 4.1.2

Raise $4$ to the power of $3$ .

$256 - 8 \cdot 64 + 17 {(4)}^{2} - 8 \cdot 4 + 16$

Step 4.1.3

Multiply $- 8$ by $64$ .

$256 - 512 + 17 {(4)}^{2} - 8 \cdot 4 + 16$

Step 4.1.4

Raise $4$ to the power of $2$ .

$256 - 512 + 17 \cdot 16 - 8 \cdot 4 + 16$

Step 4.1.5

Multiply $17$ by $16$ .

$256 - 512 + 272 - 8 \cdot 4 + 16$

Step 4.1.6

Multiply $- 8$ by $4$ .

$256 - 512 + 272 - 32 + 16$

Step 4.2

Simplify by adding and subtracting.

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Step 4.2.1

Subtract $512$ from $256$ .

$- 256 + 272 - 32 + 16$

Step 4.2.2

Add $- 256$ and $272$ .

$16 - 32 + 16$

Step 4.2.3

Subtract $32$ from $16$ .

$- 16 + 16$

Step 4.2.4

Add $- 16$ and $16$ .

$0$

Step 5

Since $4$ is a known root, divide the polynomial by $x - 4$ to find the quotient polynomial. This polynomial can then be used to find the remaining roots.

$\frac{x^{4} - 8 x^{3} + 17 x^{2} - 8 x + 16}{x - 4}$

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.

$4$	$1$	$- 8$	$17$	$- 8$	$16$

Step 6.2

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

$4$	$1$	$- 8$	$17$	$- 8$	$16$

	$1$

Step 6.3

Multiply the newest entry in the result $(1)$ by the divisor $(4)$ and place the result of $(4)$ under the next term in the dividend $(- 8)$ .

$4$	$1$	$- 8$	$17$	$- 8$	$16$
		$4$
	$1$

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.

$4$	$1$	$- 8$	$17$	$- 8$	$16$
		$4$
	$1$	$- 4$

Step 6.5

Multiply the newest entry in the result $(- 4)$ by the divisor $(4)$ and place the result of $(- 16)$ under the next term in the dividend $(17)$ .

$4$	$1$	$- 8$	$17$	$- 8$	$16$
		$4$	$- 16$
	$1$	$- 4$

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.

$4$	$1$	$- 8$	$17$	$- 8$	$16$
		$4$	$- 16$
	$1$	$- 4$	$1$

Step 6.7

Multiply the newest entry in the result $(1)$ by the divisor $(4)$ and place the result of $(4)$ under the next term in the dividend $(- 8)$ .

$4$	$1$	$- 8$	$17$	$- 8$	$16$
		$4$	$- 16$	$4$
	$1$	$- 4$	$1$

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.

$4$	$1$	$- 8$	$17$	$- 8$	$16$
		$4$	$- 16$	$4$
	$1$	$- 4$	$1$	$- 4$

Step 6.9

Multiply the newest entry in the result $(- 4)$ by the divisor $(4)$ and place the result of $(- 16)$ under the next term in the dividend $(16)$ .

$4$	$1$	$- 8$	$17$	$- 8$	$16$
		$4$	$- 16$	$4$	$- 16$
	$1$	$- 4$	$1$	$- 4$

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.

$4$	$1$	$- 8$	$17$	$- 8$	$16$
		$4$	$- 16$	$4$	$- 16$
	$1$	$- 4$	$1$	$- 4$	$0$

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.

$1 x^{3} + - 4 x^{2} + (1) x - 4$

Step 6.12

Simplify the quotient polynomial.

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

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.

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

Step 7.2

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

$(x - 4) + x^{2} (x - 4) + 1 (x - 4)$

$(x - 4) x^{2} (x - 4) + 1 (x - 4)$

Step 8

Factor the polynomial by factoring out the greatest common factor, $x - 4$ .

$(x - 4) (x - 4) (x^{2} + 1)$

Step 9

Factor the left side of the equation.

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Step 9.1

Regroup terms.

$- 8 x^{3} - 8 x + x^{4} + 17 x^{2} + 16 = 0$

Step 9.2

Factor $- 8 x$ out of $- 8 x^{3} - 8 x$ .

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Step 9.2.1

Factor $- 8 x$ out of $- 8 x^{3}$ .

$- 8 x \cdot x^{2} - 8 x + x^{4} + 17 x^{2} + 16 = 0$

Step 9.2.2

Factor $- 8 x$ out of $- 8 x$ .

$- 8 x \cdot x^{2} - 8 x \cdot 1 + x^{4} + 17 x^{2} + 16 = 0$

Step 9.2.3

Factor $- 8 x$ out of $- 8 x (x^{2}) - 8 x (1)$ .

$- 8 x (x^{2} + 1) + x^{4} + 17 x^{2} + 16 = 0$

Step 9.3

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

$- 8 x (x^{2} + 1) + {(x^{2})}^{2} + 17 x^{2} + 16 = 0$

Step 9.4

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

$- 8 x (x^{2} + 1) + u^{2} + 17 u + 16 = 0$

Step 9.5

Factor $u^{2} + 17 u + 16$ using the AC method.

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Step 9.5.1

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 $16$ and whose sum is $17$ .

$1, 16$

Step 9.5.2

Write the factored form using these integers.

$- 8 x (x^{2} + 1) + (u + 1) (u + 16) = 0$

Step 9.6

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

$- 8 x (x^{2} + 1) + (x^{2} + 1) (x^{2} + 16) = 0$

Step 9.7

Factor $x^{2} + 1$ out of $- 8 x (x^{2} + 1) + (x^{2} + 1) (x^{2} + 16)$ .

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Step 9.7.1

Factor $x^{2} + 1$ out of $- 8 x (x^{2} + 1)$ .

$(x^{2} + 1) (- 8 x) + (x^{2} + 1) (x^{2} + 16) = 0$

Step 9.7.2

Factor $x^{2} + 1$ out of $(x^{2} + 1) (- 8 x) + (x^{2} + 1) (x^{2} + 16)$ .

$(x^{2} + 1) (- 8 x + x^{2} + 16) = 0$

Step 9.8

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

$(x^{2} + 1) (- 8 u + u^{2} + 16) = 0$

Step 9.9

Factor using the perfect square rule.

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Step 9.9.1

Rearrange terms.

$(x^{2} + 1) (u^{2} - 8 u + 16) = 0$

Step 9.9.2

Rewrite $16$ as $4^{2}$ .

$(x^{2} + 1) (u^{2} - 8 u + 4^{2}) = 0$

Step 9.9.3

Check that the middle term is two times the product of the numbers being squared in the first term and third term.

$8 u = 2 \cdot u \cdot 4$

Step 9.9.4

Rewrite the polynomial.

$(x^{2} + 1) (u^{2} - 2 \cdot u \cdot 4 + 4^{2}) = 0$

Step 9.9.5

Factor using the perfect square trinomial rule $a^{2} - 2 a b + b^{2} = {(a - b)}^{2}$ , where $a = u$ and $b = 4$ .

$(x^{2} + 1) {(u - 4)}^{2} = 0$

Step 9.10

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

$(x^{2} + 1) {(x - 4)}^{2} = 0$

Step 10

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

$x^{2} + 1 = 0$

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

Step 11

Set $x^{2} + 1$ equal to $0$ and solve for $x$ .

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Step 11.1

Set $x^{2} + 1$ equal to $0$ .

$x^{2} + 1 = 0$

Step 11.2

Solve $x^{2} + 1 = 0$ for $x$ .

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Step 11.2.1

Subtract $1$ from both sides of the equation.

$x^{2} = - 1$

Step 11.2.2

Take the specified root of both sides of the equation to eliminate the exponent on the left side.

$x = \pm \sqrt{- 1}$

Step 11.2.3

Rewrite $\sqrt{- 1}$ as $i$ .

$x = \pm i$

Step 11.2.4

The complete solution is the result of both the positive and negative portions of the solution.

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Step 11.2.4.1

First, use the positive value of the $\pm$ to find the first solution.

$x = i$

Step 11.2.4.2

Next, use the negative value of the $\pm$ to find the second solution.

$x = - i$

Step 11.2.4.3

The complete solution is the result of both the positive and negative portions of the solution.

$x = i, - i$

Step 12

Set ${(x - 4)}^{2}$ equal to $0$ and solve for $x$ .

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Step 12.1

Set ${(x - 4)}^{2}$ equal to $0$ .

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

Step 12.2

Solve ${(x - 4)}^{2} = 0$ for $x$ .

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Step 12.2.1

Set the $x - 4$ equal to $0$ .

$x - 4 = 0$

Step 12.2.2

Add $4$ to both sides of the equation.

$x = 4$

Step 13

The final solution is all the values that make $(x^{2} + 1) {(x - 4)}^{2} = 0$ true.

$x = i, - i, 4$

Step 14