Precalculus Examples

$f (x) = 2 x^{4} - x^{3} - 42 x^{2} + 16 x + 160$

Step 1

Set $2 x^{4} - x^{3} - 42 x^{2} + 16 x + 160$ equal to $0$ .

$2 x^{4} - x^{3} - 42 x^{2} + 16 x + 160 = 0$

Step 2

Solve for $x$ .

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Step 2.1

Factor the left side of the equation.

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Step 2.1.1

Regroup terms.

$2 x^{4} + 16 x - x^{3} - 42 x^{2} + 160 = 0$

Step 2.1.2

Factor $2 x$ out of $2 x^{4} + 16 x$ .

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Step 2.1.2.1

Factor $2 x$ out of $2 x^{4}$ .

$2 x (x^{3}) + 16 x - x^{3} - 42 x^{2} + 160 = 0$

Step 2.1.2.2

Factor $2 x$ out of $16 x$ .

$2 x (x^{3}) + 2 x (8) - x^{3} - 42 x^{2} + 160 = 0$

Step 2.1.2.3

Factor $2 x$ out of $2 x (x^{3}) + 2 x (8)$ .

$2 x (x^{3} + 8) - x^{3} - 42 x^{2} + 160 = 0$

Step 2.1.3

Rewrite $8$ as $2^{3}$ .

$2 x (x^{3} + 2^{3}) - x^{3} - 42 x^{2} + 160 = 0$

Step 2.1.4

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

$2 x ((x + 2) (x^{2} - x \cdot 2 + 2^{2})) - x^{3} - 42 x^{2} + 160 = 0$

Step 2.1.5

Factor.

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Step 2.1.5.1

Simplify.

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Step 2.1.5.1.1

Multiply $2$ by $- 1$ .

$2 x ((x + 2) (x^{2} - 2 x + 2^{2})) - x^{3} - 42 x^{2} + 160 = 0$

Step 2.1.5.1.2

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

$2 x ((x + 2) (x^{2} - 2 x + 4)) - x^{3} - 42 x^{2} + 160 = 0$

Step 2.1.5.2

Remove unnecessary parentheses.

$2 x (x + 2) (x^{2} - 2 x + 4) - x^{3} - 42 x^{2} + 160 = 0$

Step 2.1.6

Factor $- x^{3} - 42 x^{2} + 160$ using the rational roots test.

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Step 2.1.6.1

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.

$p = \pm 1, \pm 160, \pm 2, \pm 80, \pm 4, \pm 40, \pm 5, \pm 32, \pm 8, \pm 20, \pm 10, \pm 16$

$q = \pm 1$

Step 2.1.6.2

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

$\pm 1, \pm 160, \pm 2, \pm 80, \pm 4, \pm 40, \pm 5, \pm 32, \pm 8, \pm 20, \pm 10, \pm 16$

Step 2.1.6.3

Substitute $- 2$ and simplify the expression. In this case, the expression is equal to $0$ so $- 2$ is a root of the polynomial.

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Step 2.1.6.3.1

Substitute $- 2$ into the polynomial.

$- {(- 2)}^{3} - 42 {(- 2)}^{2} + 160$

Step 2.1.6.3.2

Raise $- 2$ to the power of $3$ .

$- - 8 - 42 {(- 2)}^{2} + 160$

Step 2.1.6.3.3

Multiply $- 1$ by $- 8$ .

$8 - 42 {(- 2)}^{2} + 160$

Step 2.1.6.3.4

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

$8 - 42 \cdot 4 + 160$

Step 2.1.6.3.5

Multiply $- 42$ by $4$ .

$8 - 168 + 160$

Step 2.1.6.3.6

Subtract $168$ from $8$ .

$- 160 + 160$

Step 2.1.6.3.7

Add $- 160$ and $160$ .

$0$

Step 2.1.6.4

Since $- 2$ is a known root, divide the polynomial by $x + 2$ to find the quotient polynomial. This polynomial can then be used to find the remaining roots.

$\frac{- x^{3} - 42 x^{2} + 160}{x + 2}$

Step 2.1.6.5

Divide $- x^{3} - 42 x^{2} + 160$ by $x + 2$ .

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Step 2.1.6.5.1

Set up the polynomials to be divided. If there is not a term for every exponent, insert one with a value of $0$ .

$x$

$2$

$x^{3}$

$42 x^{2}$

$0 x$

$160$

Step 2.1.6.5.2

Divide the highest order term in the dividend $- x^{3}$ by the highest order term in divisor $x$ .

				-	$x^{2}$
	$x$	+	$2$	-	$x^{3}$	-	$42 x^{2}$	+	$0 x$	+	$160$

Step 2.1.6.5.3

Multiply the new quotient term by the divisor.

			-	$x^{2}$
$x$	+	$2$	-	$x^{3}$	-	$42 x^{2}$	+	$0 x$	+	$160$
			-	$x^{3}$	-	$2 x^{2}$

Step 2.1.6.5.4

The expression needs to be subtracted from the dividend, so change all the signs in $- x^{3} - 2 x^{2}$

			-	$x^{2}$
$x$	+	$2$	-	$x^{3}$	-	$42 x^{2}$	+	$0 x$	+	$160$
			+	$x^{3}$	+	$2 x^{2}$

Step 2.1.6.5.5

After changing the signs, add the last dividend from the multiplied polynomial to find the new dividend.

			-	$x^{2}$
$x$	+	$2$	-	$x^{3}$	-	$42 x^{2}$	+	$0 x$	+	$160$
			+	$x^{3}$	+	$2 x^{2}$
					-	$40 x^{2}$

Step 2.1.6.5.6

Pull the next terms from the original dividend down into the current dividend.

			-	$x^{2}$
$x$	+	$2$	-	$x^{3}$	-	$42 x^{2}$	+	$0 x$	+	$160$
			+	$x^{3}$	+	$2 x^{2}$
					-	$40 x^{2}$	+	$0 x$

Step 2.1.6.5.7

Divide the highest order term in the dividend $- 40 x^{2}$ by the highest order term in divisor $x$ .

			-	$x^{2}$	-	$40 x$
$x$	+	$2$	-	$x^{3}$	-	$42 x^{2}$	+	$0 x$	+	$160$
			+	$x^{3}$	+	$2 x^{2}$
					-	$40 x^{2}$	+	$0 x$

Step 2.1.6.5.8

Multiply the new quotient term by the divisor.

			-	$x^{2}$	-	$40 x$
$x$	+	$2$	-	$x^{3}$	-	$42 x^{2}$	+	$0 x$	+	$160$
			+	$x^{3}$	+	$2 x^{2}$
					-	$40 x^{2}$	+	$0 x$
					-	$40 x^{2}$	-	$80 x$

Step 2.1.6.5.9

The expression needs to be subtracted from the dividend, so change all the signs in $- 40 x^{2} - 80 x$

			-	$x^{2}$	-	$40 x$
$x$	+	$2$	-	$x^{3}$	-	$42 x^{2}$	+	$0 x$	+	$160$
			+	$x^{3}$	+	$2 x^{2}$
					-	$40 x^{2}$	+	$0 x$
					+	$40 x^{2}$	+	$80 x$

Step 2.1.6.5.10

After changing the signs, add the last dividend from the multiplied polynomial to find the new dividend.

			-	$x^{2}$	-	$40 x$
$x$	+	$2$	-	$x^{3}$	-	$42 x^{2}$	+	$0 x$	+	$160$
			+	$x^{3}$	+	$2 x^{2}$
					-	$40 x^{2}$	+	$0 x$
					+	$40 x^{2}$	+	$80 x$
							+	$80 x$

Step 2.1.6.5.11

Pull the next terms from the original dividend down into the current dividend.

			-	$x^{2}$	-	$40 x$
$x$	+	$2$	-	$x^{3}$	-	$42 x^{2}$	+	$0 x$	+	$160$
			+	$x^{3}$	+	$2 x^{2}$
					-	$40 x^{2}$	+	$0 x$
					+	$40 x^{2}$	+	$80 x$
							+	$80 x$	+	$160$

Step 2.1.6.5.12

Divide the highest order term in the dividend $80 x$ by the highest order term in divisor $x$ .

			-	$x^{2}$	-	$40 x$	+	$80$
$x$	+	$2$	-	$x^{3}$	-	$42 x^{2}$	+	$0 x$	+	$160$
			+	$x^{3}$	+	$2 x^{2}$
					-	$40 x^{2}$	+	$0 x$
					+	$40 x^{2}$	+	$80 x$
							+	$80 x$	+	$160$

Step 2.1.6.5.13

Multiply the new quotient term by the divisor.

			-	$x^{2}$	-	$40 x$	+	$80$
$x$	+	$2$	-	$x^{3}$	-	$42 x^{2}$	+	$0 x$	+	$160$
			+	$x^{3}$	+	$2 x^{2}$
					-	$40 x^{2}$	+	$0 x$
					+	$40 x^{2}$	+	$80 x$
							+	$80 x$	+	$160$
							+	$80 x$	+	$160$

Step 2.1.6.5.14

The expression needs to be subtracted from the dividend, so change all the signs in $80 x + 160$

			-	$x^{2}$	-	$40 x$	+	$80$
$x$	+	$2$	-	$x^{3}$	-	$42 x^{2}$	+	$0 x$	+	$160$
			+	$x^{3}$	+	$2 x^{2}$
					-	$40 x^{2}$	+	$0 x$
					+	$40 x^{2}$	+	$80 x$
							+	$80 x$	+	$160$
							-	$80 x$	-	$160$

Step 2.1.6.5.15

After changing the signs, add the last dividend from the multiplied polynomial to find the new dividend.

			-	$x^{2}$	-	$40 x$	+	$80$
$x$	+	$2$	-	$x^{3}$	-	$42 x^{2}$	+	$0 x$	+	$160$
			+	$x^{3}$	+	$2 x^{2}$
					-	$40 x^{2}$	+	$0 x$
					+	$40 x^{2}$	+	$80 x$
							+	$80 x$	+	$160$
							-	$80 x$	-	$160$
										$0$

Step 2.1.6.5.16

Since the remander is $0$ , the final answer is the quotient.

$- x^{2} - 40 x + 80$

Step 2.1.6.6

Write $- x^{3} - 42 x^{2} + 160$ as a set of factors.

$2 x (x + 2) (x^{2} - 2 x + 4) + (x + 2) (- x^{2} - 40 x + 80) = 0$

Step 2.1.7

Factor $x + 2$ out of $2 x (x + 2) (x^{2} - 2 x + 4) + (x + 2) (- x^{2} - 40 x + 80)$ .

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Step 2.1.7.1

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

$(x + 2) (2 x (x^{2} - 2 x + 4)) + (x + 2) (- x^{2} - 40 x + 80) = 0$

Step 2.1.7.2

Factor $x + 2$ out of $(x + 2) (2 x (x^{2} - 2 x + 4)) + (x + 2) (- x^{2} - 40 x + 80)$ .

$(x + 2) (2 x (x^{2} - 2 x + 4) - x^{2} - 40 x + 80) = 0$

Step 2.1.8

Apply the distributive property.

$(x + 2) (2 x \cdot x^{2} + 2 x (- 2 x) + 2 x \cdot 4 - x^{2} - 40 x + 80) = 0$

Step 2.1.9

Simplify.

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Step 2.1.9.1

Multiply $x$ by $x^{2}$ by adding the exponents.

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Step 2.1.9.1.1

Move $x^{2}$ .

$(x + 2) (2 (x^{2} x) + 2 x (- 2 x) + 2 x \cdot 4 - x^{2} - 40 x + 80) = 0$

Step 2.1.9.1.2

Multiply $x^{2}$ by $x$ .

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Step 2.1.9.1.2.1

Raise $x$ to the power of $1$ .

$(x + 2) (2 (x^{2} x) + 2 x (- 2 x) + 2 x \cdot 4 - x^{2} - 40 x + 80) = 0$

Step 2.1.9.1.2.2

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

$(x + 2) (2 x^{2 + 1} + 2 x (- 2 x) + 2 x \cdot 4 - x^{2} - 40 x + 80) = 0$

Step 2.1.9.1.3

Add $2$ and $1$ .

$(x + 2) (2 x^{3} + 2 x (- 2 x) + 2 x \cdot 4 - x^{2} - 40 x + 80) = 0$

Step 2.1.9.2

Rewrite using the commutative property of multiplication.

$(x + 2) (2 x^{3} + 2 \cdot (- 2 x \cdot x) + 2 x \cdot 4 - x^{2} - 40 x + 80) = 0$

Step 2.1.9.3

Multiply $4$ by $2$ .

$(x + 2) (2 x^{3} + 2 \cdot (- 2 x \cdot x) + 8 x - x^{2} - 40 x + 80) = 0$

Step 2.1.10

Simplify each term.

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Step 2.1.10.1

Multiply $x$ by $x$ by adding the exponents.

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Step 2.1.10.1.1

Move $x$ .

$(x + 2) (2 x^{3} + 2 \cdot (- 2 (x \cdot x)) + 8 x - x^{2} - 40 x + 80) = 0$

Step 2.1.10.1.2

Multiply $x$ by $x$ .

$(x + 2) (2 x^{3} + 2 \cdot (- 2 x^{2}) + 8 x - x^{2} - 40 x + 80) = 0$

Step 2.1.10.2

Multiply $2$ by $- 2$ .

$(x + 2) (2 x^{3} - 4 x^{2} + 8 x - x^{2} - 40 x + 80) = 0$

Step 2.1.11

Subtract $x^{2}$ from $- 4 x^{2}$ .

$(x + 2) (2 x^{3} - 5 x^{2} + 8 x - 40 x + 80) = 0$

Step 2.1.12

Subtract $40 x$ from $8 x$ .

$(x + 2) (2 x^{3} - 5 x^{2} - 32 x + 80) = 0$

Step 2.1.13

Factor.

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Step 2.1.13.1

Rewrite $2 x^{3} - 5 x^{2} - 32 x + 80$ in a factored form.

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Step 2.1.13.1.1

Factor out the greatest common factor from each group.

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Step 2.1.13.1.1.1

Group the first two terms and the last two terms.

$(x + 2) ((2 x^{3} - 5 x^{2}) - 32 x + 80) = 0$

Step 2.1.13.1.1.2

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

$(x + 2) (x^{2} (2 x - 5) - 16 (2 x - 5)) = 0$

Step 2.1.13.1.2

Factor the polynomial by factoring out the greatest common factor, $2 x - 5$ .

$(x + 2) ((2 x - 5) (x^{2} - 16)) = 0$

Step 2.1.13.1.3

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

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

Step 2.1.13.1.4

Factor.

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Step 2.1.13.1.4.1

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 $b = 4$ .

$(x + 2) ((2 x - 5) ((x + 4) (x - 4))) = 0$

Step 2.1.13.1.4.2

Remove unnecessary parentheses.

$(x + 2) ((2 x - 5) (x + 4) (x - 4)) = 0$

Step 2.1.13.2

Remove unnecessary parentheses.

$(x + 2) (2 x - 5) (x + 4) (x - 4) = 0$

Step 2.2

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

$x + 2 = 0$

$2 x - 5 = 0$

$x + 4 = 0$

$x - 4 = 0$

Step 2.3

Set $x + 2$ equal to $0$ and solve for $x$ .

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Step 2.3.1

Set $x + 2$ equal to $0$ .

$x + 2 = 0$

Step 2.3.2

Subtract $2$ from both sides of the equation.

$x = - 2$

Step 2.4

Set $2 x - 5$ equal to $0$ and solve for $x$ .

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Step 2.4.1

Set $2 x - 5$ equal to $0$ .

$2 x - 5 = 0$

Step 2.4.2

Solve $2 x - 5 = 0$ for $x$ .

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Step 2.4.2.1

Add $5$ to both sides of the equation.

$2 x = 5$

Step 2.4.2.2

Divide each term in $2 x = 5$ by $2$ and simplify.

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Step 2.4.2.2.1

Divide each term in $2 x = 5$ by $2$ .

$\frac{2 x}{2} = \frac{5}{2}$

Step 2.4.2.2.2

Simplify the left side.

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Step 2.4.2.2.2.1

Cancel the common factor of $2$ .

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Step 2.4.2.2.2.1.1

Cancel the common factor.

$\frac{2 x}{2} = \frac{5}{2}$

Step 2.4.2.2.2.1.2

Divide $x$ by $1$ .

$x = \frac{5}{2}$

Step 2.5

Set $x + 4$ equal to $0$ and solve for $x$ .

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Step 2.5.1

Set $x + 4$ equal to $0$ .

$x + 4 = 0$

Step 2.5.2

Subtract $4$ from both sides of the equation.

$x = - 4$

Step 2.6

Set $x - 4$ equal to $0$ and solve for $x$ .

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Step 2.6.1

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

$x - 4 = 0$

Step 2.6.2

Add $4$ to both sides of the equation.

$x = 4$

Step 2.7

The final solution is all the values that make $(x + 2) (2 x - 5) (x + 4) (x - 4) = 0$ true.

$x = - 2, \frac{5}{2}, - 4, 4$

Step 3