Precalculus Examples

$f (x) = - 4 x^{3} + 23 x^{2} - 14 x - 5$

Step 1

Set $- 4 x^{3} + 23 x^{2} - 14 x - 5$ equal to $0$ .

$- 4 x^{3} + 23 x^{2} - 14 x - 5 = 0$

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

Factor $- 4 x^{3} + 23 x^{2} - 14 x - 5$ using the rational roots test.

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Step 2.1.1.1

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.

$p = \pm 1, \pm 5$

$q = \pm 1, \pm 4, \pm 2$

Step 2.1.1.2

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

$\pm 1, \pm 0.25, \pm 0.5, \pm 5, \pm 1.25, \pm 2.5$

Step 2.1.1.3

Substitute $- 0.25$ and simplify the expression. In this case, the expression is equal to $0$ so $- 0.25$ is a root of the polynomial.

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Step 2.1.1.3.1

Substitute $- 0.25$ into the polynomial.

$- 4 {(- 0.25)}^{3} + 23 {(- 0.25)}^{2} - 14 \cdot - 0.25 - 5$

Step 2.1.1.3.2

Raise $- 0.25$ to the power of $3$ .

$- 4 \cdot - 0.015625 + 23 {(- 0.25)}^{2} - 14 \cdot - 0.25 - 5$

Step 2.1.1.3.3

Multiply $- 4$ by $- 0.015625$ .

$0.0625 + 23 {(- 0.25)}^{2} - 14 \cdot - 0.25 - 5$

Step 2.1.1.3.4

Raise $- 0.25$ to the power of $2$ .

$0.0625 + 23 \cdot 0.0625 - 14 \cdot - 0.25 - 5$

Step 2.1.1.3.5

Multiply $23$ by $0.0625$ .

$0.0625 + 1.4375 - 14 \cdot - 0.25 - 5$

Step 2.1.1.3.6

Add $0.0625$ and $1.4375$ .

$1.5 - 14 \cdot - 0.25 - 5$

Step 2.1.1.3.7

Multiply $- 14$ by $- 0.25$ .

$1.5 + 3.5 - 5$

Step 2.1.1.3.8

Add $1.5$ and $3.5$ .

$5 - 5$

Step 2.1.1.3.9

Subtract $5$ from $5$ .

$0$

Step 2.1.1.4

Since $- 0.25$ is a known root, divide the polynomial by $4 x + 1$ to find the quotient polynomial. This polynomial can then be used to find the remaining roots.

$\frac{- 4 x^{3} + 23 x^{2} - 14 x - 5}{4 x + 1}$

Step 2.1.1.5

Divide $- 4 x^{3} + 23 x^{2} - 14 x - 5$ by $4 x + 1$ .

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Step 2.1.1.5.1

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

$4 x$

$1$

$4 x^{3}$

$23 x^{2}$

$14 x$

$5$

Step 2.1.1.5.2

Divide the highest order term in the dividend $- 4 x^{3}$ by the highest order term in divisor $4 x$ .

				-	$x^{2}$
	$4 x$	+	$1$	-	$4 x^{3}$	+	$23 x^{2}$	-	$14 x$	-	$5$

Step 2.1.1.5.3

Multiply the new quotient term by the divisor.

			-	$x^{2}$
$4 x$	+	$1$	-	$4 x^{3}$	+	$23 x^{2}$	-	$14 x$	-	$5$
			-	$4 x^{3}$	-	$x^{2}$

Step 2.1.1.5.4

The expression needs to be subtracted from the dividend, so change all the signs in $- 4 x^{3} - x^{2}$

			-	$x^{2}$
$4 x$	+	$1$	-	$4 x^{3}$	+	$23 x^{2}$	-	$14 x$	-	$5$
			+	$4 x^{3}$	+	$x^{2}$

Step 2.1.1.5.5

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

			-	$x^{2}$
$4 x$	+	$1$	-	$4 x^{3}$	+	$23 x^{2}$	-	$14 x$	-	$5$
			+	$4 x^{3}$	+	$x^{2}$
					+	$24 x^{2}$

Step 2.1.1.5.6

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

			-	$x^{2}$
$4 x$	+	$1$	-	$4 x^{3}$	+	$23 x^{2}$	-	$14 x$	-	$5$
			+	$4 x^{3}$	+	$x^{2}$
					+	$24 x^{2}$	-	$14 x$

Step 2.1.1.5.7

Divide the highest order term in the dividend $24 x^{2}$ by the highest order term in divisor $4 x$ .

			-	$x^{2}$	+	$6 x$
$4 x$	+	$1$	-	$4 x^{3}$	+	$23 x^{2}$	-	$14 x$	-	$5$
			+	$4 x^{3}$	+	$x^{2}$
					+	$24 x^{2}$	-	$14 x$

Step 2.1.1.5.8

Multiply the new quotient term by the divisor.

			-	$x^{2}$	+	$6 x$
$4 x$	+	$1$	-	$4 x^{3}$	+	$23 x^{2}$	-	$14 x$	-	$5$
			+	$4 x^{3}$	+	$x^{2}$
					+	$24 x^{2}$	-	$14 x$
					+	$24 x^{2}$	+	$6 x$

Step 2.1.1.5.9

The expression needs to be subtracted from the dividend, so change all the signs in $24 x^{2} + 6 x$

			-	$x^{2}$	+	$6 x$
$4 x$	+	$1$	-	$4 x^{3}$	+	$23 x^{2}$	-	$14 x$	-	$5$
			+	$4 x^{3}$	+	$x^{2}$
					+	$24 x^{2}$	-	$14 x$
					-	$24 x^{2}$	-	$6 x$

Step 2.1.1.5.10

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

			-	$x^{2}$	+	$6 x$
$4 x$	+	$1$	-	$4 x^{3}$	+	$23 x^{2}$	-	$14 x$	-	$5$
			+	$4 x^{3}$	+	$x^{2}$
					+	$24 x^{2}$	-	$14 x$
					-	$24 x^{2}$	-	$6 x$
							-	$20 x$

Step 2.1.1.5.11

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

			-	$x^{2}$	+	$6 x$
$4 x$	+	$1$	-	$4 x^{3}$	+	$23 x^{2}$	-	$14 x$	-	$5$
			+	$4 x^{3}$	+	$x^{2}$
					+	$24 x^{2}$	-	$14 x$
					-	$24 x^{2}$	-	$6 x$
							-	$20 x$	-	$5$

Step 2.1.1.5.12

Divide the highest order term in the dividend $- 20 x$ by the highest order term in divisor $4 x$ .

			-	$x^{2}$	+	$6 x$	-	$5$
$4 x$	+	$1$	-	$4 x^{3}$	+	$23 x^{2}$	-	$14 x$	-	$5$
			+	$4 x^{3}$	+	$x^{2}$
					+	$24 x^{2}$	-	$14 x$
					-	$24 x^{2}$	-	$6 x$
							-	$20 x$	-	$5$

Step 2.1.1.5.13

Multiply the new quotient term by the divisor.

			-	$x^{2}$	+	$6 x$	-	$5$
$4 x$	+	$1$	-	$4 x^{3}$	+	$23 x^{2}$	-	$14 x$	-	$5$
			+	$4 x^{3}$	+	$x^{2}$
					+	$24 x^{2}$	-	$14 x$
					-	$24 x^{2}$	-	$6 x$
							-	$20 x$	-	$5$
							-	$20 x$	-	$5$

Step 2.1.1.5.14

The expression needs to be subtracted from the dividend, so change all the signs in $- 20 x - 5$

			-	$x^{2}$	+	$6 x$	-	$5$
$4 x$	+	$1$	-	$4 x^{3}$	+	$23 x^{2}$	-	$14 x$	-	$5$
			+	$4 x^{3}$	+	$x^{2}$
					+	$24 x^{2}$	-	$14 x$
					-	$24 x^{2}$	-	$6 x$
							-	$20 x$	-	$5$
							+	$20 x$	+	$5$

Step 2.1.1.5.15

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

			-	$x^{2}$	+	$6 x$	-	$5$
$4 x$	+	$1$	-	$4 x^{3}$	+	$23 x^{2}$	-	$14 x$	-	$5$
			+	$4 x^{3}$	+	$x^{2}$
					+	$24 x^{2}$	-	$14 x$
					-	$24 x^{2}$	-	$6 x$
							-	$20 x$	-	$5$
							+	$20 x$	+	$5$
										$0$

Step 2.1.1.5.16

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

$- x^{2} + 6 x - 5$

Step 2.1.1.6

Write $- 4 x^{3} + 23 x^{2} - 14 x - 5$ as a set of factors.

$(4 x + 1) (- x^{2} + 6 x - 5) = 0$

Step 2.1.2

Factor by grouping.

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Step 2.1.2.1

Factor by grouping.

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Step 2.1.2.1.1

For a polynomial of the form $a x^{2} + b x + c$ , rewrite the middle term as a sum of two terms whose product is $a \cdot c = - 1 \cdot - 5 = 5$ and whose sum is $b = 6$ .

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Step 2.1.2.1.1.1

Factor $6$ out of $6 x$ .

$(4 x + 1) (- x^{2} + 6 (x) - 5) = 0$

Step 2.1.2.1.1.2

Rewrite $6$ as $1$ plus $5$

$(4 x + 1) (- x^{2} + (1 + 5) x - 5) = 0$

Step 2.1.2.1.1.3

Apply the distributive property.

$(4 x + 1) (- x^{2} + 1 x + 5 x - 5) = 0$

Step 2.1.2.1.1.4

Multiply $x$ by $1$ .

$(4 x + 1) (- x^{2} + x + 5 x - 5) = 0$

Step 2.1.2.1.2

Factor out the greatest common factor from each group.

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Step 2.1.2.1.2.1

Group the first two terms and the last two terms.

$(4 x + 1) ((- x^{2} + x) + 5 x - 5) = 0$

Step 2.1.2.1.2.2

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

$(4 x + 1) (x (- x + 1) - 5 (- x + 1)) = 0$

Step 2.1.2.1.3

Factor the polynomial by factoring out the greatest common factor, $- x + 1$ .

$(4 x + 1) ((- x + 1) (x - 5)) = 0$

Step 2.1.2.2

Remove unnecessary parentheses.

$(4 x + 1) (- x + 1) (x - 5) = 0$

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 $0$ .

$4 x + 1 = 0$

$- x + 1 = 0$

$x - 5 = 0$

Step 2.3

Set $4 x + 1$ equal to $0$ and solve for $x$ .

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Step 2.3.1

Set $4 x + 1$ equal to $0$ .

$4 x + 1 = 0$

Step 2.3.2

Solve $4 x + 1 = 0$ for $x$ .

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Step 2.3.2.1

Subtract $1$ from both sides of the equation.

$4 x = - 1$

Step 2.3.2.2

Divide each term in $4 x = - 1$ by $4$ and simplify.

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Step 2.3.2.2.1

Divide each term in $4 x = - 1$ by $4$ .

$\frac{4 x}{4} = \frac{- 1}{4}$

Step 2.3.2.2.2

Simplify the left side.

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Step 2.3.2.2.2.1

Cancel the common factor of $4$ .

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Step 2.3.2.2.2.1.1

Cancel the common factor.

$\frac{4 x}{4} = \frac{- 1}{4}$

Step 2.3.2.2.2.1.2

Divide $x$ by $1$ .

$x = \frac{- 1}{4}$

Step 2.3.2.2.3

Simplify the right side.

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Step 2.3.2.2.3.1

Move the negative in front of the fraction.

$x = - \frac{1}{4}$

Step 2.4

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

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Step 2.4.1

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

$- x + 1 = 0$

Step 2.4.2

Solve $- x + 1 = 0$ for $x$ .

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Step 2.4.2.1

Subtract $1$ from both sides of the equation.

$- x = - 1$

Step 2.4.2.2

Divide each term in $- x = - 1$ by $- 1$ and simplify.

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Step 2.4.2.2.1

Divide each term in $- x = - 1$ by $- 1$ .

$\frac{- x}{- 1} = \frac{- 1}{- 1}$

Step 2.4.2.2.2

Simplify the left side.

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Step 2.4.2.2.2.1

Dividing two negative values results in a positive value.

$\frac{x}{1} = \frac{- 1}{- 1}$

Step 2.4.2.2.2.2

Divide $x$ by $1$ .

$x = \frac{- 1}{- 1}$

Step 2.4.2.2.3

Simplify the right side.

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Step 2.4.2.2.3.1

Divide $- 1$ by $- 1$ .

$x = 1$

Step 2.5

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

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Step 2.5.1

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

$x - 5 = 0$

Step 2.5.2

Add $5$ to both sides of the equation.

$x = 5$

Step 2.6

The final solution is all the values that make $(4 x + 1) (- x + 1) (x - 5) = 0$ true.

$x = - \frac{1}{4}, 1, 5$

Step 3