Precalculus Examples

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

Step 1

Simplify each term.

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Step 1.1

Apply the distributive property.

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

Step 1.2

Multiply $2$ by $- 1$ .

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

Step 1.3

Multiply $- 1$ by $- 5$ .

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

Step 2

Factor the left side of the equation.

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Step 2.1

Use the Binomial Theorem.

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

Step 2.2

Simplify each term.

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Step 2.2.1

Apply the product rule to $2 x$ .

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

Step 2.2.2

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

$8 x^{3} + 3 {(2 x)}^{2} \cdot - 5 + 3 (2 x) {(- 5)}^{2} + {(- 5)}^{3} - 2 x + 5 = 0$

Step 2.2.3

Apply the product rule to $2 x$ .

$8 x^{3} + 3 (2^{2} x^{2}) \cdot - 5 + 3 (2 x) {(- 5)}^{2} + {(- 5)}^{3} - 2 x + 5 = 0$

Step 2.2.4

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

$8 x^{3} + 3 (4 x^{2}) \cdot - 5 + 3 (2 x) {(- 5)}^{2} + {(- 5)}^{3} - 2 x + 5 = 0$

Step 2.2.5

Multiply $4$ by $3$ .

$8 x^{3} + 12 x^{2} \cdot - 5 + 3 (2 x) {(- 5)}^{2} + {(- 5)}^{3} - 2 x + 5 = 0$

Step 2.2.6

Multiply $- 5$ by $12$ .

$8 x^{3} - 60 x^{2} + 3 (2 x) {(- 5)}^{2} + {(- 5)}^{3} - 2 x + 5 = 0$

Step 2.2.7

Multiply $2$ by $3$ .

$8 x^{3} - 60 x^{2} + 6 x {(- 5)}^{2} + {(- 5)}^{3} - 2 x + 5 = 0$

Step 2.2.8

Raise $- 5$ to the power of $2$ .

$8 x^{3} - 60 x^{2} + 6 x \cdot 25 + {(- 5)}^{3} - 2 x + 5 = 0$

Step 2.2.9

Multiply $25$ by $6$ .

$8 x^{3} - 60 x^{2} + 150 x + {(- 5)}^{3} - 2 x + 5 = 0$

Step 2.2.10

Raise $- 5$ to the power of $3$ .

$8 x^{3} - 60 x^{2} + 150 x - 125 - 2 x + 5 = 0$

Step 2.3

Subtract $2 x$ from $150 x$ .

$8 x^{3} - 60 x^{2} + 148 x - 125 + 5 = 0$

Step 2.4

Add $- 125$ and $5$ .

$8 x^{3} - 60 x^{2} + 148 x - 120 = 0$

Step 2.5

Rewrite $8 x^{3} - 60 x^{2} + 148 x - 120$ in a factored form.

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Step 2.5.1

Factor $4$ out of $8 x^{3} - 60 x^{2} + 148 x - 120$ .

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Step 2.5.1.1

Factor $4$ out of $8 x^{3}$ .

$4 (2 x^{3}) - 60 x^{2} + 148 x - 120 = 0$

Step 2.5.1.2

Factor $4$ out of $- 60 x^{2}$ .

$4 (2 x^{3}) + 4 (- 15 x^{2}) + 148 x - 120 = 0$

Step 2.5.1.3

Factor $4$ out of $148 x$ .

$4 (2 x^{3}) + 4 (- 15 x^{2}) + 4 (37 x) - 120 = 0$

Step 2.5.1.4

Factor $4$ out of $- 120$ .

$4 (2 x^{3}) + 4 (- 15 x^{2}) + 4 (37 x) + 4 (- 30) = 0$

Step 2.5.1.5

Factor $4$ out of $4 (2 x^{3}) + 4 (- 15 x^{2})$ .

$4 (2 x^{3} - 15 x^{2}) + 4 (37 x) + 4 (- 30) = 0$

Step 2.5.1.6

Factor $4$ out of $4 (2 x^{3} - 15 x^{2}) + 4 (37 x)$ .

$4 (2 x^{3} - 15 x^{2} + 37 x) + 4 (- 30) = 0$

Step 2.5.1.7

Factor $4$ out of $4 (2 x^{3} - 15 x^{2} + 37 x) + 4 (- 30)$ .

$4 (2 x^{3} - 15 x^{2} + 37 x - 30) = 0$

Step 2.5.2

Factor $2 x^{3} - 15 x^{2} + 37 x - 30$ using the rational roots test.

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Step 2.5.2.1

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.

$p = \pm 1, \pm 30, \pm 2, \pm 15, \pm 3, \pm 10, \pm 5, \pm 6$

$q = \pm 1, \pm 2$

Step 2.5.2.2

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

$\pm 1, \pm 0.5, \pm 30, \pm 15, \pm 2, \pm 7.5, \pm 3, \pm 1.5, \pm 10, \pm 5, \pm 2.5, \pm 6$

Step 2.5.2.3

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 2.5.2.3.1

Substitute $2$ into the polynomial.

$2 \cdot 2^{3} - 15 \cdot 2^{2} + 37 \cdot 2 - 30$

Step 2.5.2.3.2

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

$2 \cdot 8 - 15 \cdot 2^{2} + 37 \cdot 2 - 30$

Step 2.5.2.3.3

Multiply $2$ by $8$ .

$16 - 15 \cdot 2^{2} + 37 \cdot 2 - 30$

Step 2.5.2.3.4

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

$16 - 15 \cdot 4 + 37 \cdot 2 - 30$

Step 2.5.2.3.5

Multiply $- 15$ by $4$ .

$16 - 60 + 37 \cdot 2 - 30$

Step 2.5.2.3.6

Subtract $60$ from $16$ .

$- 44 + 37 \cdot 2 - 30$

Step 2.5.2.3.7

Multiply $37$ by $2$ .

$- 44 + 74 - 30$

Step 2.5.2.3.8

Add $- 44$ and $74$ .

$30 - 30$

Step 2.5.2.3.9

Subtract $30$ from $30$ .

$0$

Step 2.5.2.4

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.

$\frac{2 x^{3} - 15 x^{2} + 37 x - 30}{x - 2}$

Step 2.5.2.5

Divide $2 x^{3} - 15 x^{2} + 37 x - 30$ by $x - 2$ .

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Step 2.5.2.5.1

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

$x$

$2$

$2 x^{3}$

$15 x^{2}$

$37 x$

$30$

Step 2.5.2.5.2

Divide the highest order term in the dividend $2 x^{3}$ by the highest order term in divisor $x$ .

					$2 x^{2}$
	$x$	-	$2$		$2 x^{3}$	-	$15 x^{2}$	+	$37 x$	-	$30$

Step 2.5.2.5.3

Multiply the new quotient term by the divisor.

				$2 x^{2}$
$x$	-	$2$		$2 x^{3}$	-	$15 x^{2}$	+	$37 x$	-	$30$
			+	$2 x^{3}$	-	$4 x^{2}$

Step 2.5.2.5.4

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

				$2 x^{2}$
$x$	-	$2$		$2 x^{3}$	-	$15 x^{2}$	+	$37 x$	-	$30$
			-	$2 x^{3}$	+	$4 x^{2}$

Step 2.5.2.5.5

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

				$2 x^{2}$
$x$	-	$2$		$2 x^{3}$	-	$15 x^{2}$	+	$37 x$	-	$30$
			-	$2 x^{3}$	+	$4 x^{2}$
					-	$11 x^{2}$

Step 2.5.2.5.6

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

				$2 x^{2}$
$x$	-	$2$		$2 x^{3}$	-	$15 x^{2}$	+	$37 x$	-	$30$
			-	$2 x^{3}$	+	$4 x^{2}$
					-	$11 x^{2}$	+	$37 x$

Step 2.5.2.5.7

Divide the highest order term in the dividend $- 11 x^{2}$ by the highest order term in divisor $x$ .

				$2 x^{2}$	-	$11 x$
$x$	-	$2$		$2 x^{3}$	-	$15 x^{2}$	+	$37 x$	-	$30$
			-	$2 x^{3}$	+	$4 x^{2}$
					-	$11 x^{2}$	+	$37 x$

Step 2.5.2.5.8

Multiply the new quotient term by the divisor.

				$2 x^{2}$	-	$11 x$
$x$	-	$2$		$2 x^{3}$	-	$15 x^{2}$	+	$37 x$	-	$30$
			-	$2 x^{3}$	+	$4 x^{2}$
					-	$11 x^{2}$	+	$37 x$
					-	$11 x^{2}$	+	$22 x$

Step 2.5.2.5.9

The expression needs to be subtracted from the dividend, so change all the signs in $- 11 x^{2} + 22 x$

				$2 x^{2}$	-	$11 x$
$x$	-	$2$		$2 x^{3}$	-	$15 x^{2}$	+	$37 x$	-	$30$
			-	$2 x^{3}$	+	$4 x^{2}$
					-	$11 x^{2}$	+	$37 x$
					+	$11 x^{2}$	-	$22 x$

Step 2.5.2.5.10

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

				$2 x^{2}$	-	$11 x$
$x$	-	$2$		$2 x^{3}$	-	$15 x^{2}$	+	$37 x$	-	$30$
			-	$2 x^{3}$	+	$4 x^{2}$
					-	$11 x^{2}$	+	$37 x$
					+	$11 x^{2}$	-	$22 x$
							+	$15 x$

Step 2.5.2.5.11

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

				$2 x^{2}$	-	$11 x$
$x$	-	$2$		$2 x^{3}$	-	$15 x^{2}$	+	$37 x$	-	$30$
			-	$2 x^{3}$	+	$4 x^{2}$
					-	$11 x^{2}$	+	$37 x$
					+	$11 x^{2}$	-	$22 x$
							+	$15 x$	-	$30$

Step 2.5.2.5.12

Divide the highest order term in the dividend $15 x$ by the highest order term in divisor $x$ .

				$2 x^{2}$	-	$11 x$	+	$15$
$x$	-	$2$		$2 x^{3}$	-	$15 x^{2}$	+	$37 x$	-	$30$
			-	$2 x^{3}$	+	$4 x^{2}$
					-	$11 x^{2}$	+	$37 x$
					+	$11 x^{2}$	-	$22 x$
							+	$15 x$	-	$30$

Step 2.5.2.5.13

Multiply the new quotient term by the divisor.

				$2 x^{2}$	-	$11 x$	+	$15$
$x$	-	$2$		$2 x^{3}$	-	$15 x^{2}$	+	$37 x$	-	$30$
			-	$2 x^{3}$	+	$4 x^{2}$
					-	$11 x^{2}$	+	$37 x$
					+	$11 x^{2}$	-	$22 x$
							+	$15 x$	-	$30$
							+	$15 x$	-	$30$

Step 2.5.2.5.14

The expression needs to be subtracted from the dividend, so change all the signs in $15 x - 30$

				$2 x^{2}$	-	$11 x$	+	$15$
$x$	-	$2$		$2 x^{3}$	-	$15 x^{2}$	+	$37 x$	-	$30$
			-	$2 x^{3}$	+	$4 x^{2}$
					-	$11 x^{2}$	+	$37 x$
					+	$11 x^{2}$	-	$22 x$
							+	$15 x$	-	$30$
							-	$15 x$	+	$30$

Step 2.5.2.5.15

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

				$2 x^{2}$	-	$11 x$	+	$15$
$x$	-	$2$		$2 x^{3}$	-	$15 x^{2}$	+	$37 x$	-	$30$
			-	$2 x^{3}$	+	$4 x^{2}$
					-	$11 x^{2}$	+	$37 x$
					+	$11 x^{2}$	-	$22 x$
							+	$15 x$	-	$30$
							-	$15 x$	+	$30$
										$0$

Step 2.5.2.5.16

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

$2 x^{2} - 11 x + 15$

Step 2.5.2.6

Write $2 x^{3} - 15 x^{2} + 37 x - 30$ as a set of factors.

$4 ((x - 2) (2 x^{2} - 11 x + 15)) = 0$

Step 2.5.3

Factor.

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Step 2.5.3.1

Factor by grouping.

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Step 2.5.3.1.1

Factor by grouping.

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Step 2.5.3.1.1.1

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 $a \cdot c = 2 \cdot 15 = 30$ and whose sum is $b = - 11$ .

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Step 2.5.3.1.1.1.1

Factor $- 11$ out of $- 11 x$ .

$4 ((x - 2) (2 x^{2} - 11 x + 15)) = 0$

Step 2.5.3.1.1.1.2

Rewrite $- 11$ as $- 5$ plus $- 6$

$4 ((x - 2) (2 x^{2} + (- 5 - 6) x + 15)) = 0$

Step 2.5.3.1.1.1.3

Apply the distributive property.

$4 ((x - 2) (2 x^{2} - 5 x - 6 x + 15)) = 0$

Step 2.5.3.1.1.2

Factor out the greatest common factor from each group.

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Step 2.5.3.1.1.2.1

Group the first two terms and the last two terms.

$4 ((x - 2) ((2 x^{2} - 5 x) - 6 x + 15)) = 0$

Step 2.5.3.1.1.2.2

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

$4 ((x - 2) (x (2 x - 5) - 3 (2 x - 5))) = 0$

Step 2.5.3.1.1.3

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

$4 ((x - 2) ((2 x - 5) (x - 3))) = 0$

Step 2.5.3.1.2

Remove unnecessary parentheses.

$4 ((x - 2) (2 x - 5) (x - 3)) = 0$

Step 2.5.3.2

Remove unnecessary parentheses.

$4 (x - 2) (2 x - 5) (x - 3) = 0$

Step 3

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

$x - 2 = 0$

$2 x - 5 = 0$

$x - 3 = 0$

Step 4

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

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Step 4.1

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

$x - 2 = 0$

Step 4.2

Add $2$ to both sides of the equation.

$x = 2$

Step 5

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

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Step 5.1

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

$2 x - 5 = 0$

Step 5.2

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

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Step 5.2.1

Add $5$ to both sides of the equation.

$2 x = 5$

Step 5.2.2

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

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Step 5.2.2.1

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

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

Step 5.2.2.2

Simplify the left side.

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Step 5.2.2.2.1

Cancel the common factor of $2$ .

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Step 5.2.2.2.1.1

Cancel the common factor.

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

Step 5.2.2.2.1.2

Divide $x$ by $1$ .

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

Step 6

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

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Step 6.1

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

$x - 3 = 0$

Step 6.2

Add $3$ to both sides of the equation.

$x = 3$

Step 7

The final solution is all the values that make $4 (x - 2) (2 x - 5) (x - 3) = 0$ true.

$x = 2, \frac{5}{2}, 3$

Step 8

The result can be shown in multiple forms.

Exact Form:

$x = 2, \frac{5}{2}, 3$

Decimal Form:

$x = 2, 2.5, 3$

Mixed Number Form:

$x = 2, 2 \frac{1}{2}, 3$