Precalculus Examples

$f (x) = 4 x^{5} + 12 x^{4} - 11 x^{3} - 42 x^{2} + 7 x + 30$

Step 1

Set $4 x^{5} + 12 x^{4} - 11 x^{3} - 42 x^{2} + 7 x + 30$ equal to $0$ .

$4 x^{5} + 12 x^{4} - 11 x^{3} - 42 x^{2} + 7 x + 30 = 0$

Step 2

Solve for $x$ .

Tap for more steps...

Step 2.1

Factor the left side of the equation.

Tap for more steps...

Step 2.1.1

Regroup terms.

$4 x^{5} - 11 x^{3} + 7 x + 12 x^{4} - 42 x^{2} + 30 = 0$

Step 2.1.2

Factor $x$ out of $4 x^{5} - 11 x^{3} + 7 x$ .

Tap for more steps...

Step 2.1.2.1

Factor $x$ out of $4 x^{5}$ .

$x (4 x^{4}) - 11 x^{3} + 7 x + 12 x^{4} - 42 x^{2} + 30 = 0$

Step 2.1.2.2

Factor $x$ out of $- 11 x^{3}$ .

$x (4 x^{4}) + x (- 11 x^{2}) + 7 x + 12 x^{4} - 42 x^{2} + 30 = 0$

Step 2.1.2.3

Factor $x$ out of $7 x$ .

$x (4 x^{4}) + x (- 11 x^{2}) + x \cdot 7 + 12 x^{4} - 42 x^{2} + 30 = 0$

Step 2.1.2.4

Factor $x$ out of $x (4 x^{4}) + x (- 11 x^{2})$ .

$x (4 x^{4} - 11 x^{2}) + x \cdot 7 + 12 x^{4} - 42 x^{2} + 30 = 0$

Step 2.1.2.5

Factor $x$ out of $x (4 x^{4} - 11 x^{2}) + x \cdot 7$ .

$x (4 x^{4} - 11 x^{2} + 7) + 12 x^{4} - 42 x^{2} + 30 = 0$

Step 2.1.3

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

$x (4 {(x^{2})}^{2} - 11 x^{2} + 7) + 12 x^{4} - 42 x^{2} + 30 = 0$

Step 2.1.4

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

$x (4 u^{2} - 11 u + 7) + 12 x^{4} - 42 x^{2} + 30 = 0$

Step 2.1.5

Factor by grouping.

Tap for more steps...

Step 2.1.5.1

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 $a \cdot c = 4 \cdot 7 = 28$ and whose sum is $b = - 11$ .

Tap for more steps...

Step 2.1.5.1.1

Factor $- 11$ out of $- 11 u$ .

$x (4 u^{2} - 11 u + 7) + 12 x^{4} - 42 x^{2} + 30 = 0$

Step 2.1.5.1.2

Rewrite $- 11$ as $- 4$ plus $- 7$

$x (4 u^{2} + (- 4 - 7) u + 7) + 12 x^{4} - 42 x^{2} + 30 = 0$

Step 2.1.5.1.3

Apply the distributive property.

$x (4 u^{2} - 4 u - 7 u + 7) + 12 x^{4} - 42 x^{2} + 30 = 0$

Step 2.1.5.2

Factor out the greatest common factor from each group.

Tap for more steps...

Step 2.1.5.2.1

Group the first two terms and the last two terms.

$x ((4 u^{2} - 4 u) - 7 u + 7) + 12 x^{4} - 42 x^{2} + 30 = 0$

Step 2.1.5.2.2

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

$x (4 u (u - 1) - 7 (u - 1)) + 12 x^{4} - 42 x^{2} + 30 = 0$

Step 2.1.5.3

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

$x ((u - 1) (4 u - 7)) + 12 x^{4} - 42 x^{2} + 30 = 0$

Step 2.1.6

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

$x ((x^{2} - 1) (4 x^{2} - 7)) + 12 x^{4} - 42 x^{2} + 30 = 0$

Step 2.1.7

Rewrite $1$ as $1^{2}$ .

$x ((x^{2} - 1^{2}) (4 x^{2} - 7)) + 12 x^{4} - 42 x^{2} + 30 = 0$

Step 2.1.8

Factor.

Tap for more steps...

Step 2.1.8.1

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 $b = 1$ .

$x ((x + 1) (x - 1) (4 x^{2} - 7)) + 12 x^{4} - 42 x^{2} + 30 = 0$

Step 2.1.8.2

Remove unnecessary parentheses.

$x (x + 1) (x - 1) (4 x^{2} - 7) + 12 x^{4} - 42 x^{2} + 30 = 0$

Step 2.1.9

Factor $6$ out of $12 x^{4} - 42 x^{2} + 30$ .

Tap for more steps...

Step 2.1.9.1

Factor $6$ out of $12 x^{4}$ .

$x (x + 1) (x - 1) (4 x^{2} - 7) + 6 (2 x^{4}) - 42 x^{2} + 30 = 0$

Step 2.1.9.2

Factor $6$ out of $- 42 x^{2}$ .

$x (x + 1) (x - 1) (4 x^{2} - 7) + 6 (2 x^{4}) + 6 (- 7 x^{2}) + 30 = 0$

Step 2.1.9.3

Factor $6$ out of $30$ .

$x (x + 1) (x - 1) (4 x^{2} - 7) + 6 (2 x^{4}) + 6 (- 7 x^{2}) + 6 (5) = 0$

Step 2.1.9.4

Factor $6$ out of $6 (2 x^{4}) + 6 (- 7 x^{2})$ .

$x (x + 1) (x - 1) (4 x^{2} - 7) + 6 (2 x^{4} - 7 x^{2}) + 6 (5) = 0$

Step 2.1.9.5

Factor $6$ out of $6 (2 x^{4} - 7 x^{2}) + 6 (5)$ .

$x (x + 1) (x - 1) (4 x^{2} - 7) + 6 (2 x^{4} - 7 x^{2} + 5) = 0$

Step 2.1.10

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

$x (x + 1) (x - 1) (4 x^{2} - 7) + 6 (2 {(x^{2})}^{2} - 7 x^{2} + 5) = 0$

Step 2.1.11

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

$x (x + 1) (x - 1) (4 x^{2} - 7) + 6 (2 u^{2} - 7 u + 5) = 0$

Step 2.1.12

Factor by grouping.

Tap for more steps...

Step 2.1.12.1

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 $a \cdot c = 2 \cdot 5 = 10$ and whose sum is $b = - 7$ .

Tap for more steps...

Step 2.1.12.1.1

Factor $- 7$ out of $- 7 u$ .

$x (x + 1) (x - 1) (4 x^{2} - 7) + 6 (2 u^{2} - 7 u + 5) = 0$

Step 2.1.12.1.2

Rewrite $- 7$ as $- 2$ plus $- 5$

$x (x + 1) (x - 1) (4 x^{2} - 7) + 6 (2 u^{2} + (- 2 - 5) u + 5) = 0$

Step 2.1.12.1.3

Apply the distributive property.

$x (x + 1) (x - 1) (4 x^{2} - 7) + 6 (2 u^{2} - 2 u - 5 u + 5) = 0$

Step 2.1.12.2

Factor out the greatest common factor from each group.

Tap for more steps...

Step 2.1.12.2.1

Group the first two terms and the last two terms.

$x (x + 1) (x - 1) (4 x^{2} - 7) + 6 ((2 u^{2} - 2 u) - 5 u + 5) = 0$

Step 2.1.12.2.2

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

$x (x + 1) (x - 1) (4 x^{2} - 7) + 6 (2 u (u - 1) - 5 (u - 1)) = 0$

Step 2.1.12.3

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

$x (x + 1) (x - 1) (4 x^{2} - 7) + 6 ((u - 1) (2 u - 5)) = 0$

Step 2.1.13

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

$x (x + 1) (x - 1) (4 x^{2} - 7) + 6 ((x^{2} - 1) (2 x^{2} - 5)) = 0$

Step 2.1.14

Rewrite $1$ as $1^{2}$ .

$x (x + 1) (x - 1) (4 x^{2} - 7) + 6 ((x^{2} - 1^{2}) (2 x^{2} - 5)) = 0$

Step 2.1.15

Factor.

Tap for more steps...

Step 2.1.15.1

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 $b = 1$ .

$x (x + 1) (x - 1) (4 x^{2} - 7) + 6 ((x + 1) (x - 1) (2 x^{2} - 5)) = 0$

Step 2.1.15.2

Remove unnecessary parentheses.

$x (x + 1) (x - 1) (4 x^{2} - 7) + 6 (x + 1) (x - 1) (2 x^{2} - 5) = 0$

Step 2.1.16

Factor $(x + 1) (x - 1)$ out of $x (x + 1) (x - 1) (4 x^{2} - 7) + 6 (x + 1) (x - 1) (2 x^{2} - 5)$ .

Tap for more steps...

Step 2.1.16.1

Factor $(x + 1) (x - 1)$ out of $x (x + 1) (x - 1) (4 x^{2} - 7)$ .

$(x + 1) (x - 1) ((x) (4 x^{2} - 7)) + 6 (x + 1) (x - 1) (2 x^{2} - 5) = 0$

Step 2.1.16.2

Factor $(x + 1) (x - 1)$ out of $6 (x + 1) (x - 1) (2 x^{2} - 5)$ .

$(x + 1) (x - 1) ((x) (4 x^{2} - 7)) + (x + 1) (x - 1) ((6) (2 x^{2} - 5)) = 0$

Step 2.1.16.3

Factor $(x + 1) (x - 1)$ out of $(x + 1) (x - 1) ((x) (4 x^{2} - 7)) + (x + 1) (x - 1) ((6) (2 x^{2} - 5))$ .

$(x + 1) (x - 1) ((x) (4 x^{2} - 7) + (6) (2 x^{2} - 5)) = 0$

Step 2.1.17

Apply the distributive property.

$(x + 1) (x - 1) (x (4 x^{2}) + x \cdot - 7 + (6) (2 x^{2} - 5)) = 0$

Step 2.1.18

Rewrite using the commutative property of multiplication.

$(x + 1) (x - 1) (4 x \cdot x^{2} + x \cdot - 7 + (6) (2 x^{2} - 5)) = 0$

Step 2.1.19

Move $- 7$ to the left of $x$ .

$(x + 1) (x - 1) (4 x \cdot x^{2} - 7 \cdot x + (6) (2 x^{2} - 5)) = 0$

Step 2.1.20

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

Tap for more steps...

Step 2.1.20.1

Move $x^{2}$ .

$(x + 1) (x - 1) (4 (x^{2} x) - 7 \cdot x + (6) (2 x^{2} - 5)) = 0$

Step 2.1.20.2

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

Tap for more steps...

Step 2.1.20.2.1

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

$(x + 1) (x - 1) (4 (x^{2} x) - 7 \cdot x + (6) (2 x^{2} - 5)) = 0$

Step 2.1.20.2.2

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

$(x + 1) (x - 1) (4 x^{2 + 1} - 7 \cdot x + (6) (2 x^{2} - 5)) = 0$

Step 2.1.20.3

Add $2$ and $1$ .

$(x + 1) (x - 1) (4 x^{3} - 7 \cdot x + (6) (2 x^{2} - 5)) = 0$

$(x + 1) (x - 1) (4 x^{3} - 7 x + (6) (2 x^{2} - 5)) = 0$

Step 2.1.21

Apply the distributive property.

$(x + 1) (x - 1) (4 x^{3} - 7 x + 6 (2 x^{2}) + 6 \cdot - 5) = 0$

Step 2.1.22

Multiply $2$ by $6$ .

$(x + 1) (x - 1) (4 x^{3} - 7 x + 12 x^{2} + 6 \cdot - 5) = 0$

Step 2.1.23

Multiply $6$ by $- 5$ .

$(x + 1) (x - 1) (4 x^{3} - 7 x + 12 x^{2} - 30) = 0$

Step 2.1.24

Reorder terms.

$(x + 1) (x - 1) (4 x^{3} + 12 x^{2} - 7 x - 30) = 0$

Step 2.1.25

Factor.

Tap for more steps...

Step 2.1.25.1

Rewrite $4 x^{3} + 12 x^{2} - 7 x - 30$ in a factored form.

Tap for more steps...

Step 2.1.25.1.1

Factor $4 x^{3} + 12 x^{2} - 7 x - 30$ using the rational roots test.

Tap for more steps...

Step 2.1.25.1.1.1

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.

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

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

Step 2.1.25.1.1.2

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

$\pm 1, \pm 0.25, \pm 0.5, \pm 30, \pm 7.5, \pm 15, \pm 2, \pm 3.75, \pm 3, \pm 0.75, \pm 1.5, \pm 10, \pm 2.5, \pm 5, \pm 1.25, \pm 6$

Step 2.1.25.1.1.3

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

Tap for more steps...

Step 2.1.25.1.1.3.1

Substitute $- 2$ into the polynomial.

$4 {(- 2)}^{3} + 12 {(- 2)}^{2} - 7 \cdot - 2 - 30$

Step 2.1.25.1.1.3.2

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

$4 \cdot - 8 + 12 {(- 2)}^{2} - 7 \cdot - 2 - 30$

Step 2.1.25.1.1.3.3

Multiply $4$ by $- 8$ .

$- 32 + 12 {(- 2)}^{2} - 7 \cdot - 2 - 30$

Step 2.1.25.1.1.3.4

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

$- 32 + 12 \cdot 4 - 7 \cdot - 2 - 30$

Step 2.1.25.1.1.3.5

Multiply $12$ by $4$ .

$- 32 + 48 - 7 \cdot - 2 - 30$

Step 2.1.25.1.1.3.6

Add $- 32$ and $48$ .

$16 - 7 \cdot - 2 - 30$

Step 2.1.25.1.1.3.7

Multiply $- 7$ by $- 2$ .

$16 + 14 - 30$

Step 2.1.25.1.1.3.8

Add $16$ and $14$ .

$30 - 30$

Step 2.1.25.1.1.3.9

Subtract $30$ from $30$ .

$0$

Step 2.1.25.1.1.4

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.

$\frac{4 x^{3} + 12 x^{2} - 7 x - 30}{x + 2}$

Step 2.1.25.1.1.5

Divide $4 x^{3} + 12 x^{2} - 7 x - 30$ by $x + 2$ .

Tap for more steps...

Step 2.1.25.1.1.5.1

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

$x$

$2$

$4 x^{3}$

$12 x^{2}$

$7 x$

$30$

Step 2.1.25.1.1.5.2

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

					$4 x^{2}$
	$x$	+	$2$		$4 x^{3}$	+	$12 x^{2}$	-	$7 x$	-	$30$

Step 2.1.25.1.1.5.3

Multiply the new quotient term by the divisor.

				$4 x^{2}$
$x$	+	$2$		$4 x^{3}$	+	$12 x^{2}$	-	$7 x$	-	$30$
			+	$4 x^{3}$	+	$8 x^{2}$

Step 2.1.25.1.1.5.4

The expression needs to be subtracted from the dividend, so change all the signs in $4 x^{3} + 8 x^{2}$

				$4 x^{2}$
$x$	+	$2$		$4 x^{3}$	+	$12 x^{2}$	-	$7 x$	-	$30$
			-	$4 x^{3}$	-	$8 x^{2}$

Step 2.1.25.1.1.5.5

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

				$4 x^{2}$
$x$	+	$2$		$4 x^{3}$	+	$12 x^{2}$	-	$7 x$	-	$30$
			-	$4 x^{3}$	-	$8 x^{2}$
					+	$4 x^{2}$

Step 2.1.25.1.1.5.6

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

				$4 x^{2}$
$x$	+	$2$		$4 x^{3}$	+	$12 x^{2}$	-	$7 x$	-	$30$
			-	$4 x^{3}$	-	$8 x^{2}$
					+	$4 x^{2}$	-	$7 x$

Step 2.1.25.1.1.5.7

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

				$4 x^{2}$	+	$4 x$
$x$	+	$2$		$4 x^{3}$	+	$12 x^{2}$	-	$7 x$	-	$30$
			-	$4 x^{3}$	-	$8 x^{2}$
					+	$4 x^{2}$	-	$7 x$

Step 2.1.25.1.1.5.8

Multiply the new quotient term by the divisor.

				$4 x^{2}$	+	$4 x$
$x$	+	$2$		$4 x^{3}$	+	$12 x^{2}$	-	$7 x$	-	$30$
			-	$4 x^{3}$	-	$8 x^{2}$
					+	$4 x^{2}$	-	$7 x$
					+	$4 x^{2}$	+	$8 x$

Step 2.1.25.1.1.5.9

The expression needs to be subtracted from the dividend, so change all the signs in $4 x^{2} + 8 x$

				$4 x^{2}$	+	$4 x$
$x$	+	$2$		$4 x^{3}$	+	$12 x^{2}$	-	$7 x$	-	$30$
			-	$4 x^{3}$	-	$8 x^{2}$
					+	$4 x^{2}$	-	$7 x$
					-	$4 x^{2}$	-	$8 x$

Step 2.1.25.1.1.5.10

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

				$4 x^{2}$	+	$4 x$
$x$	+	$2$		$4 x^{3}$	+	$12 x^{2}$	-	$7 x$	-	$30$
			-	$4 x^{3}$	-	$8 x^{2}$
					+	$4 x^{2}$	-	$7 x$
					-	$4 x^{2}$	-	$8 x$
							-	$15 x$

Step 2.1.25.1.1.5.11

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

				$4 x^{2}$	+	$4 x$
$x$	+	$2$		$4 x^{3}$	+	$12 x^{2}$	-	$7 x$	-	$30$
			-	$4 x^{3}$	-	$8 x^{2}$
					+	$4 x^{2}$	-	$7 x$
					-	$4 x^{2}$	-	$8 x$
							-	$15 x$	-	$30$

Step 2.1.25.1.1.5.12

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

				$4 x^{2}$	+	$4 x$	-	$15$
$x$	+	$2$		$4 x^{3}$	+	$12 x^{2}$	-	$7 x$	-	$30$
			-	$4 x^{3}$	-	$8 x^{2}$
					+	$4 x^{2}$	-	$7 x$
					-	$4 x^{2}$	-	$8 x$
							-	$15 x$	-	$30$

Step 2.1.25.1.1.5.13

Multiply the new quotient term by the divisor.

				$4 x^{2}$	+	$4 x$	-	$15$
$x$	+	$2$		$4 x^{3}$	+	$12 x^{2}$	-	$7 x$	-	$30$
			-	$4 x^{3}$	-	$8 x^{2}$
					+	$4 x^{2}$	-	$7 x$
					-	$4 x^{2}$	-	$8 x$
							-	$15 x$	-	$30$
							-	$15 x$	-	$30$

Step 2.1.25.1.1.5.14

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

				$4 x^{2}$	+	$4 x$	-	$15$
$x$	+	$2$		$4 x^{3}$	+	$12 x^{2}$	-	$7 x$	-	$30$
			-	$4 x^{3}$	-	$8 x^{2}$
					+	$4 x^{2}$	-	$7 x$
					-	$4 x^{2}$	-	$8 x$
							-	$15 x$	-	$30$
							+	$15 x$	+	$30$

Step 2.1.25.1.1.5.15

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

				$4 x^{2}$	+	$4 x$	-	$15$
$x$	+	$2$		$4 x^{3}$	+	$12 x^{2}$	-	$7 x$	-	$30$
			-	$4 x^{3}$	-	$8 x^{2}$
					+	$4 x^{2}$	-	$7 x$
					-	$4 x^{2}$	-	$8 x$
							-	$15 x$	-	$30$
							+	$15 x$	+	$30$
										$0$

Step 2.1.25.1.1.5.16

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

$4 x^{2} + 4 x - 15$

Step 2.1.25.1.1.6

Write $4 x^{3} + 12 x^{2} - 7 x - 30$ as a set of factors.

$(x + 1) (x - 1) ((x + 2) (4 x^{2} + 4 x - 15)) = 0$

Step 2.1.25.1.2

Factor by grouping.

Tap for more steps...

Step 2.1.25.1.2.1

Factor by grouping.

Tap for more steps...

Step 2.1.25.1.2.1.1

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 $a \cdot c = 4 \cdot - 15 = - 60$ and whose sum is $b = 4$ .

Tap for more steps...

Step 2.1.25.1.2.1.1.1

Factor $4$ out of $4 x$ .

$(x + 1) (x - 1) ((x + 2) (4 x^{2} + 4 (x) - 15)) = 0$

Step 2.1.25.1.2.1.1.2

Rewrite $4$ as $- 6$ plus $10$

$(x + 1) (x - 1) ((x + 2) (4 x^{2} + (- 6 + 10) x - 15)) = 0$

Step 2.1.25.1.2.1.1.3

Apply the distributive property.

$(x + 1) (x - 1) ((x + 2) (4 x^{2} - 6 x + 10 x - 15)) = 0$

Step 2.1.25.1.2.1.2

Factor out the greatest common factor from each group.

Tap for more steps...

Step 2.1.25.1.2.1.2.1

Group the first two terms and the last two terms.

$(x + 1) (x - 1) ((x + 2) ((4 x^{2} - 6 x) + 10 x - 15)) = 0$

Step 2.1.25.1.2.1.2.2

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

$(x + 1) (x - 1) ((x + 2) (2 x (2 x - 3) + 5 (2 x - 3))) = 0$

Step 2.1.25.1.2.1.3

Factor the polynomial by factoring out the greatest common factor, $2 x - 3$ .

$(x + 1) (x - 1) ((x + 2) ((2 x - 3) (2 x + 5))) = 0$

Step 2.1.25.1.2.2

Remove unnecessary parentheses.

$(x + 1) (x - 1) ((x + 2) (2 x - 3) (2 x + 5)) = 0$

Step 2.1.25.2

Remove unnecessary parentheses.

$(x + 1) (x - 1) (x + 2) (2 x - 3) (2 x + 5) = 0$

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 $0$ .

$x + 1 = 0$

$x - 1 = 0$

$x + 2 = 0$

$2 x - 3 = 0$

$2 x + 5 = 0$

Step 2.3

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

Tap for more steps...

Step 2.3.1

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

$x + 1 = 0$

Step 2.3.2

Subtract $1$ from both sides of the equation.

$x = - 1$

Step 2.4

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

Tap for more steps...

Step 2.4.1

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

$x - 1 = 0$

Step 2.4.2

Add $1$ to both sides of the equation.

$x = 1$

Step 2.5

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

Tap for more steps...

Step 2.5.1

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

$x + 2 = 0$

Step 2.5.2

Subtract $2$ from both sides of the equation.

$x = - 2$

Step 2.6

Set $2 x - 3$ equal to $0$ and solve for $x$ .

Tap for more steps...

Step 2.6.1

Set $2 x - 3$ equal to $0$ .

$2 x - 3 = 0$

Step 2.6.2

Solve $2 x - 3 = 0$ for $x$ .

Tap for more steps...

Step 2.6.2.1

Add $3$ to both sides of the equation.

$2 x = 3$

Step 2.6.2.2

Divide each term in $2 x = 3$ by $2$ and simplify.

Tap for more steps...

Step 2.6.2.2.1

Divide each term in $2 x = 3$ by $2$ .

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

Step 2.6.2.2.2

Simplify the left side.

Tap for more steps...

Step 2.6.2.2.2.1

Cancel the common factor of $2$ .

Tap for more steps...

Step 2.6.2.2.2.1.1

Cancel the common factor.

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

Step 2.6.2.2.2.1.2

Divide $x$ by $1$ .

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

Step 2.7

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

Tap for more steps...

Step 2.7.1

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

$2 x + 5 = 0$

Step 2.7.2

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

Tap for more steps...

Step 2.7.2.1

Subtract $5$ from both sides of the equation.

$2 x = - 5$

Step 2.7.2.2

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

Tap for more steps...

Step 2.7.2.2.1

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

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

Step 2.7.2.2.2

Simplify the left side.

Tap for more steps...

Step 2.7.2.2.2.1

Cancel the common factor of $2$ .

Tap for more steps...

Step 2.7.2.2.2.1.1

Cancel the common factor.

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

Step 2.7.2.2.2.1.2

Divide $x$ by $1$ .

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

Step 2.7.2.2.3

Simplify the right side.

Tap for more steps...

Step 2.7.2.2.3.1

Move the negative in front of the fraction.

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

Step 2.8

The final solution is all the values that make $(x + 1) (x - 1) (x + 2) (2 x - 3) (2 x + 5) = 0$ true.

$x = - 1, 1, - 2, \frac{3}{2}, - \frac{5}{2}$

Step 3