Find the Roots (Zeros) P(x)=x^5-x^4-5x^3+x^2+8x+4

$P (x) = x^{5} - x^{4} - 5 x^{3} + x^{2} + 8 x + 4$

Set $x^{5} - x^{4} - 5 x^{3} + x^{2} + 8 x + 4$ equal to $0$ .

$x^{5} - x^{4} - 5 x^{3} + x^{2} + 8 x + 4 = 0$

Solve for $x$ .

Factor the left side of the equation.

Regroup terms.

$x^{5} + x^{2} - x^{4} - 5 x^{3} + 8 x + 4 = 0$

Factor $x^{2}$ out of $x^{5} + x^{2}$ .

Factor $x^{2}$ out of $x^{5}$ .

$x^{2} x^{3} + x^{2} - x^{4} - 5 x^{3} + 8 x + 4 = 0$

Multiply by $1$ .

$x^{2} x^{3} + x^{2} \cdot 1 - x^{4} - 5 x^{3} + 8 x + 4 = 0$

Factor $x^{2}$ out of $x^{2} x^{3} + x^{2} \cdot 1$ .

$x^{2} (x^{3} + 1) - x^{4} - 5 x^{3} + 8 x + 4 = 0$

Rewrite $1$ as $1^{3}$ .

$x^{2} (x^{3} + 1^{3}) - x^{4} - 5 x^{3} + 8 x + 4 = 0$

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 $b = 1$ .

$x^{2} ((x + 1) (x^{2} - x \cdot 1 + 1^{2})) - x^{4} - 5 x^{3} + 8 x + 4 = 0$

Factor.

Simplify.

Multiply $- 1$ by $1$ .

$x^{2} ((x + 1) (x^{2} - x + 1^{2})) - x^{4} - 5 x^{3} + 8 x + 4 = 0$

One to any power is one.

$x^{2} ((x + 1) (x^{2} - x + 1)) - x^{4} - 5 x^{3} + 8 x + 4 = 0$

Remove unnecessary parentheses.

$x^{2} (x + 1) (x^{2} - x + 1) - x^{4} - 5 x^{3} + 8 x + 4 = 0$

Factor $- x^{4} - 5 x^{3} + 8 x + 4$ using the rational roots test.

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.

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

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

$\pm 1, \pm 4, \pm 2$

Substitute $- 1$ and simplify the expression. In this case, the expression is equal to $0$ so $- 1$ is a root of the polynomial.

Substitute $- 1$ into the polynomial.

$- {(- 1)}^{4} - 5 {(- 1)}^{3} + 8 \cdot - 1 + 4$

Raise $- 1$ to the power of $4$ .

$- 1 \cdot 1 - 5 {(- 1)}^{3} + 8 \cdot - 1 + 4$

Multiply $- 1$ by $1$ .

$- 1 - 5 {(- 1)}^{3} + 8 \cdot - 1 + 4$

Raise $- 1$ to the power of $3$ .

$- 1 - 5 \cdot - 1 + 8 \cdot - 1 + 4$

Multiply $- 5$ by $- 1$ .

$- 1 + 5 + 8 \cdot - 1 + 4$

Add $- 1$ and $5$ .

$4 + 8 \cdot - 1 + 4$

Multiply $8$ by $- 1$ .

$4 - 8 + 4$

Subtract $8$ from $4$ .

$- 4 + 4$

Add $- 4$ and $4$ .

$0$

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

$\frac{- x^{4} - 5 x^{3} + 8 x + 4}{x + 1}$

Divide $- x^{4} - 5 x^{3} + 8 x + 4$ by $x + 1$ .

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

$x$

$1$

$x^{4}$

$5 x^{3}$

$0 x^{2}$

$8 x$

$4$

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

$x^{3}$

$x$

$1$

$x^{4}$

$5 x^{3}$

$0 x^{2}$

$8 x$

$4$

Multiply the new quotient term by the divisor.

$x^{3}$

$x$

$1$

$x^{4}$

$5 x^{3}$

$0 x^{2}$

$8 x$

$4$

$x^{4}$

$x^{3}$

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

$x^{3}$

$x$

$1$

$x^{4}$

$5 x^{3}$

$0 x^{2}$

$8 x$

$4$

$x^{4}$

$x^{3}$

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

$x^{3}$

$x$

$1$

$x^{4}$

$5 x^{3}$

$0 x^{2}$

$8 x$

$4$

$x^{4}$

$x^{3}$

$4 x^{3}$

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

$x^{3}$

$x$

$1$

$x^{4}$

$5 x^{3}$

$0 x^{2}$

$8 x$

$4$

$x^{4}$

$x^{3}$

$4 x^{3}$

$0 x^{2}$

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

$x^{3}$

$4 x^{2}$

$x$

$1$

$x^{4}$

$5 x^{3}$

$0 x^{2}$

$8 x$

$4$

$x^{4}$

$x^{3}$

$4 x^{3}$

$0 x^{2}$

Multiply the new quotient term by the divisor.

$x^{3}$

$4 x^{2}$

$x$

$1$

$x^{4}$

$5 x^{3}$

$0 x^{2}$

$8 x$

$4$

$x^{4}$

$x^{3}$

$4 x^{3}$

$0 x^{2}$

$4 x^{3}$

$4 x^{2}$

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

$x^{3}$

$4 x^{2}$

$x$

$1$

$x^{4}$

$5 x^{3}$

$0 x^{2}$

$8 x$

$4$

$x^{4}$

$x^{3}$

$4 x^{3}$

$0 x^{2}$

$4 x^{3}$

$4 x^{2}$

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

$x^{3}$

$4 x^{2}$

$x$

$1$

$x^{4}$

$5 x^{3}$

$0 x^{2}$

$8 x$

$4$

$x^{4}$

$x^{3}$

$4 x^{3}$

$0 x^{2}$

$4 x^{3}$

$4 x^{2}$

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

$x^{3}$

$4 x^{2}$

$x$

$1$

$x^{4}$

$5 x^{3}$

$0 x^{2}$

$8 x$

$4$

$x^{4}$

$x^{3}$

$4 x^{3}$

$0 x^{2}$

$4 x^{3}$

$4 x^{2}$

$8 x$

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

$x^{3}$

$4 x^{2}$

$4 x$

$x$

$1$

$x^{4}$

$5 x^{3}$

$0 x^{2}$

$8 x$

$4$

$x^{4}$

$x^{3}$

$4 x^{3}$

$0 x^{2}$

$4 x^{3}$

$4 x^{2}$

$8 x$

Multiply the new quotient term by the divisor.

$x^{3}$

$4 x^{2}$

$4 x$

$x$

$1$

$x^{4}$

$5 x^{3}$

$0 x^{2}$

$8 x$

$4$

$x^{4}$

$x^{3}$

$4 x^{3}$

$0 x^{2}$

$4 x^{3}$

$4 x^{2}$

$8 x$

$4 x^{2}$

$4 x$

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

$x^{3}$

$4 x^{2}$

$4 x$

$x$

$1$

$x^{4}$

$5 x^{3}$

$0 x^{2}$

$8 x$

$4$

$x^{4}$

$x^{3}$

$4 x^{3}$

$0 x^{2}$

$4 x^{3}$

$4 x^{2}$

$8 x$

$4 x^{2}$

$4 x$

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

$x^{3}$

$4 x^{2}$

$4 x$

$x$

$1$

$x^{4}$

$5 x^{3}$

$0 x^{2}$

$8 x$

$4$

$x^{4}$

$x^{3}$

$4 x^{3}$

$0 x^{2}$

$4 x^{3}$

$4 x^{2}$

$8 x$

$4 x^{2}$

$4 x$

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

$x^{3}$

$4 x^{2}$

$4 x$

$x$

$1$

$x^{4}$

$5 x^{3}$

$0 x^{2}$

$8 x$

$4$

$x^{4}$

$x^{3}$

$4 x^{3}$

$0 x^{2}$

$4 x^{3}$

$4 x^{2}$

$8 x$

$4 x^{2}$

$4 x$

$4$

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

$x^{3}$

$4 x^{2}$

$4 x$

$4$

$x$

$1$

$x^{4}$

$5 x^{3}$

$0 x^{2}$

$8 x$

$4$

$x^{4}$

$x^{3}$

$4 x^{3}$

$0 x^{2}$

$4 x^{3}$

$4 x^{2}$

$8 x$

$4 x^{2}$

$4 x$

$4$

Multiply the new quotient term by the divisor.

$x^{3}$

$4 x^{2}$

$4 x$

$4$

$x$

$1$

$x^{4}$

$5 x^{3}$

$0 x^{2}$

$8 x$

$4$

$x^{4}$

$x^{3}$

$4 x^{3}$

$0 x^{2}$

$4 x^{3}$

$4 x^{2}$

$8 x$

$4 x^{2}$

$4 x$

$4$

$4 x$

$4$

The expression needs to be subtracted from the dividend, so change all the signs in $4 x + 4$

$x^{3}$

$4 x^{2}$

$4 x$

$4$

$x$

$1$

$x^{4}$

$5 x^{3}$

$0 x^{2}$

$8 x$

$4$

$x^{4}$

$x^{3}$

$4 x^{3}$

$0 x^{2}$

$4 x^{3}$

$4 x^{2}$

$8 x$

$4 x^{2}$

$4 x$

$4$

$4 x$

$4$

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

$x^{3}$

$4 x^{2}$

$4 x$

$4$

$x$

$1$

$x^{4}$

$5 x^{3}$

$0 x^{2}$

$8 x$

$4$

$x^{4}$

$x^{3}$

$4 x^{3}$

$0 x^{2}$

$4 x^{3}$

$4 x^{2}$

$8 x$

$4 x^{2}$

$4 x$

$4$

$4 x$

$4$

$0$

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

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

Write $- x^{4} - 5 x^{3} + 8 x + 4$ as a set of factors.

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

Factor $x + 1$ out of $x^{2} (x + 1) (x^{2} - x + 1) + (x + 1) (- x^{3} - 4 x^{2} + 4 x + 4)$ .

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

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

Factor $x + 1$ out of $(x + 1) (x^{2} (x^{2} - x + 1)) + (x + 1) (- x^{3} - 4 x^{2} + 4 x + 4)$ .

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

Apply the distributive property.

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

Simplify.

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

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

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

Add $2$ and $2$ .

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

Rewrite using the commutative property of multiplication.

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

Multiply $x^{2}$ by $1$ .

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

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

Move $x$ .

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

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

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

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

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

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

Add $1$ and $2$ .

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

Subtract $x^{3}$ from $- x^{3}$ .

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

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

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

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

$x + 1 = 0$

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

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

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

$x + 1 = 0$

Subtract $1$ from both sides of the equation.

$x = - 1$

Set $x^{4} - 2 x^{3} - 3 x^{2} + 4 x + 4$ equal to $0$ and solve for $x$ .

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

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

Solve $x^{4} - 2 x^{3} - 3 x^{2} + 4 x + 4 = 0$ for $x$ .

Factor the left side of the equation.

Regroup terms.

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

Factor $x^{3}$ out of $x^{4} - 2 x^{3}$ .

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

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

Factor $x^{3}$ out of $- 2 x^{3}$ .

$x^{3} x + x^{3} \cdot - 2 - 3 x^{2} + 4 x + 4 = 0$

Factor $x^{3}$ out of $x^{3} x + x^{3} \cdot - 2$ .

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

Factor by grouping.

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 $a \cdot c = - 3 \cdot 4 = - 12$ and whose sum is $b = 4$ .

Factor $4$ out of $4 x$ .

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

Rewrite $4$ as $- 2$ plus $6$

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

Apply the distributive property.

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

Factor out the greatest common factor from each group.

Group the first two terms and the last two terms.

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

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

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

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

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

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

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

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

Factor $x - 2$ out of $(- 3 x - 2) (x - 2)$ .

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

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

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

Factor.

Rewrite $x^{3} - 3 x - 2$ in a factored form.

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

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.

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

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

$\pm 1, \pm 2$

Substitute $- 1$ and simplify the expression. In this case, the expression is equal to $0$ so $- 1$ is a root of the polynomial.

Substitute $- 1$ into the polynomial.

${(- 1)}^{3} - 3 \cdot - 1 - 2$

Raise $- 1$ to the power of $3$ .

$- 1 - 3 \cdot - 1 - 2$

Multiply $- 3$ by $- 1$ .

$- 1 + 3 - 2$

Add $- 1$ and $3$ .

$2 - 2$

Subtract $2$ from $2$ .

$0$

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

$\frac{x^{3} - 3 x - 2}{x + 1}$

Divide $x^{3} - 3 x - 2$ by $x + 1$ .

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

$x$

$1$

$x^{3}$

$0 x^{2}$

$3 x$

$2$

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

					$x^{2}$
	$x$	+	$1$		$x^{3}$	+	$0 x^{2}$	-	$3 x$	-	$2$

Multiply the new quotient term by the divisor.

				$x^{2}$
$x$	+	$1$		$x^{3}$	+	$0 x^{2}$	-	$3 x$	-	$2$
			+	$x^{3}$	+	$x^{2}$

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

				$x^{2}$
$x$	+	$1$		$x^{3}$	+	$0 x^{2}$	-	$3 x$	-	$2$
			-	$x^{3}$	-	$x^{2}$

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

				$x^{2}$
$x$	+	$1$		$x^{3}$	+	$0 x^{2}$	-	$3 x$	-	$2$
			-	$x^{3}$	-	$x^{2}$
					-	$x^{2}$

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

				$x^{2}$
$x$	+	$1$		$x^{3}$	+	$0 x^{2}$	-	$3 x$	-	$2$
			-	$x^{3}$	-	$x^{2}$
					-	$x^{2}$	-	$3 x$

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

				$x^{2}$	-	$x$
$x$	+	$1$		$x^{3}$	+	$0 x^{2}$	-	$3 x$	-	$2$
			-	$x^{3}$	-	$x^{2}$
					-	$x^{2}$	-	$3 x$

Multiply the new quotient term by the divisor.

				$x^{2}$	-	$x$
$x$	+	$1$		$x^{3}$	+	$0 x^{2}$	-	$3 x$	-	$2$
			-	$x^{3}$	-	$x^{2}$
					-	$x^{2}$	-	$3 x$
					-	$x^{2}$	-	$x$

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

				$x^{2}$	-	$x$
$x$	+	$1$		$x^{3}$	+	$0 x^{2}$	-	$3 x$	-	$2$
			-	$x^{3}$	-	$x^{2}$
					-	$x^{2}$	-	$3 x$
					+	$x^{2}$	+	$x$

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

				$x^{2}$	-	$x$
$x$	+	$1$		$x^{3}$	+	$0 x^{2}$	-	$3 x$	-	$2$
			-	$x^{3}$	-	$x^{2}$
					-	$x^{2}$	-	$3 x$
					+	$x^{2}$	+	$x$
							-	$2 x$

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

				$x^{2}$	-	$x$
$x$	+	$1$		$x^{3}$	+	$0 x^{2}$	-	$3 x$	-	$2$
			-	$x^{3}$	-	$x^{2}$
					-	$x^{2}$	-	$3 x$
					+	$x^{2}$	+	$x$
							-	$2 x$	-	$2$

Divide the highest order term in the dividend $- 2 x$ by the highest order term in divisor $x$ .

				$x^{2}$	-	$x$	-	$2$
$x$	+	$1$		$x^{3}$	+	$0 x^{2}$	-	$3 x$	-	$2$
			-	$x^{3}$	-	$x^{2}$
					-	$x^{2}$	-	$3 x$
					+	$x^{2}$	+	$x$
							-	$2 x$	-	$2$

Multiply the new quotient term by the divisor.

				$x^{2}$	-	$x$	-	$2$
$x$	+	$1$		$x^{3}$	+	$0 x^{2}$	-	$3 x$	-	$2$
			-	$x^{3}$	-	$x^{2}$
					-	$x^{2}$	-	$3 x$
					+	$x^{2}$	+	$x$
							-	$2 x$	-	$2$
							-	$2 x$	-	$2$

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

				$x^{2}$	-	$x$	-	$2$
$x$	+	$1$		$x^{3}$	+	$0 x^{2}$	-	$3 x$	-	$2$
			-	$x^{3}$	-	$x^{2}$
					-	$x^{2}$	-	$3 x$
					+	$x^{2}$	+	$x$
							-	$2 x$	-	$2$
							+	$2 x$	+	$2$

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

				$x^{2}$	-	$x$	-	$2$
$x$	+	$1$		$x^{3}$	+	$0 x^{2}$	-	$3 x$	-	$2$
			-	$x^{3}$	-	$x^{2}$
					-	$x^{2}$	-	$3 x$
					+	$x^{2}$	+	$x$
							-	$2 x$	-	$2$
							+	$2 x$	+	$2$
										$0$

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

$x^{2} - x - 2$

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

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

Factor $x^{2} - x - 2$ using the AC method.

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 $- 2$ and whose sum is $- 1$ .

$- 2, 1$

Write the factored form using these integers.

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

Remove unnecessary parentheses.

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

Remove unnecessary parentheses.

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

Combine exponents.

Raise $x - 2$ to the power of $1$ .

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

Raise $x - 2$ to the power of $1$ .

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

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

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

Add $1$ and $1$ .

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

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

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

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

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

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

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

Add $1$ and $1$ .

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

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

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

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

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

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

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

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

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

$x - 2 = 0$

Add $2$ to both sides of the equation.

$x = 2$

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

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

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

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

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

$x + 1 = 0$

Subtract $1$ from both sides of the equation.

$x = - 1$

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

$x = 2, - 1$

The final solution is all the values that make $(x + 1) (x^{4} - 2 x^{3} - 3 x^{2} + 4 x + 4) = 0$ true.

$x = - 1, 2$

Find the Roots (Zeros) P(x)=x^5-x^4-5x^3+x^2+8x+4

Related Questions

Support LCMGCF.com by disabling your adblocker.