Find the Roots (Zeros) P(x)=x^5-4x^4-x^3+10x^2+2x-4

$P (x) = x^{5} - 4 x^{4} - x^{3} + 10 x^{2} + 2 x - 4$

Step 1

Set $x^{5} - 4 x^{4} - x^{3} + 10 x^{2} + 2 x - 4$ equal to $0$ .

$x^{5} - 4 x^{4} - x^{3} + 10 x^{2} + 2 x - 4 = 0$

Step 2

Solve for $x$ .

Factor the left side of the equation.

Regroup terms.

$x^{5} - x^{3} - 4 x^{4} + 10 x^{2} + 2 x - 4 = 0$

Factor $x^{3}$ out of $x^{5} - x^{3}$ .

Factor $x^{3}$ out of $x^{5}$ .

$x^{3} x^{2} - x^{3} - 4 x^{4} + 10 x^{2} + 2 x - 4 = 0$

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

$x^{3} x^{2} + x^{3} \cdot - 1 - 4 x^{4} + 10 x^{2} + 2 x - 4 = 0$

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

$x^{3} (x^{2} - 1) - 4 x^{4} + 10 x^{2} + 2 x - 4 = 0$

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

$x^{3} (x^{2} - 1^{2}) - 4 x^{4} + 10 x^{2} + 2 x - 4 = 0$

Factor.

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 $b = 1$ .

$x^{3} ((x + 1) (x - 1)) - 4 x^{4} + 10 x^{2} + 2 x - 4 = 0$

Remove unnecessary parentheses.

$x^{3} (x + 1) (x - 1) - 4 x^{4} + 10 x^{2} + 2 x - 4 = 0$

Factor $2$ out of $- 4 x^{4} + 10 x^{2} + 2 x - 4$ .

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

$x^{3} (x + 1) (x - 1) + 2 (- 2 x^{4}) + 10 x^{2} + 2 x - 4 = 0$

Factor $2$ out of $10 x^{2}$ .

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

Factor $2$ out of $2 x$ .

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

Factor $2$ out of $- 4$ .

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

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

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

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

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

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

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

Factor.

Factor $- 2 x^{4} + 5 x^{2} + x - 2$ 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.

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

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

$\pm 1, \pm 0.5, \pm 2$

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.

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

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

$- 2 \cdot 1 + 5 {(- 1)}^{2} - 1 - 2$

Multiply $- 2$ by $1$ .

$- 2 + 5 {(- 1)}^{2} - 1 - 2$

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

$- 2 + 5 \cdot 1 - 1 - 2$

Multiply $5$ by $1$ .

$- 2 + 5 - 1 - 2$

Add $- 2$ and $5$ .

$3 - 1 - 2$

Subtract $1$ from $3$ .

$2 - 2$

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.

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

Divide $- 2 x^{4} + 5 x^{2} + x - 2$ 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 $0$ .

$x$

$1$

$2 x^{4}$

$0 x^{3}$

$5 x^{2}$

$x$

$2$

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

$2 x^{3}$

$x$

$1$

$2 x^{4}$

$0 x^{3}$

$5 x^{2}$

$x$

$2$

Multiply the new quotient term by the divisor.

$2 x^{3}$

$x$

$1$

$2 x^{4}$

$0 x^{3}$

$5 x^{2}$

$x$

$2$

$2 x^{4}$

$2 x^{3}$

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

$2 x^{3}$

$x$

$1$

$2 x^{4}$

$0 x^{3}$

$5 x^{2}$

$x$

$2$

$2 x^{4}$

$2 x^{3}$

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

$2 x^{3}$

$x$

$1$

$2 x^{4}$

$0 x^{3}$

$5 x^{2}$

$x$

$2$

$2 x^{4}$

$2 x^{3}$

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

$2 x^{3}$

$x$

$1$

$2 x^{4}$

$0 x^{3}$

$5 x^{2}$

$x$

$2$

$2 x^{4}$

$2 x^{3}$

$5 x^{2}$

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

$2 x^{3}$

$2 x^{2}$

$x$

$1$

$2 x^{4}$

$0 x^{3}$

$5 x^{2}$

$x$

$2$

$2 x^{4}$

$2 x^{3}$

$5 x^{2}$

Multiply the new quotient term by the divisor.

$2 x^{3}$

$2 x^{2}$

$x$

$1$

$2 x^{4}$

$0 x^{3}$

$5 x^{2}$

$x$

$2$

$2 x^{4}$

$2 x^{3}$

$5 x^{2}$

$2 x^{3}$

$2 x^{2}$

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

$2 x^{3}$

$2 x^{2}$

$x$

$1$

$2 x^{4}$

$0 x^{3}$

$5 x^{2}$

$x$

$2$

$2 x^{4}$

$2 x^{3}$

$5 x^{2}$

$2 x^{3}$

$2 x^{2}$

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

$2 x^{3}$

$2 x^{2}$

$x$

$1$

$2 x^{4}$

$0 x^{3}$

$5 x^{2}$

$x$

$2$

$2 x^{4}$

$2 x^{3}$

$5 x^{2}$

$2 x^{3}$

$2 x^{2}$

$3 x^{2}$

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

$2 x^{3}$

$2 x^{2}$

$x$

$1$

$2 x^{4}$

$0 x^{3}$

$5 x^{2}$

$x$

$2$

$2 x^{4}$

$2 x^{3}$

$5 x^{2}$

$2 x^{3}$

$2 x^{2}$

$3 x^{2}$

$x$

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

$2 x^{3}$

$2 x^{2}$

$3 x$

$x$

$1$

$2 x^{4}$

$0 x^{3}$

$5 x^{2}$

$x$

$2$

$2 x^{4}$

$2 x^{3}$

$5 x^{2}$

$2 x^{3}$

$2 x^{2}$

$3 x^{2}$

$x$

Multiply the new quotient term by the divisor.

$2 x^{3}$

$2 x^{2}$

$3 x$

$x$

$1$

$2 x^{4}$

$0 x^{3}$

$5 x^{2}$

$x$

$2$

$2 x^{4}$

$2 x^{3}$

$5 x^{2}$

$2 x^{3}$

$2 x^{2}$

$3 x^{2}$

$x$

$3 x^{2}$

$3 x$

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

$2 x^{3}$

$2 x^{2}$

$3 x$

$x$

$1$

$2 x^{4}$

$0 x^{3}$

$5 x^{2}$

$x$

$2$

$2 x^{4}$

$2 x^{3}$

$5 x^{2}$

$2 x^{3}$

$2 x^{2}$

$3 x^{2}$

$x$

$3 x^{2}$

$3 x$

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

$2 x^{3}$

$2 x^{2}$

$3 x$

$x$

$1$

$2 x^{4}$

$0 x^{3}$

$5 x^{2}$

$x$

$2$

$2 x^{4}$

$2 x^{3}$

$5 x^{2}$

$2 x^{3}$

$2 x^{2}$

$3 x^{2}$

$x$

$3 x^{2}$

$3 x$

$2 x$

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

$2 x^{3}$

$2 x^{2}$

$3 x$

$x$

$1$

$2 x^{4}$

$0 x^{3}$

$5 x^{2}$

$x$

$2$

$2 x^{4}$

$2 x^{3}$

$5 x^{2}$

$2 x^{3}$

$2 x^{2}$

$3 x^{2}$

$x$

$3 x^{2}$

$3 x$

$2 x$

$2$

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

$2 x^{3}$

$2 x^{2}$

$3 x$

$2$

$x$

$1$

$2 x^{4}$

$0 x^{3}$

$5 x^{2}$

$x$

$2$

$2 x^{4}$

$2 x^{3}$

$5 x^{2}$

$2 x^{3}$

$2 x^{2}$

$3 x^{2}$

$x$

$3 x^{2}$

$3 x$

$2 x$

$2$

Multiply the new quotient term by the divisor.

$2 x^{3}$

$2 x^{2}$

$3 x$

$2$

$x$

$1$

$2 x^{4}$

$0 x^{3}$

$5 x^{2}$

$x$

$2$

$2 x^{4}$

$2 x^{3}$

$5 x^{2}$

$2 x^{3}$

$2 x^{2}$

$3 x^{2}$

$x$

$3 x^{2}$

$3 x$

$2 x$

$2$

$2 x$

$2$

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

$2 x^{3}$

$2 x^{2}$

$3 x$

$2$

$x$

$1$

$2 x^{4}$

$0 x^{3}$

$5 x^{2}$

$x$

$2$

$2 x^{4}$

$2 x^{3}$

$5 x^{2}$

$2 x^{3}$

$2 x^{2}$

$3 x^{2}$

$x$

$3 x^{2}$

$3 x$

$2 x$

$2$

$2 x$

$2$

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

$2 x^{3}$

$2 x^{2}$

$3 x$

$2$

$x$

$1$

$2 x^{4}$

$0 x^{3}$

$5 x^{2}$

$x$

$2$

$2 x^{4}$

$2 x^{3}$

$5 x^{2}$

$2 x^{3}$

$2 x^{2}$

$3 x^{2}$

$x$

$3 x^{2}$

$3 x$

$2 x$

$2$

$2 x$

$2$

$0$

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

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

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

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

Remove unnecessary parentheses.

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

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

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

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

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

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

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

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

Apply the distributive property.

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

Multiply $x^{3}$ by $x$ by adding the exponents.

Multiply $x^{3}$ by $x$ .

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

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

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

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

Add $3$ and $1$ .

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

Move $- 1$ to the left of $x^{3}$ .

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

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

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

Apply the distributive property.

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

Simplify.

Multiply $- 2$ by $2$ .

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

Multiply $2$ by $2$ .

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

Multiply $3$ by $2$ .

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

Multiply $2$ by $- 2$ .

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

Subtract $4 x^{3}$ from $- x^{3}$ .

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

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

$x + 1 = 0$

$x^{4} - 5 x^{3} + 4 x^{2} + 6 x - 4 = 0$

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 $x^{4} - 5 x^{3} + 4 x^{2} + 6 x - 4$ equal to $0$ and solve for $x$ .

Set $x^{4} - 5 x^{3} + 4 x^{2} + 6 x - 4$ equal to $0$ .

$x^{4} - 5 x^{3} + 4 x^{2} + 6 x - 4 = 0$

Solve $x^{4} - 5 x^{3} + 4 x^{2} + 6 x - 4 = 0$ for $x$ .

Factor the left side of the equation.

Regroup terms.

$4 x^{2} - 4 + x^{4} - 5 x^{3} + 6 x = 0$

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

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

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

Factor $4$ out of $- 4$ .

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

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

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

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

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

Factor.

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 $b = 1$ .

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

Remove unnecessary parentheses.

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

Factor $x$ out of $x^{4} - 5 x^{3} + 6 x$ .

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

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

Factor $x$ out of $- 5 x^{3}$ .

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

Factor $x$ out of $6 x$ .

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

Factor $x$ out of $x \cdot x^{3} + x (- 5 x^{2})$ .

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

Factor $x$ out of $x (x^{3} - 5 x^{2}) + x \cdot 6$ .

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

Factor.

Factor $x^{3} - 5 x^{2} + 6$ 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.

$p = \pm 1, \pm 6, \pm 2, \pm 3 q = \pm 1$

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

$\pm 1, \pm 6, \pm 2, \pm 3$

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.

${(- 1)}^{3} - 5 {(- 1)}^{2} + 6$

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

$- 1 - 5 {(- 1)}^{2} + 6$

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

$- 1 - 5 \cdot 1 + 6$

Multiply $- 5$ by $1$ .

$- 1 - 5 + 6$

Subtract $5$ from $- 1$ .

$- 6 + 6$

Add $- 6$ and $6$ .

$0$

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.

$\frac{x^{3} - 5 x^{2} + 6}{x + 1}$

Divide $x^{3} - 5 x^{2} + 6$ 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 $0$ .

$x$

$1$

$x^{3}$

$5 x^{2}$

$0 x$

$6$

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

					$x^{2}$
	$x$	+	$1$		$x^{3}$	-	$5 x^{2}$	+	$0 x$	+	$6$

Multiply the new quotient term by the divisor.

				$x^{2}$
$x$	+	$1$		$x^{3}$	-	$5 x^{2}$	+	$0 x$	+	$6$
			+	$x^{3}$	+	$x^{2}$

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

				$x^{2}$
$x$	+	$1$		$x^{3}$	-	$5 x^{2}$	+	$0 x$	+	$6$
			-	$x^{3}$	-	$x^{2}$

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

				$x^{2}$
$x$	+	$1$		$x^{3}$	-	$5 x^{2}$	+	$0 x$	+	$6$
			-	$x^{3}$	-	$x^{2}$
					-	$6 x^{2}$

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

				$x^{2}$
$x$	+	$1$		$x^{3}$	-	$5 x^{2}$	+	$0 x$	+	$6$
			-	$x^{3}$	-	$x^{2}$
					-	$6 x^{2}$	+	$0 x$

Divide the highest order term in the dividend $- 6 x^{2}$ by the highest order term in divisor $x$ .

				$x^{2}$	-	$6 x$
$x$	+	$1$		$x^{3}$	-	$5 x^{2}$	+	$0 x$	+	$6$
			-	$x^{3}$	-	$x^{2}$
					-	$6 x^{2}$	+	$0 x$

Multiply the new quotient term by the divisor.

				$x^{2}$	-	$6 x$
$x$	+	$1$		$x^{3}$	-	$5 x^{2}$	+	$0 x$	+	$6$
			-	$x^{3}$	-	$x^{2}$
					-	$6 x^{2}$	+	$0 x$
					-	$6 x^{2}$	-	$6 x$

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

				$x^{2}$	-	$6 x$
$x$	+	$1$		$x^{3}$	-	$5 x^{2}$	+	$0 x$	+	$6$
			-	$x^{3}$	-	$x^{2}$
					-	$6 x^{2}$	+	$0 x$
					+	$6 x^{2}$	+	$6 x$

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

				$x^{2}$	-	$6 x$
$x$	+	$1$		$x^{3}$	-	$5 x^{2}$	+	$0 x$	+	$6$
			-	$x^{3}$	-	$x^{2}$
					-	$6 x^{2}$	+	$0 x$
					+	$6 x^{2}$	+	$6 x$
							+	$6 x$

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

				$x^{2}$	-	$6 x$
$x$	+	$1$		$x^{3}$	-	$5 x^{2}$	+	$0 x$	+	$6$
			-	$x^{3}$	-	$x^{2}$
					-	$6 x^{2}$	+	$0 x$
					+	$6 x^{2}$	+	$6 x$
							+	$6 x$	+	$6$

Divide the highest order term in the dividend $6 x$ by the highest order term in divisor $x$ .

				$x^{2}$	-	$6 x$	+	$6$
$x$	+	$1$		$x^{3}$	-	$5 x^{2}$	+	$0 x$	+	$6$
			-	$x^{3}$	-	$x^{2}$
					-	$6 x^{2}$	+	$0 x$
					+	$6 x^{2}$	+	$6 x$
							+	$6 x$	+	$6$

Multiply the new quotient term by the divisor.

				$x^{2}$	-	$6 x$	+	$6$
$x$	+	$1$		$x^{3}$	-	$5 x^{2}$	+	$0 x$	+	$6$
			-	$x^{3}$	-	$x^{2}$
					-	$6 x^{2}$	+	$0 x$
					+	$6 x^{2}$	+	$6 x$
							+	$6 x$	+	$6$
							+	$6 x$	+	$6$

The expression needs to be subtracted from the dividend, so change all the signs in $6 x + 6$

				$x^{2}$	-	$6 x$	+	$6$
$x$	+	$1$		$x^{3}$	-	$5 x^{2}$	+	$0 x$	+	$6$
			-	$x^{3}$	-	$x^{2}$
					-	$6 x^{2}$	+	$0 x$
					+	$6 x^{2}$	+	$6 x$
							+	$6 x$	+	$6$
							-	$6 x$	-	$6$

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

				$x^{2}$	-	$6 x$	+	$6$
$x$	+	$1$		$x^{3}$	-	$5 x^{2}$	+	$0 x$	+	$6$
			-	$x^{3}$	-	$x^{2}$
					-	$6 x^{2}$	+	$0 x$
					+	$6 x^{2}$	+	$6 x$
							+	$6 x$	+	$6$
							-	$6 x$	-	$6$
										$0$

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

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

Write $x^{3} - 5 x^{2} + 6$ as a set of factors.

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

Remove unnecessary parentheses.

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

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

Factor $x + 1$ out of $4 (x + 1) (x - 1)$ .

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

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

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

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

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

Apply the distributive property.

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

Multiply $4$ by $- 1$ .

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

Apply the distributive property.

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

Simplify.

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

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

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

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

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

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

Add $1$ and $2$ .

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

Rewrite using the commutative property of multiplication.

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

Move $6$ to the left of $x$ .

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

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

Move $x$ .

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

Multiply $x$ by $x$ .

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

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

Add $4 x$ and $6 x$ .

$(x + 1) (- 4 + x^{3} - 6 x^{2} + 10 x) = 0$

Reorder terms.

$(x + 1) (x^{3} - 6 x^{2} + 10 x - 4) = 0$

Factor.

Factor $x^{3} - 6 x^{2} + 10 x - 4$ 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 $2$ and simplify the expression. In this case, the expression is equal to $0$ so $2$ is a root of the polynomial.

Substitute $2$ into the polynomial.

$2^{3} - 6 \cdot 2^{2} + 10 \cdot 2 - 4$

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

$8 - 6 \cdot 2^{2} + 10 \cdot 2 - 4$

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

$8 - 6 \cdot 4 + 10 \cdot 2 - 4$

Multiply $- 6$ by $4$ .

$8 - 24 + 10 \cdot 2 - 4$

Subtract $24$ from $8$ .

$- 16 + 10 \cdot 2 - 4$

Multiply $10$ by $2$ .

$- 16 + 20 - 4$

Add $- 16$ and $20$ .

$4 - 4$

Subtract $4$ from $4$ .

$0$

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.

$\frac{x^{3} - 6 x^{2} + 10 x - 4}{x - 2}$

Divide $x^{3} - 6 x^{2} + 10 x - 4$ by $x - 2$ .

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

$x$

$2$

$x^{3}$

$6 x^{2}$

$10 x$

$4$

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

					$x^{2}$
	$x$	-	$2$		$x^{3}$	-	$6 x^{2}$	+	$10 x$	-	$4$

Multiply the new quotient term by the divisor.

				$x^{2}$
$x$	-	$2$		$x^{3}$	-	$6 x^{2}$	+	$10 x$	-	$4$
			+	$x^{3}$	-	$2 x^{2}$

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

				$x^{2}$
$x$	-	$2$		$x^{3}$	-	$6 x^{2}$	+	$10 x$	-	$4$
			-	$x^{3}$	+	$2 x^{2}$

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

				$x^{2}$
$x$	-	$2$		$x^{3}$	-	$6 x^{2}$	+	$10 x$	-	$4$
			-	$x^{3}$	+	$2 x^{2}$
					-	$4 x^{2}$

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

				$x^{2}$
$x$	-	$2$		$x^{3}$	-	$6 x^{2}$	+	$10 x$	-	$4$
			-	$x^{3}$	+	$2 x^{2}$
					-	$4 x^{2}$	+	$10 x$

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

				$x^{2}$	-	$4 x$
$x$	-	$2$		$x^{3}$	-	$6 x^{2}$	+	$10 x$	-	$4$
			-	$x^{3}$	+	$2 x^{2}$
					-	$4 x^{2}$	+	$10 x$

Multiply the new quotient term by the divisor.

				$x^{2}$	-	$4 x$
$x$	-	$2$		$x^{3}$	-	$6 x^{2}$	+	$10 x$	-	$4$
			-	$x^{3}$	+	$2 x^{2}$
					-	$4 x^{2}$	+	$10 x$
					-	$4 x^{2}$	+	$8 x$

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

				$x^{2}$	-	$4 x$
$x$	-	$2$		$x^{3}$	-	$6 x^{2}$	+	$10 x$	-	$4$
			-	$x^{3}$	+	$2 x^{2}$
					-	$4 x^{2}$	+	$10 x$
					+	$4 x^{2}$	-	$8 x$

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

				$x^{2}$	-	$4 x$
$x$	-	$2$		$x^{3}$	-	$6 x^{2}$	+	$10 x$	-	$4$
			-	$x^{3}$	+	$2 x^{2}$
					-	$4 x^{2}$	+	$10 x$
					+	$4 x^{2}$	-	$8 x$
							+	$2 x$

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

				$x^{2}$	-	$4 x$
$x$	-	$2$		$x^{3}$	-	$6 x^{2}$	+	$10 x$	-	$4$
			-	$x^{3}$	+	$2 x^{2}$
					-	$4 x^{2}$	+	$10 x$
					+	$4 x^{2}$	-	$8 x$
							+	$2 x$	-	$4$

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

				$x^{2}$	-	$4 x$	+	$2$
$x$	-	$2$		$x^{3}$	-	$6 x^{2}$	+	$10 x$	-	$4$
			-	$x^{3}$	+	$2 x^{2}$
					-	$4 x^{2}$	+	$10 x$
					+	$4 x^{2}$	-	$8 x$
							+	$2 x$	-	$4$

Multiply the new quotient term by the divisor.

				$x^{2}$	-	$4 x$	+	$2$
$x$	-	$2$		$x^{3}$	-	$6 x^{2}$	+	$10 x$	-	$4$
			-	$x^{3}$	+	$2 x^{2}$
					-	$4 x^{2}$	+	$10 x$
					+	$4 x^{2}$	-	$8 x$
							+	$2 x$	-	$4$
							+	$2 x$	-	$4$

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

				$x^{2}$	-	$4 x$	+	$2$
$x$	-	$2$		$x^{3}$	-	$6 x^{2}$	+	$10 x$	-	$4$
			-	$x^{3}$	+	$2 x^{2}$
					-	$4 x^{2}$	+	$10 x$
					+	$4 x^{2}$	-	$8 x$
							+	$2 x$	-	$4$
							-	$2 x$	+	$4$

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

				$x^{2}$	-	$4 x$	+	$2$
$x$	-	$2$		$x^{3}$	-	$6 x^{2}$	+	$10 x$	-	$4$
			-	$x^{3}$	+	$2 x^{2}$
					-	$4 x^{2}$	+	$10 x$
					+	$4 x^{2}$	-	$8 x$
							+	$2 x$	-	$4$
							-	$2 x$	+	$4$
										$0$

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

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

Write $x^{3} - 6 x^{2} + 10 x - 4$ as a set of factors.

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

Remove unnecessary parentheses.

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

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

$x + 1 = 0$

$x - 2 = 0$

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

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

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

$x - 2 = 0$

Add $2$ to both sides of the equation.

$x = 2$

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

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

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

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

Use the quadratic formula to find the solutions.

$\frac{- b \pm \sqrt{b^{2} - 4 (a c)}}{2 a}$

Substitute the values $a = 1$ , $b = - 4$ , and $c = 2$ into the quadratic formula and solve for $x$ .

$\frac{4 \pm \sqrt{{(- 4)}^{2} - 4 \cdot (1 \cdot 2)}}{2 \cdot 1}$

Simplify.

Simplify the numerator.

Raise $- 4$ to the power of $2$ .

$x = \frac{4 \pm \sqrt{16 - 4 \cdot 1 \cdot 2}}{2 \cdot 1}$

Multiply $- 4 \cdot 1 \cdot 2$ .

Multiply $- 4$ by $1$ .

$x = \frac{4 \pm \sqrt{16 - 4 \cdot 2}}{2 \cdot 1}$

Multiply $- 4$ by $2$ .

$x = \frac{4 \pm \sqrt{16 - 8}}{2 \cdot 1}$

Subtract $8$ from $16$ .

$x = \frac{4 \pm \sqrt{8}}{2 \cdot 1}$

Rewrite $8$ as $2^{2} \cdot 2$ .

Factor $4$ out of $8$ .

$x = \frac{4 \pm \sqrt{4 (2)}}{2 \cdot 1}$

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

$x = \frac{4 \pm \sqrt{2^{2} \cdot 2}}{2 \cdot 1}$

Pull terms out from under the radical.

$x = \frac{4 \pm 2 \sqrt{2}}{2 \cdot 1}$

Multiply $2$ by $1$ .

$x = \frac{4 \pm 2 \sqrt{2}}{2}$

Simplify $\frac{4 \pm 2 \sqrt{2}}{2}$ .

$x = 2 \pm \sqrt{2}$

The final answer is the combination of both solutions.

$x = 2 + \sqrt{2}, 2 - \sqrt{2}$

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

$x = - 1, 2, 2 + \sqrt{2}, 2 - \sqrt{2}$

The final solution is all the values that make $(x + 1) (x^{4} - 5 x^{3} + 4 x^{2} + 6 x - 4) = 0$ true.

$x = - 1, 2, 2 + \sqrt{2}, 2 - \sqrt{2}$

Step 3

The result can be shown in multiple forms.

Exact Form:

$x = - 1, 2, 2 + \sqrt{2}, 2 - \sqrt{2}$

Decimal Form:

$x = - 1, 2, 3.41421356 \dots, 0.58578643 \dots$

Step 4

Find the Roots (Zeros) P(x)=x^5-4x^4-x^3+10x^2+2x-4

Related Questions

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