Find the Roots (Zeros) x^3+5x^2-x-7=x^2+6x+3

$x^{3} + 5 x^{2} - x - 7 = x^{2} + 6 x + 3$

Move all terms containing $x$ to the left side of the equation.

Subtract $x^{2}$ from both sides of the equation.

$x^{3} + 5 x^{2} - x - 7 - x^{2} = 6 x + 3$

Subtract $6 x$ from both sides of the equation.

$x^{3} + 5 x^{2} - x - 7 - x^{2} - 6 x = 3$

Subtract $x^{2}$ from $5 x^{2}$ .

$x^{3} + 4 x^{2} - x - 7 - 6 x = 3$

Subtract $6 x$ from $- x$ .

$x^{3} + 4 x^{2} - 7 x - 7 = 3$

Subtract $3$ from both sides of the equation.

$x^{3} + 4 x^{2} - 7 x - 7 - 3 = 0$

Subtract $3$ from $- 7$ .

$x^{3} + 4 x^{2} - 7 x - 10 = 0$

Factor the left side of the equation.

Factor $x^{3} + 4 x^{2} - 7 x - 10$ 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.

$p = \pm 1, \pm 10, \pm 2, \pm 5 q = \pm 1$

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

$\pm 1, \pm 10, \pm 2, \pm 5$

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.

${(- 1)}^{3} + 4 {(- 1)}^{2} - 7 \cdot - 1 - 10$

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

$- 1 + 4 {(- 1)}^{2} - 7 \cdot - 1 - 10$

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

$- 1 + 4 \cdot 1 - 7 \cdot - 1 - 10$

Multiply $4$ by $1$ .

$- 1 + 4 - 7 \cdot - 1 - 10$

Add $- 1$ and $4$ .

$3 - 7 \cdot - 1 - 10$

Multiply $- 7$ by $- 1$ .

$3 + 7 - 10$

Add $3$ and $7$ .

$10 - 10$

Subtract $10$ from $10$ .

$0$

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.

$\frac{x^{3} + 4 x^{2} - 7 x - 10}{x + 1}$

Divide $x^{3} + 4 x^{2} - 7 x - 10$ 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 $0$ .

$x$

$1$

$x^{3}$

$4 x^{2}$

$7 x$

$10$

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

					$x^{2}$
	$x$	+	$1$		$x^{3}$	+	$4 x^{2}$	-	$7 x$	-	$10$

Multiply the new quotient term by the divisor.

				$x^{2}$
$x$	+	$1$		$x^{3}$	+	$4 x^{2}$	-	$7 x$	-	$10$
			+	$x^{3}$	+	$x^{2}$

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

				$x^{2}$
$x$	+	$1$		$x^{3}$	+	$4 x^{2}$	-	$7 x$	-	$10$
			-	$x^{3}$	-	$x^{2}$

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

				$x^{2}$
$x$	+	$1$		$x^{3}$	+	$4 x^{2}$	-	$7 x$	-	$10$
			-	$x^{3}$	-	$x^{2}$
					+	$3 x^{2}$

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

				$x^{2}$
$x$	+	$1$		$x^{3}$	+	$4 x^{2}$	-	$7 x$	-	$10$
			-	$x^{3}$	-	$x^{2}$
					+	$3 x^{2}$	-	$7 x$

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

				$x^{2}$	+	$3 x$
$x$	+	$1$		$x^{3}$	+	$4 x^{2}$	-	$7 x$	-	$10$
			-	$x^{3}$	-	$x^{2}$
					+	$3 x^{2}$	-	$7 x$

Multiply the new quotient term by the divisor.

				$x^{2}$	+	$3 x$
$x$	+	$1$		$x^{3}$	+	$4 x^{2}$	-	$7 x$	-	$10$
			-	$x^{3}$	-	$x^{2}$
					+	$3 x^{2}$	-	$7 x$
					+	$3 x^{2}$	+	$3 x$

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

				$x^{2}$	+	$3 x$
$x$	+	$1$		$x^{3}$	+	$4 x^{2}$	-	$7 x$	-	$10$
			-	$x^{3}$	-	$x^{2}$
					+	$3 x^{2}$	-	$7 x$
					-	$3 x^{2}$	-	$3 x$

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

				$x^{2}$	+	$3 x$
$x$	+	$1$		$x^{3}$	+	$4 x^{2}$	-	$7 x$	-	$10$
			-	$x^{3}$	-	$x^{2}$
					+	$3 x^{2}$	-	$7 x$
					-	$3 x^{2}$	-	$3 x$
							-	$10 x$

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

				$x^{2}$	+	$3 x$
$x$	+	$1$		$x^{3}$	+	$4 x^{2}$	-	$7 x$	-	$10$
			-	$x^{3}$	-	$x^{2}$
					+	$3 x^{2}$	-	$7 x$
					-	$3 x^{2}$	-	$3 x$
							-	$10 x$	-	$10$

Divide the highest order term in the dividend $- 10 x$ by the highest order term in divisor $x$ .

				$x^{2}$	+	$3 x$	-	$10$
$x$	+	$1$		$x^{3}$	+	$4 x^{2}$	-	$7 x$	-	$10$
			-	$x^{3}$	-	$x^{2}$
					+	$3 x^{2}$	-	$7 x$
					-	$3 x^{2}$	-	$3 x$
							-	$10 x$	-	$10$

Multiply the new quotient term by the divisor.

				$x^{2}$	+	$3 x$	-	$10$
$x$	+	$1$		$x^{3}$	+	$4 x^{2}$	-	$7 x$	-	$10$
			-	$x^{3}$	-	$x^{2}$
					+	$3 x^{2}$	-	$7 x$
					-	$3 x^{2}$	-	$3 x$
							-	$10 x$	-	$10$
							-	$10 x$	-	$10$

The expression needs to be subtracted from the dividend, so change all the signs in $- 10 x - 10$

				$x^{2}$	+	$3 x$	-	$10$
$x$	+	$1$		$x^{3}$	+	$4 x^{2}$	-	$7 x$	-	$10$
			-	$x^{3}$	-	$x^{2}$
					+	$3 x^{2}$	-	$7 x$
					-	$3 x^{2}$	-	$3 x$
							-	$10 x$	-	$10$
							+	$10 x$	+	$10$

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

				$x^{2}$	+	$3 x$	-	$10$
$x$	+	$1$		$x^{3}$	+	$4 x^{2}$	-	$7 x$	-	$10$
			-	$x^{3}$	-	$x^{2}$
					+	$3 x^{2}$	-	$7 x$
					-	$3 x^{2}$	-	$3 x$
							-	$10 x$	-	$10$
							+	$10 x$	+	$10$
										$0$

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

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

Write $x^{3} + 4 x^{2} - 7 x - 10$ as a set of factors.

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

Factor $x^{2} + 3 x - 10$ 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 $- 10$ and whose sum is $3$ .

$- 2, 5$

Write the factored form using these integers.

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

Remove unnecessary parentheses.

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

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

$x + 1 = 0$

$x - 2 = 0$

$x + 5 = 0$

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

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

$x + 5 = 0$

Subtract $5$ from both sides of the equation.

$x = - 5$

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

$x = - 1, 2, - 5$

Find the Roots (Zeros) x^3+5x^2-x-7=x^2+6x+3

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