Solve for a (x+4)(ax^2+bx+c)=-2x^3-7x^2+3x-4

$(x + 4) (a x^{2} + b x + c) = - 2 x^{3} - 7 x^{2} + 3 x - 4$

Divide each term in $(x + 4) (a x^{2} + b x + c) = - 2 x^{3} - 7 x^{2} + 3 x - 4$ by $x + 4$ and simplify.

Divide each term in $(x + 4) (a x^{2} + b x + c) = - 2 x^{3} - 7 x^{2} + 3 x - 4$ by $x + 4$ .

$\frac{(x + 4) (a x^{2} + b x + c)}{x + 4} = \frac{- 2 x^{3}}{x + 4} + \frac{- 7 x^{2}}{x + 4} + \frac{3 x}{x + 4} + \frac{- 4}{x + 4}$

Simplify the left side.

Cancel the common factor of $x + 4$ .

Cancel the common factor.

$\frac{(x + 4) (a x^{2} + b x + c)}{x + 4} = \frac{- 2 x^{3}}{x + 4} + \frac{- 7 x^{2}}{x + 4} + \frac{3 x}{x + 4} + \frac{- 4}{x + 4}$

Divide $a x^{2} + b x + c$ by $1$ .

$a x^{2} + b x + c = \frac{- 2 x^{3}}{x + 4} + \frac{- 7 x^{2}}{x + 4} + \frac{3 x}{x + 4} + \frac{- 4}{x + 4}$

Simplify the right side.

Combine the numerators over the common denominator.

$a x^{2} + b x + c = \frac{- 2 x^{3} - 7 x^{2} + 3 x - 4}{x + 4}$

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

$a x^{2} + b x + c = \frac{- (2 x^{3}) - 7 x^{2} + 3 x - 4}{x + 4}$

Factor $- 1$ out of $- 7 x^{2}$ .

$a x^{2} + b x + c = \frac{- (2 x^{3}) - (7 x^{2}) + 3 x - 4}{x + 4}$

Factor $- 1$ out of $- (2 x^{3}) - (7 x^{2})$ .

$a x^{2} + b x + c = \frac{- (2 x^{3} + 7 x^{2}) + 3 x - 4}{x + 4}$

Factor $- 1$ out of $3 x$ .

$a x^{2} + b x + c = \frac{- (2 x^{3} + 7 x^{2}) - (- 3 x) - 4}{x + 4}$

Factor $- 1$ out of $- (2 x^{3} + 7 x^{2}) - (- 3 x)$ .

$a x^{2} + b x + c = \frac{- (2 x^{3} + 7 x^{2} - 3 x) - 4}{x + 4}$

Rewrite $- 4$ as $- 1 (4)$ .

$a x^{2} + b x + c = \frac{- (2 x^{3} + 7 x^{2} - 3 x) - 1 (4)}{x + 4}$

Factor $- 1$ out of $- (2 x^{3} + 7 x^{2} - 3 x) - 1 (4)$ .

$a x^{2} + b x + c = \frac{- (2 x^{3} + 7 x^{2} - 3 x + 4)}{x + 4}$

Simplify the expression.

Rewrite $- (2 x^{3} + 7 x^{2} - 3 x + 4)$ as $- 1 (2 x^{3} + 7 x^{2} - 3 x + 4)$ .

$a x^{2} + b x + c = \frac{- 1 (2 x^{3} + 7 x^{2} - 3 x + 4)}{x + 4}$

Move the negative in front of the fraction.

$a x^{2} + b x + c = - \frac{2 x^{3} + 7 x^{2} - 3 x + 4}{x + 4}$

Move all terms not containing $a$ to the right side of the equation.

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

$a x^{2} + c = - \frac{2 x^{3} + 7 x^{2} - 3 x + 4}{x + 4} - b x$

Subtract $c$ from both sides of the equation.

$a x^{2} = - \frac{2 x^{3} + 7 x^{2} - 3 x + 4}{x + 4} - b x - c$

Simplify each term.

Split the fraction $\frac{2 x^{3} + 7 x^{2} - 3 x + 4}{x + 4}$ into two fractions.

$a x^{2} = - (\frac{2 x^{3} + 7 x^{2} - 3 x}{x + 4} + \frac{4}{x + 4}) - b x - c$

Factor $x$ out of $2 x^{3} + 7 x^{2} - 3 x$ .

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

$a x^{2} = - (\frac{x (2 x^{2}) + 7 x^{2} - 3 x}{x + 4} + \frac{4}{x + 4}) - b x - c$

Factor $x$ out of $7 x^{2}$ .

$a x^{2} = - (\frac{x (2 x^{2}) + x (7 x) - 3 x}{x + 4} + \frac{4}{x + 4}) - b x - c$

Factor $x$ out of $- 3 x$ .

$a x^{2} = - (\frac{x (2 x^{2}) + x (7 x) + x \cdot - 3}{x + 4} + \frac{4}{x + 4}) - b x - c$

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

$a x^{2} = - (\frac{x (2 x^{2} + 7 x) + x \cdot - 3}{x + 4} + \frac{4}{x + 4}) - b x - c$

Factor $x$ out of $x (2 x^{2} + 7 x) + x \cdot - 3$ .

$a x^{2} = - (\frac{x (2 x^{2} + 7 x - 3)}{x + 4} + \frac{4}{x + 4}) - b x - c$

Apply the distributive property.

$a x^{2} = - \frac{x (2 x^{2} + 7 x - 3)}{x + 4} - \frac{4}{x + 4} - b x - c$

Divide each term in $a x^{2} = - \frac{x (2 x^{2} + 7 x - 3)}{x + 4} - \frac{4}{x + 4} - b x - c$ by $x^{2}$ and simplify.

Divide each term in $a x^{2} = - \frac{x (2 x^{2} + 7 x - 3)}{x + 4} - \frac{4}{x + 4} - b x - c$ by $x^{2}$ .

$\frac{a x^{2}}{x^{2}} = \frac{- \frac{x (2 x^{2} + 7 x - 3)}{x + 4}}{x^{2}} + \frac{- \frac{4}{x + 4}}{x^{2}} + \frac{- b x}{x^{2}} + \frac{- c}{x^{2}}$

Simplify the left side.

Cancel the common factor of $x^{2}$ .

Cancel the common factor.

$\frac{a x^{2}}{x^{2}} = \frac{- \frac{x (2 x^{2} + 7 x - 3)}{x + 4}}{x^{2}} + \frac{- \frac{4}{x + 4}}{x^{2}} + \frac{- b x}{x^{2}} + \frac{- c}{x^{2}}$

Divide $a$ by $1$ .

$a = \frac{- \frac{x (2 x^{2} + 7 x - 3)}{x + 4}}{x^{2}} + \frac{- \frac{4}{x + 4}}{x^{2}} + \frac{- b x}{x^{2}} + \frac{- c}{x^{2}}$

Simplify the right side.

Simplify terms.

Combine the numerators over the common denominator.

$a = \frac{- \frac{x (2 x^{2} + 7 x - 3)}{x + 4} - \frac{4}{x + 4} - b x - c}{x^{2}}$

Combine the numerators over the common denominator.

$a = \frac{\frac{- x (2 x^{2} + 7 x - 3) - 4}{x + 4} - b x - c}{x^{2}}$

Simplify each term.

Simplify the numerator.

Apply the distributive property.

$a = \frac{\frac{- x (2 x^{2}) - x (7 x) - x \cdot - 3 - 4}{x + 4} - b x - c}{x^{2}}$

Simplify.

Rewrite using the commutative property of multiplication.

$a = \frac{\frac{- 1 \cdot 2 x \cdot x^{2} - x (7 x) - x \cdot - 3 - 4}{x + 4} - b x - c}{x^{2}}$

Rewrite using the commutative property of multiplication.

$a = \frac{\frac{- 1 \cdot 2 x \cdot x^{2} - 1 \cdot 7 x \cdot x - x \cdot - 3 - 4}{x + 4} - b x - c}{x^{2}}$

Multiply $- 3$ by $- 1$ .

$a = \frac{\frac{- 1 \cdot 2 x \cdot x^{2} - 1 \cdot 7 x \cdot x + 3 x - 4}{x + 4} - b x - c}{x^{2}}$

Simplify each term.

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

Move $x^{2}$ .

$a = \frac{\frac{- 1 \cdot 2 (x^{2} x) - 1 \cdot 7 x \cdot x + 3 x - 4}{x + 4} - b x - c}{x^{2}}$

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

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

$a = \frac{\frac{- 1 \cdot 2 (x^{2} x^{1}) - 1 \cdot 7 x \cdot x + 3 x - 4}{x + 4} - b x - c}{x^{2}}$

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

$a = \frac{\frac{- 1 \cdot 2 x^{2 + 1} - 1 \cdot 7 x \cdot x + 3 x - 4}{x + 4} - b x - c}{x^{2}}$

Add $2$ and $1$ .

$a = \frac{\frac{- 1 \cdot 2 x^{3} - 1 \cdot 7 x \cdot x + 3 x - 4}{x + 4} - b x - c}{x^{2}}$

Multiply $- 1$ by $2$ .

$a = \frac{\frac{- 2 x^{3} - 1 \cdot 7 x \cdot x + 3 x - 4}{x + 4} - b x - c}{x^{2}}$

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

Move $x$ .

$a = \frac{\frac{- 2 x^{3} - 1 \cdot 7 (x \cdot x) + 3 x - 4}{x + 4} - b x - c}{x^{2}}$

Multiply $x$ by $x$ .

$a = \frac{\frac{- 2 x^{3} - 1 \cdot 7 x^{2} + 3 x - 4}{x + 4} - b x - c}{x^{2}}$

Multiply $- 1$ by $7$ .

$a = \frac{\frac{- 2 x^{3} - 7 x^{2} + 3 x - 4}{x + 4} - b x - c}{x^{2}}$

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

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

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

$\pm 1, \pm 0.5, \pm 4, \pm 2$

Substitute $- 4$ and simplify the expression. In this case, the expression is equal to $0$ so $- 4$ is a root of the polynomial.

Substitute $- 4$ into the polynomial.

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

Raise $- 4$ to the power of $3$ .

$- 2 \cdot - 64 - 7 {(- 4)}^{2} + 3 \cdot - 4 - 4$

Multiply $- 2$ by $- 64$ .

$128 - 7 {(- 4)}^{2} + 3 \cdot - 4 - 4$

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

$128 - 7 \cdot 16 + 3 \cdot - 4 - 4$

Multiply $- 7$ by $16$ .

$128 - 112 + 3 \cdot - 4 - 4$

Subtract $112$ from $128$ .

$16 + 3 \cdot - 4 - 4$

Multiply $3$ by $- 4$ .

$16 - 12 - 4$

Subtract $12$ from $16$ .

$4 - 4$

Subtract $4$ from $4$ .

$0$

Since $- 4$ is a known root, divide the polynomial by $x + 4$ to find the quotient polynomial. This polynomial can then be used to find the remaining roots.

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

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

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

$x$

$4$

$2 x^{3}$

$7 x^{2}$

$3 x$

$4$

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

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

Multiply the new quotient term by the divisor.

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

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

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

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

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

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

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

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

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

Multiply the new quotient term by the divisor.

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

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

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

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

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

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

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

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

			-	$2 x^{2}$	+	$x$	-	$1$
$x$	+	$4$	-	$2 x^{3}$	-	$7 x^{2}$	+	$3 x$	-	$4$
			+	$2 x^{3}$	+	$8 x^{2}$
					+	$x^{2}$	+	$3 x$
					-	$x^{2}$	-	$4 x$
							-	$x$	-	$4$

Multiply the new quotient term by the divisor.

			-	$2 x^{2}$	+	$x$	-	$1$
$x$	+	$4$	-	$2 x^{3}$	-	$7 x^{2}$	+	$3 x$	-	$4$
			+	$2 x^{3}$	+	$8 x^{2}$
					+	$x^{2}$	+	$3 x$
					-	$x^{2}$	-	$4 x$
							-	$x$	-	$4$
							-	$x$	-	$4$

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

			-	$2 x^{2}$	+	$x$	-	$1$
$x$	+	$4$	-	$2 x^{3}$	-	$7 x^{2}$	+	$3 x$	-	$4$
			+	$2 x^{3}$	+	$8 x^{2}$
					+	$x^{2}$	+	$3 x$
					-	$x^{2}$	-	$4 x$
							-	$x$	-	$4$
							+	$x$	+	$4$

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

			-	$2 x^{2}$	+	$x$	-	$1$
$x$	+	$4$	-	$2 x^{3}$	-	$7 x^{2}$	+	$3 x$	-	$4$
			+	$2 x^{3}$	+	$8 x^{2}$
					+	$x^{2}$	+	$3 x$
					-	$x^{2}$	-	$4 x$
							-	$x$	-	$4$
							+	$x$	+	$4$
										$0$

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

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

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

$a = \frac{\frac{(x + 4) (- 2 x^{2} + x - 1)}{x + 4} - b x - c}{x^{2}}$

Cancel the common factor of $x + 4$ .

Cancel the common factor.

$a = \frac{\frac{(x + 4) (- 2 x^{2} + x - 1)}{x + 4} - b x - c}{x^{2}}$

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

$a = \frac{- 2 x^{2} + x - 1 - b x - c}{x^{2}}$

Simplify with factoring out.

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

$a = \frac{- (2 x^{2}) + x - 1 - b x - c}{x^{2}}$

Factor $- 1$ out of $x$ .

$a = \frac{- (2 x^{2}) - 1 (- x) - 1 - b x - c}{x^{2}}$

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

$a = \frac{- (2 x^{2} - x) - 1 - b x - c}{x^{2}}$

Rewrite $- 1$ as $- 1 (1)$ .

$a = \frac{- (2 x^{2} - x) - 1 (1) - b x - c}{x^{2}}$

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

$a = \frac{- (2 x^{2} - x + 1) - b x - c}{x^{2}}$

Factor $- 1$ out of $- b x$ .

$a = \frac{- (2 x^{2} - x + 1) - (b x) - c}{x^{2}}$

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

$a = \frac{- (2 x^{2} - x + 1 + b x) - c}{x^{2}}$

Factor $- 1$ out of $- c$ .

$a = \frac{- (2 x^{2} - x + 1 + b x) - (c)}{x^{2}}$

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

$a = \frac{- (2 x^{2} - x + 1 + b x + c)}{x^{2}}$

Simplify the expression.

Rewrite $- (2 x^{2} - x + 1 + b x + c)$ as $- 1 (2 x^{2} - x + 1 + b x + c)$ .

$a = \frac{- 1 (2 x^{2} - x + 1 + b x + c)}{x^{2}}$

Move the negative in front of the fraction.

$a = - \frac{2 x^{2} - x + 1 + b x + c}{x^{2}}$

Solve for a (x+4)(ax^2+bx+c)=-2x^3-7x^2+3x-4

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