Solve (x-1)(2x+8)/x+4=0

$$\frac { ( x - 1 ) ( 2 x + 8 ) } { x + 4 } = 0$$

Solve for x

$$x=1$$

Steps Using Factoring By Grouping

Variable x cannot be equal to -4 since division by zero is not defined. Multiply both sides of the equation by x+4.

$$\left(x-1\right)\left(2x+8\right)=0$$

Use the distributive property to multiply x-1 by 2x+8 and combine like terms.

$$2x^{2}+6x-8=0$$

Divide both sides by 2.

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

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as x^{2}+ax+bx-4. To find a and b, set up a system to be solved.

$$a+b=3 ab=1\left(-4\right)=-4$$

Since ab is negative, a and b have the opposite signs. Since a+b is positive, the positive number has greater absolute value than the negative. List all such integer pairs that give product -4.

$$-1,4 -2,2$$

Calculate the sum for each pair.

$$-1+4=3 -2+2=0$$

The solution is the pair that gives sum 3.

$$a=-1 b=4$$

Rewrite x^{2}+3x-4 as \left(x^{2}-x\right)+\left(4x-4\right).

$$\left(x^{2}-x\right)+\left(4x-4\right)$$

Factor out x in the first and 4 in the second group.

$$x\left(x-1\right)+4\left(x-1\right)$$

Factor out common term x-1 by using distributive property.

$$\left(x-1\right)\left(x+4\right)$$

To find equation solutions, solve x-1=0 and x+4=0.

$$x=1 x=-4$$

Variable x cannot be equal to -4.

$$x=1$$

Steps Using the Quadratic Formula

Variable x cannot be equal to -4 since division by zero is not defined. Multiply both sides of the equation by x+4.

$$\left(x-1\right)\left(2x+8\right)=0$$

Use the distributive property to multiply x-1 by 2x+8 and combine like terms.

$$2x^{2}+6x-8=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 2 for a, 6 for b, and -8 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-6±\sqrt{6^{2}-4\times 2\left(-8\right)}}{2\times 2}$$

Square 6.

$$x=\frac{-6±\sqrt{36-4\times 2\left(-8\right)}}{2\times 2}$$

Multiply -4 times 2.

$$x=\frac{-6±\sqrt{36-8\left(-8\right)}}{2\times 2}$$

Multiply -8 times -8.

$$x=\frac{-6±\sqrt{36+64}}{2\times 2}$$

Add 36 to 64.

$$x=\frac{-6±\sqrt{100}}{2\times 2}$$

Take the square root of 100.

$$x=\frac{-6±10}{2\times 2}$$

Multiply 2 times 2.

$$x=\frac{-6±10}{4}$$

Now solve the equation x=\frac{-6±10}{4} when ± is plus. Add -6 to 10.

$$x=\frac{4}{4}$$

Divide 4 by 4.

$$x=1$$

Now solve the equation x=\frac{-6±10}{4} when ± is minus. Subtract 10 from -6.

$$x=-\frac{16}{4}$$

Divide -16 by 4.

$$x=-4$$

The equation is now solved.

$$x=1 x=-4$$

Variable x cannot be equal to -4.

$$x=1$$

Steps for Completing the Square

Variable x cannot be equal to -4 since division by zero is not defined. Multiply both sides of the equation by x+4.

$$\left(x-1\right)\left(2x+8\right)=0$$

Use the distributive property to multiply x-1 by 2x+8 and combine like terms.

$$2x^{2}+6x-8=0$$

Add 8 to both sides. Anything plus zero gives itself.

$$2x^{2}+6x=8$$

Divide both sides by 2.

$$\frac{2x^{2}+6x}{2}=\frac{8}{2}$$

Dividing by 2 undoes the multiplication by 2.

$$x^{2}+\frac{6}{2}x=\frac{8}{2}$$

Divide 6 by 2.

$$x^{2}+3x=\frac{8}{2}$$

Divide 8 by 2.

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

Divide 3, the coefficient of the x term, by 2 to get \frac{3}{2}. Then add the square of \frac{3}{2} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+3x+\left(\frac{3}{2}\right)^{2}=4+\left(\frac{3}{2}\right)^{2}$$

Square \frac{3}{2} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+3x+\frac{9}{4}=4+\frac{9}{4}$$

Add 4 to \frac{9}{4}.

$$x^{2}+3x+\frac{9}{4}=\frac{25}{4}$$

Factor x^{2}+3x+\frac{9}{4}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x+\frac{3}{2}\right)^{2}=\frac{25}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{3}{2}\right)^{2}}=\sqrt{\frac{25}{4}}$$

Simplify.

$$x+\frac{3}{2}=\frac{5}{2} x+\frac{3}{2}=-\frac{5}{2}$$

Subtract \frac{3}{2} from both sides of the equation.

$$x=1 x=-4$$

Variable x cannot be equal to -4.

$$x=1$$