Solve (2x+3)(x^2-16)+(x-4)(x+40)=2(x-4)(x^2-16)

$$( 2 x + 3 ) ( x ^ { 2 } - 16 ) + ( x - 4 ) ( x + 40 ) = 2 ( x - 4 ) ( x ^ { 2 } - 16 )$$

Solve for x

$$x=-7$$

$$x=4$$

Steps Using Factoring By Grouping

Use the distributive property to multiply 2x+3 by x^{2}-16.

$$2x^{3}-32x+3x^{2}-48+\left(x-4\right)\left(x+40\right)=2\left(x-4\right)\left(x^{2}-16\right)$$

Use the distributive property to multiply x-4 by x+40 and combine like terms.

$$2x^{3}-32x+3x^{2}-48+x^{2}+36x-160=2\left(x-4\right)\left(x^{2}-16\right)$$

Combine 3x^{2} and x^{2} to get 4x^{2}.

$$2x^{3}-32x+4x^{2}-48+36x-160=2\left(x-4\right)\left(x^{2}-16\right)$$

Combine -32x and 36x to get 4x.

$$2x^{3}+4x+4x^{2}-48-160=2\left(x-4\right)\left(x^{2}-16\right)$$

Subtract 160 from -48 to get -208.

$$2x^{3}+4x+4x^{2}-208=2\left(x-4\right)\left(x^{2}-16\right)$$

Use the distributive property to multiply 2 by x-4.

$$2x^{3}+4x+4x^{2}-208=\left(2x-8\right)\left(x^{2}-16\right)$$

Use the distributive property to multiply 2x-8 by x^{2}-16.

$$2x^{3}+4x+4x^{2}-208=2x^{3}-32x-8x^{2}+128$$

Subtract 2x^{3} from both sides.

$$2x^{3}+4x+4x^{2}-208-2x^{3}=-32x-8x^{2}+128$$

Combine 2x^{3} and -2x^{3} to get 0.

$$4x+4x^{2}-208=-32x-8x^{2}+128$$

Add 32x to both sides.

$$4x+4x^{2}-208+32x=-8x^{2}+128$$

Combine 4x and 32x to get 36x.

$$36x+4x^{2}-208=-8x^{2}+128$$

Add 8x^{2} to both sides.

$$36x+4x^{2}-208+8x^{2}=128$$

Combine 4x^{2} and 8x^{2} to get 12x^{2}.

$$36x+12x^{2}-208=128$$

Subtract 128 from both sides.

$$36x+12x^{2}-208-128=0$$

Subtract 128 from -208 to get -336.

$$36x+12x^{2}-336=0$$

Divide both sides by 12.

$$3x+x^{2}-28=0$$

Rearrange the polynomial to put it in standard form. Place the terms in order from highest to lowest power.

$$x^{2}+3x-28=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as x^{2}+ax+bx-28. To find a and b, set up a system to be solved.

$$a+b=3 ab=1\left(-28\right)=-28$$

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 -28.

$$-1,28 -2,14 -4,7$$

Calculate the sum for each pair.

$$-1+28=27 -2+14=12 -4+7=3$$

The solution is the pair that gives sum 3.

$$a=-4 b=7$$

Rewrite x^{2}+3x-28 as \left(x^{2}-4x\right)+\left(7x-28\right).

$$\left(x^{2}-4x\right)+\left(7x-28\right)$$

Factor out x in the first and 7 in the second group.

$$x\left(x-4\right)+7\left(x-4\right)$$

Factor out common term x-4 by using distributive property.

$$\left(x-4\right)\left(x+7\right)$$

To find equation solutions, solve x-4=0 and x+7=0.

$$x=4 x=-7$$

Steps Using the Quadratic Formula

Use the distributive property to multiply 2x+3 by x^{2}-16.

$$2x^{3}-32x+3x^{2}-48+\left(x-4\right)\left(x+40\right)=2\left(x-4\right)\left(x^{2}-16\right)$$

Use the distributive property to multiply x-4 by x+40 and combine like terms.

$$2x^{3}-32x+3x^{2}-48+x^{2}+36x-160=2\left(x-4\right)\left(x^{2}-16\right)$$

Combine 3x^{2} and x^{2} to get 4x^{2}.

$$2x^{3}-32x+4x^{2}-48+36x-160=2\left(x-4\right)\left(x^{2}-16\right)$$

Combine -32x and 36x to get 4x.

$$2x^{3}+4x+4x^{2}-48-160=2\left(x-4\right)\left(x^{2}-16\right)$$

Subtract 160 from -48 to get -208.

$$2x^{3}+4x+4x^{2}-208=2\left(x-4\right)\left(x^{2}-16\right)$$

Use the distributive property to multiply 2 by x-4.

$$2x^{3}+4x+4x^{2}-208=\left(2x-8\right)\left(x^{2}-16\right)$$

Use the distributive property to multiply 2x-8 by x^{2}-16.

$$2x^{3}+4x+4x^{2}-208=2x^{3}-32x-8x^{2}+128$$

Subtract 2x^{3} from both sides.

$$2x^{3}+4x+4x^{2}-208-2x^{3}=-32x-8x^{2}+128$$

Combine 2x^{3} and -2x^{3} to get 0.

$$4x+4x^{2}-208=-32x-8x^{2}+128$$

Add 32x to both sides.

$$4x+4x^{2}-208+32x=-8x^{2}+128$$

Combine 4x and 32x to get 36x.

$$36x+4x^{2}-208=-8x^{2}+128$$

Add 8x^{2} to both sides.

$$36x+4x^{2}-208+8x^{2}=128$$

Combine 4x^{2} and 8x^{2} to get 12x^{2}.

$$36x+12x^{2}-208=128$$

Subtract 128 from both sides.

$$36x+12x^{2}-208-128=0$$

Subtract 128 from -208 to get -336.

$$36x+12x^{2}-336=0$$

All equations of the form ax^{2}+bx+c=0 can be solved using the quadratic formula: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. The quadratic formula gives two solutions, one when ± is addition and one when it is subtraction.

$$12x^{2}+36x-336=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 12 for a, 36 for b, and -336 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-36±\sqrt{36^{2}-4\times 12\left(-336\right)}}{2\times 12}$$

Square 36.

$$x=\frac{-36±\sqrt{1296-4\times 12\left(-336\right)}}{2\times 12}$$

Multiply -4 times 12.

$$x=\frac{-36±\sqrt{1296-48\left(-336\right)}}{2\times 12}$$

Multiply -48 times -336.

$$x=\frac{-36±\sqrt{1296+16128}}{2\times 12}$$

Add 1296 to 16128.

$$x=\frac{-36±\sqrt{17424}}{2\times 12}$$

Take the square root of 17424.

$$x=\frac{-36±132}{2\times 12}$$

Multiply 2 times 12.

$$x=\frac{-36±132}{24}$$

Now solve the equation x=\frac{-36±132}{24} when ± is plus. Add -36 to 132.

$$x=\frac{96}{24}$$

Divide 96 by 24.

$$x=4$$

Now solve the equation x=\frac{-36±132}{24} when ± is minus. Subtract 132 from -36.

$$x=-\frac{168}{24}$$

Divide -168 by 24.

$$x=-7$$

The equation is now solved.

$$x=4 x=-7$$

Steps for Completing the Square

Use the distributive property to multiply 2x+3 by x^{2}-16.

$$2x^{3}-32x+3x^{2}-48+\left(x-4\right)\left(x+40\right)=2\left(x-4\right)\left(x^{2}-16\right)$$

Use the distributive property to multiply x-4 by x+40 and combine like terms.

$$2x^{3}-32x+3x^{2}-48+x^{2}+36x-160=2\left(x-4\right)\left(x^{2}-16\right)$$

Combine 3x^{2} and x^{2} to get 4x^{2}.

$$2x^{3}-32x+4x^{2}-48+36x-160=2\left(x-4\right)\left(x^{2}-16\right)$$

Combine -32x and 36x to get 4x.

$$2x^{3}+4x+4x^{2}-48-160=2\left(x-4\right)\left(x^{2}-16\right)$$

Subtract 160 from -48 to get -208.

$$2x^{3}+4x+4x^{2}-208=2\left(x-4\right)\left(x^{2}-16\right)$$

Use the distributive property to multiply 2 by x-4.

$$2x^{3}+4x+4x^{2}-208=\left(2x-8\right)\left(x^{2}-16\right)$$

Use the distributive property to multiply 2x-8 by x^{2}-16.

$$2x^{3}+4x+4x^{2}-208=2x^{3}-32x-8x^{2}+128$$

Subtract 2x^{3} from both sides.

$$2x^{3}+4x+4x^{2}-208-2x^{3}=-32x-8x^{2}+128$$

Combine 2x^{3} and -2x^{3} to get 0.

$$4x+4x^{2}-208=-32x-8x^{2}+128$$

Add 32x to both sides.

$$4x+4x^{2}-208+32x=-8x^{2}+128$$

Combine 4x and 32x to get 36x.

$$36x+4x^{2}-208=-8x^{2}+128$$

Add 8x^{2} to both sides.

$$36x+4x^{2}-208+8x^{2}=128$$

Combine 4x^{2} and 8x^{2} to get 12x^{2}.

$$36x+12x^{2}-208=128$$

Add 208 to both sides.

$$36x+12x^{2}=128+208$$

Add 128 and 208 to get 336.

$$36x+12x^{2}=336$$

Quadratic equations such as this one can be solved by completing the square. In order to complete the square, the equation must first be in the form x^{2}+bx=c.

$$12x^{2}+36x=336$$

Divide both sides by 12.

$$\frac{12x^{2}+36x}{12}=\frac{336}{12}$$

Dividing by 12 undoes the multiplication by 12.

$$x^{2}+\frac{36}{12}x=\frac{336}{12}$$

Divide 36 by 12.

$$x^{2}+3x=\frac{336}{12}$$

Divide 336 by 12.

$$x^{2}+3x=28$$

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.

$$x^{2}+3x+\left(\frac{3}{2}\right)^{2}=28+\left(\frac{3}{2}\right)^{2}$$

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

$$x^{2}+3x+\frac{9}{4}=28+\frac{9}{4}$$

Add 28 to \frac{9}{4}.

$$x^{2}+3x+\frac{9}{4}=\frac{121}{4}$$

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}.

$$\left(x+\frac{3}{2}\right)^{2}=\frac{121}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{3}{2}\right)^{2}}=\sqrt{\frac{121}{4}}$$

Simplify.

$$x+\frac{3}{2}=\frac{11}{2} x+\frac{3}{2}=-\frac{11}{2}$$

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

$$x=4 x=-7$$