Solve ax^2-2ax-8a+bx^2-2bx-8b

$$a x ^ { 2 } - 2 a x - 8 a + b x ^ { 2 } - 2 b x - 8 b$$

Factor

$$\left(x-4\right)\left(x+2\right)\left(a+b\right)$$

Solution Steps

Do the grouping ax^{2}-2ax-8a+bx^{2}-2bx-8b=\left(ax^{2}-2ax-8a\right)+\left(bx^{2}-2bx-8b\right), and factor out a in the first and b in the second group.

$$a\left(x^{2}-2x-8\right)+b\left(x^{2}-2x-8\right)$$

Factor out common term x^{2}-2x-8 by using distributive property.

$$\left(x^{2}-2x-8\right)\left(a+b\right)$$

Consider x^{2}-2x-8. Factor the expression by grouping. First, the expression needs to be rewritten as x^{2}+px+qx-8. To find p and q, set up a system to be solved.

$$p+q=-2 pq=1\left(-8\right)=-8$$

Since pq is negative, p and q have the opposite signs. Since p+q is negative, the negative number has greater absolute value than the positive. List all such integer pairs that give product -8.

$$1,-8 2,-4$$

Calculate the sum for each pair.

$$1-8=-7 2-4=-2$$

The solution is the pair that gives sum -2.

$$p=-4 q=2$$

Rewrite x^{2}-2x-8 as \left(x^{2}-4x\right)+\left(2x-8\right).

$$\left(x^{2}-4x\right)+\left(2x-8\right)$$

Factor out x in the first and 2 in the second group.

$$x\left(x-4\right)+2\left(x-4\right)$$

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

$$\left(x-4\right)\left(x+2\right)$$

Rewrite the complete factored expression.

$$\left(x-4\right)\left(x+2\right)\left(a+b\right)$$

Evaluate

$$\left(x-4\right)\left(x+2\right)\left(a+b\right)$$