Solve 4ax^2-5b+20bx^2-a

$$4 a x ^ { 2 } - 5 b + 20 b x ^ { 2 } - a$$

Factor

$$\left(2x-1\right)\left(2x+1\right)\left(a+5b\right)$$

Solution Steps

Do the grouping 4ax^{2}-5b+20bx^{2}-a=\left(4ax^{2}-a\right)+\left(20bx^{2}-5b\right), and factor out a in the first and 5b in the second group.

$$a\left(4x^{2}-1\right)+5b\left(4x^{2}-1\right)$$

Factor out common term 4x^{2}-1 by using distributive property.

$$\left(4x^{2}-1\right)\left(a+5b\right)$$

Consider 4x^{2}-1. Rewrite 4x^{2}-1 as \left(2x\right)^{2}-1^{2}. The difference of squares can be factored using the rule: p^{2}-q^{2}=\left(p-q\right)\left(p+q\right).

$$\left(2x-1\right)\left(2x+1\right)$$

Rewrite the complete factored expression.

$$\left(2x-1\right)\left(2x+1\right)\left(a+5b\right)$$

Evaluate

$$\left(a+5b\right)\left(4x^{2}-1\right)$$