Solve 28x^3-20x^2-7x+5

$$28 x ^ { 3 } - 20 x ^ { 2 } - 7 x + 5$$

Factor

$$\left(2x-1\right)\left(7x-5\right)\left(2x+1\right)$$

Solution Steps

Do the grouping 28x^{3}-20x^{2}-7x+5=\left(28x^{3}-20x^{2}\right)+\left(-7x+5\right), and factor out 4x^{2} in the first and -1 in the second group.

$$4x^{2}\left(7x-5\right)-\left(7x-5\right)$$

Factor out common term 7x-5 by using distributive property.

$$\left(7x-5\right)\left(4x^{2}-1\right)$$

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: a^{2}-b^{2}=\left(a-b\right)\left(a+b\right).

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

Rewrite the complete factored expression.

$$\left(7x-5\right)\left(2x-1\right)\left(2x+1\right)$$

Evaluate

$$\left(7x-5\right)\left(4x^{2}-1\right)$$