Solve 64ax^2-16ax-3a+64x^2-16x-3

$$64 a x ^ { 2 } - 16 a x - 3 a + 64 x ^ { 2 } - 16 x - 3$$

Factor

$$\left(8x-3\right)\left(a+1\right)\left(8x+1\right)$$

Solution Steps

Do the grouping 64ax^{2}-16ax-3a+64x^{2}-16x-3=\left(64ax^{2}-16ax-3a\right)+\left(64x^{2}-16x-3\right), and factor out a in 64ax^{2}-16ax-3a.

$$a\left(64x^{2}-16x-3\right)+64x^{2}-16x-3$$

Factor out common term 64x^{2}-16x-3 by using distributive property.

$$\left(64x^{2}-16x-3\right)\left(a+1\right)$$

Consider 64x^{2}-16x-3. Factor the expression by grouping. First, the expression needs to be rewritten as 64x^{2}+px+qx-3. To find p and q, set up a system to be solved.

$$p+q=-16 pq=64\left(-3\right)=-192$$

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 -192.

$$1,-192 2,-96 3,-64 4,-48 6,-32 8,-24 12,-16$$

Calculate the sum for each pair.

$$1-192=-191 2-96=-94 3-64=-61 4-48=-44 6-32=-26 8-24=-16 12-16=-4$$

The solution is the pair that gives sum -16.

$$p=-24 q=8$$

Rewrite 64x^{2}-16x-3 as \left(64x^{2}-24x\right)+\left(8x-3\right).

$$\left(64x^{2}-24x\right)+\left(8x-3\right)$$

Factor out 8x in 64x^{2}-24x.

$$8x\left(8x-3\right)+8x-3$$

Factor out common term 8x-3 by using distributive property.

$$\left(8x-3\right)\left(8x+1\right)$$

Rewrite the complete factored expression.

$$\left(8x-3\right)\left(a+1\right)\left(8x+1\right)$$

Evaluate

$$\left(8x-3\right)\left(a+1\right)\left(8x+1\right)$$