Solve 16x^2+1/4+5x

$$16 x ^ { 2 } + \frac { 1 } { 4 } + 5 x$$

Factor

$$\frac{\left(4x+1\right)\left(16x+1\right)}{4}$$

Solution Steps

Factor out \frac{1}{4}.

$$\frac{64x^{2}+1+20x}{4}$$

Consider 64x^{2}+1+20x. Rearrange the polynomial to put it in standard form. Place the terms in order from highest to lowest power.

$$64x^{2}+20x+1$$

Factor the expression by grouping. First, the expression needs to be rewritten as 64x^{2}+ax+bx+1. To find a and b, set up a system to be solved.

$$a+b=20 ab=64\times 1=64$$

Since ab is positive, a and b have the same sign. Since a+b is positive, a and b are both positive. List all such integer pairs that give product 64.

$$1,64 2,32 4,16 8,8$$

Calculate the sum for each pair.

$$1+64=65 2+32=34 4+16=20 8+8=16$$

The solution is the pair that gives sum 20.

$$a=4 b=16$$

Rewrite 64x^{2}+20x+1 as \left(64x^{2}+4x\right)+\left(16x+1\right).

$$\left(64x^{2}+4x\right)+\left(16x+1\right)$$

Factor out 4x in 64x^{2}+4x.

$$4x\left(16x+1\right)+16x+1$$

Factor out common term 16x+1 by using distributive property.

$$\left(16x+1\right)\left(4x+1\right)$$

Rewrite the complete factored expression.

$$\frac{\left(16x+1\right)\left(4x+1\right)}{4}$$

Evaluate

$$16x^{2}+5x+\frac{1}{4}$$