Solve 8x^2+6x+1

$$8 { x }^{ 2 } +6x+1$$

Factor

$$\left(2x+1\right)\left(4x+1\right)$$

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as 8x^{2}+ax+bx+1. To find a and b, set up a system to be solved.

$$a+b=6 ab=8\times 1=8$$

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 8.

$$1,8 2,4$$

Calculate the sum for each pair.

$$1+8=9 2+4=6$$

The solution is the pair that gives sum 6.

$$a=2 b=4$$

Rewrite 8x^{2}+6x+1 as \left(8x^{2}+2x\right)+\left(4x+1\right).

$$\left(8x^{2}+2x\right)+\left(4x+1\right)$$

Factor out 2x in 8x^{2}+2x.

$$2x\left(4x+1\right)+4x+1$$

Factor out common term 4x+1 by using distributive property.

$$\left(4x+1\right)\left(2x+1\right)$$

Steps Using the Quadratic Formula

Quadratic polynomial can be factored using the transformation ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right), where x_{1} and x_{2} are the solutions of the quadratic equation ax^{2}+bx+c=0.

$$8x^{2}+6x+1=0$$

All equations of the form ax^{2}+bx+c=0 can be solved using the quadratic formula: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. The quadratic formula gives two solutions, one when ± is addition and one when it is subtraction.

$$x=\frac{-6±\sqrt{6^{2}-4\times 8}}{2\times 8}$$

Square 6.

$$x=\frac{-6±\sqrt{36-4\times 8}}{2\times 8}$$

Multiply -4 times 8.

$$x=\frac{-6±\sqrt{36-32}}{2\times 8}$$

Add 36 to -32.

$$x=\frac{-6±\sqrt{4}}{2\times 8}$$

Take the square root of 4.

$$x=\frac{-6±2}{2\times 8}$$

Multiply 2 times 8.

$$x=\frac{-6±2}{16}$$

Now solve the equation x=\frac{-6±2}{16} when ± is plus. Add -6 to 2.

$$x=-\frac{4}{16}$$

Reduce the fraction \frac{-4}{16} to lowest terms by extracting and canceling out 4.

$$x=-\frac{1}{4}$$

Now solve the equation x=\frac{-6±2}{16} when ± is minus. Subtract 2 from -6.

$$x=-\frac{8}{16}$$

Reduce the fraction \frac{-8}{16} to lowest terms by extracting and canceling out 8.

$$x=-\frac{1}{2}$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute -\frac{1}{4} for x_{1} and -\frac{1}{2} for x_{2}.

$$8x^{2}+6x+1=8\left(x-\left(-\frac{1}{4}\right)\right)\left(x-\left(-\frac{1}{2}\right)\right)$$

Simplify all the expressions of the form p-\left(-q\right) to p+q.

$$8x^{2}+6x+1=8\left(x+\frac{1}{4}\right)\left(x+\frac{1}{2}\right)$$

Add \frac{1}{4} to x by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$8x^{2}+6x+1=8\times \frac{4x+1}{4}\left(x+\frac{1}{2}\right)$$

Add \frac{1}{2} to x by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$8x^{2}+6x+1=8\times \frac{4x+1}{4}\times \frac{2x+1}{2}$$

Multiply \frac{4x+1}{4} times \frac{2x+1}{2} by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.

$$8x^{2}+6x+1=8\times \frac{\left(4x+1\right)\left(2x+1\right)}{4\times 2}$$

Multiply 4 times 2.

$$8x^{2}+6x+1=8\times \frac{\left(4x+1\right)\left(2x+1\right)}{8}$$

Cancel out 8, the greatest common factor in 8 and 8.

$$8x^{2}+6x+1=\left(4x+1\right)\left(2x+1\right)$$

Evaluate

$$\left(2x+1\right)\left(4x+1\right)$$