Solve 8a^2+4a=

$$8 a ^ { 2 } + 4 a =$$

Factor

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

Solution Steps

Factor out 4.

$$4\left(2a^{2}+a\right)$$

Consider 2a^{2}+a. Factor out a.

$$a\left(2a+1\right)$$

Rewrite the complete factored expression.

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

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.

$$8a^{2}+4a=0$$

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.

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

Take the square root of 4^{2}.

$$a=\frac{-4±4}{2\times 8}$$

Multiply 2 times 8.

$$a=\frac{-4±4}{16}$$

Now solve the equation a=\frac{-4±4}{16} when ± is plus. Add -4 to 4.

$$a=\frac{0}{16}$$

Divide 0 by 16.

$$a=0$$

Now solve the equation a=\frac{-4±4}{16} when ± is minus. Subtract 4 from -4.

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

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

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

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute 0 for x_{1} and -\frac{1}{2} for x_{2}.

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

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

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

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

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

Cancel out 2, the greatest common factor in 8 and 2.

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

Evaluate

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