Solve (16a^2+8a)

$$( 16 a ^ { 2 } + 8 a )$$

Factor

$$8a\left(2a+1\right)$$

Solution Steps

Factor out 8.

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

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

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

Rewrite the complete factored expression.

$$8a\left(2a+1\right)$$

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.

$$16a^{2}+8a=0$$

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.

$$a=\frac{-8±\sqrt{8^{2}}}{2\times 16}$$

Take the square root of 8^{2}.

$$a=\frac{-8±8}{2\times 16}$$

Multiply 2 times 16.

$$a=\frac{-8±8}{32}$$

Now solve the equation a=\frac{-8±8}{32} when ± is plus. Add -8 to 8.

$$a=\frac{0}{32}$$

Divide 0 by 32.

$$a=0$$

Now solve the equation a=\frac{-8±8}{32} when ± is minus. Subtract 8 from -8.

$$a=-\frac{16}{32}$$

Reduce the fraction \frac{-16}{32} to lowest terms by extracting and canceling out 16.

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

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}.

$$16a^{2}+8a=16a\left(a-\left(-\frac{1}{2}\right)\right)$$

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

$$16a^{2}+8a=16a\left(a+\frac{1}{2}\right)$$

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

$$16a^{2}+8a=16a\times \frac{2a+1}{2}$$

Cancel out 2, the greatest common factor in 16 and 2.

$$16a^{2}+8a=8a\left(2a+1\right)$$

Evaluate

$$8a\left(2a+1\right)$$