Solve 4a^2+8a

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

Factor

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

Solution Steps

Factor out 4.

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

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

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

Rewrite the complete factored expression.

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

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.

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

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.

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

Take the square root of 8^{2}.

$$a=\frac{-8±8}{2\times 4}$$

Multiply 2 times 4.

$$a=\frac{-8±8}{8}$$

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

$$a=\frac{0}{8}$$

Divide 0 by 8.

$$a=0$$

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

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

Divide -16 by 8.

$$a=-2$$

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 -2 for x_{2}.

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

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

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

Evaluate

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