Solve 16x^2+8x

$$16 x ^ { 2 } + 8 x$$

Factor

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

Solution Steps

Factor out 8.

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

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

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

Rewrite the complete factored expression.

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

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.

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

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.

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

Take the square root of 8^{2}.

$$x=\frac{-8±8}{2\times 16}$$

Multiply 2 times 16.

$$x=\frac{-8±8}{32}$$

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

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

Divide 0 by 32.

$$x=0$$

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

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

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

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

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

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

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

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

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

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

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

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

Evaluate

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