Solve 12x^2+8x

$$12 { x }^{ 2 } +8x$$

Factor

$$4x\left(3x+2\right)$$

Solution Steps

Factor out 4.

$$4\left(3x^{2}+2x\right)$$

Consider 3x^{2}+2x. Factor out x.

$$x\left(3x+2\right)$$

Rewrite the complete factored expression.

$$4x\left(3x+2\right)$$

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.

$$12x^{2}+8x=0$$

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.

$$x=\frac{-8±\sqrt{8^{2}}}{2\times 12}$$

Take the square root of 8^{2}.

$$x=\frac{-8±8}{2\times 12}$$

Multiply 2 times 12.

$$x=\frac{-8±8}{24}$$

Now solve the equation x=\frac{-8±8}{24} when ± is plus. Add -8 to 8.

$$x=\frac{0}{24}$$

Divide 0 by 24.

$$x=0$$

Now solve the equation x=\frac{-8±8}{24} when ± is minus. Subtract 8 from -8.

$$x=-\frac{16}{24}$$

Reduce the fraction \frac{-16}{24} to lowest terms by extracting and canceling out 8.

$$x=-\frac{2}{3}$$

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{2}{3} for x_{2}.

$$12x^{2}+8x=12x\left(x-\left(-\frac{2}{3}\right)\right)$$

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

$$12x^{2}+8x=12x\left(x+\frac{2}{3}\right)$$

Add \frac{2}{3} to x by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$12x^{2}+8x=12x\times \frac{3x+2}{3}$$

Cancel out 3, the greatest common factor in 12 and 3.

$$12x^{2}+8x=4x\left(3x+2\right)$$

Evaluate

$$4x\left(3x+2\right)$$