Solve 6x^2+8x

$$6 { x }^{ 2 } +8x$$

Factor

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

Solution Steps

Factor out 2.

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

Consider 3x^{2}+4x. Factor out x.

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

Rewrite the complete factored expression.

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

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.

$$6x^{2}+8x=0$$

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.

$$x=\frac{-8±\sqrt{8^{2}}}{2\times 6}$$

Take the square root of 8^{2}.

$$x=\frac{-8±8}{2\times 6}$$

Multiply 2 times 6.

$$x=\frac{-8±8}{12}$$

Now solve the equation x=\frac{-8±8}{12} when ± is plus. Add -8 to 8.

$$x=\frac{0}{12}$$

Divide 0 by 12.

$$x=0$$

Now solve the equation x=\frac{-8±8}{12} when ± is minus. Subtract 8 from -8.

$$x=-\frac{16}{12}$$

Reduce the fraction \frac{-16}{12} to lowest terms by extracting and canceling out 4.

$$x=-\frac{4}{3}$$

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{4}{3} for x_{2}.

$$6x^{2}+8x=6x\left(x-\left(-\frac{4}{3}\right)\right)$$

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

$$6x^{2}+8x=6x\left(x+\frac{4}{3}\right)$$

Add \frac{4}{3} to x by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$6x^{2}+8x=6x\times \frac{3x+4}{3}$$

Cancel out 3, the greatest common factor in 6 and 3.

$$6x^{2}+8x=2x\left(3x+4\right)$$

Evaluate

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