Solve 8x^2+6x

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

Factor

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

Solution Steps

Factor out 2.

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

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

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

Rewrite the complete factored expression.

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

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.

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

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.

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

Take the square root of 6^{2}.

$$x=\frac{-6±6}{2\times 8}$$

Multiply 2 times 8.

$$x=\frac{-6±6}{16}$$

Now solve the equation x=\frac{-6±6}{16} when ± is plus. Add -6 to 6.

$$x=\frac{0}{16}$$

Divide 0 by 16.

$$x=0$$

Now solve the equation x=\frac{-6±6}{16} when ± is minus. Subtract 6 from -6.

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

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

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

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

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

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

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

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

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

Cancel out 4, the greatest common factor in 8 and 4.

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

Evaluate

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