Solve 8x^2+88x

$$8 { x }^{ 2 } +88x$$

Factor

$$8x\left(x+11\right)$$

Solution Steps

Factor out 8.

$$8\left(x^{2}+11x\right)$$

Consider x^{2}+11x. Factor out x.

$$x\left(x+11\right)$$

Rewrite the complete factored expression.

$$8x\left(x+11\right)$$

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.

$$8x^{2}+88x=0$$

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.

$$x=\frac{-88±\sqrt{88^{2}}}{2\times 8}$$

Take the square root of 88^{2}.

$$x=\frac{-88±88}{2\times 8}$$

Multiply 2 times 8.

$$x=\frac{-88±88}{16}$$

Now solve the equation x=\frac{-88±88}{16} when ± is plus. Add -88 to 88.

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

Divide 0 by 16.

$$x=0$$

Now solve the equation x=\frac{-88±88}{16} when ± is minus. Subtract 88 from -88.

$$x=-\frac{176}{16}$$

Divide -176 by 16.

$$x=-11$$

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 -11 for x_{2}.

$$8x^{2}+88x=8x\left(x-\left(-11\right)\right)$$

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

$$8x^{2}+88x=8x\left(x+11\right)$$

Evaluate

$$8x\left(x+11\right)$$