Solve 8x^2-40x=

$$8 x ^ { 2 } - 40 x =$$

Factor

$$8x\left(x-5\right)$$

Solution Steps

Factor out 8.

$$8\left(x^{2}-5x\right)$$

Consider x^{2}-5x. Factor out x.

$$x\left(x-5\right)$$

Rewrite the complete factored expression.

$$8x\left(x-5\right)$$

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.

$$8x^{2}-40x=0$$

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.

$$x=\frac{-\left(-40\right)±\sqrt{\left(-40\right)^{2}}}{2\times 8}$$

Take the square root of \left(-40\right)^{2}.

$$x=\frac{-\left(-40\right)±40}{2\times 8}$$

The opposite of -40 is 40.

$$x=\frac{40±40}{2\times 8}$$

Multiply 2 times 8.

$$x=\frac{40±40}{16}$$

Now solve the equation x=\frac{40±40}{16} when ± is plus. Add 40 to 40.

$$x=\frac{80}{16}$$

Divide 80 by 16.

$$x=5$$

Now solve the equation x=\frac{40±40}{16} when ± is minus. Subtract 40 from 40.

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

Divide 0 by 16.

$$x=0$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute 5 for x_{1} and 0 for x_{2}.

$$8x^{2}-40x=8\left(x-5\right)x$$

Evaluate

$$8x\left(x-5\right)$$