Solve 5x^2+20x

$$5 { x }^{ 2 } +20x$$

Factor

$$5x\left(x+4\right)$$

Solution Steps

Factor out 5.

$$5\left(x^{2}+4x\right)$$

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

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

Rewrite the complete factored expression.

$$5x\left(x+4\right)$$

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.

$$5x^{2}+20x=0$$

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.

$$x=\frac{-20±\sqrt{20^{2}}}{2\times 5}$$

Take the square root of 20^{2}.

$$x=\frac{-20±20}{2\times 5}$$

Multiply 2 times 5.

$$x=\frac{-20±20}{10}$$

Now solve the equation x=\frac{-20±20}{10} when ± is plus. Add -20 to 20.

$$x=\frac{0}{10}$$

Divide 0 by 10.

$$x=0$$

Now solve the equation x=\frac{-20±20}{10} when ± is minus. Subtract 20 from -20.

$$x=-\frac{40}{10}$$

Divide -40 by 10.

$$x=-4$$

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 -4 for x_{2}.

$$5x^{2}+20x=5x\left(x-\left(-4\right)\right)$$

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

$$5x^{2}+20x=5x\left(x+4\right)$$

Evaluate

$$5x\left(x+4\right)$$