Solve 10x^2+20x

$$10 x ^ { 2 } + 20 x$$

Factor

$$10x\left(x+2\right)$$

Solution Steps

Factor out 10.

$$10\left(x^{2}+2x\right)$$

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

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

Rewrite the complete factored expression.

$$10x\left(x+2\right)$$

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.

$$10x^{2}+20x=0$$

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.

$$x=\frac{-20±\sqrt{20^{2}}}{2\times 10}$$

Take the square root of 20^{2}.

$$x=\frac{-20±20}{2\times 10}$$

Multiply 2 times 10.

$$x=\frac{-20±20}{20}$$

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

$$x=\frac{0}{20}$$

Divide 0 by 20.

$$x=0$$

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

$$x=-\frac{40}{20}$$

Divide -40 by 20.

$$x=-2$$

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 -2 for x_{2}.

$$10x^{2}+20x=10x\left(x-\left(-2\right)\right)$$

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

$$10x^{2}+20x=10x\left(x+2\right)$$

Evaluate

$$10x\left(x+2\right)$$