Solve 120x^2+1800x

$$120 { x }^{ 2 } +1800x$$

Factor

$$120x\left(x+15\right)$$

Solution Steps

Factor out 120.

$$120\left(x^{2}+15x\right)$$

Consider x^{2}+15x. Factor out x.

$$x\left(x+15\right)$$

Rewrite the complete factored expression.

$$120x\left(x+15\right)$$

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.

$$120x^{2}+1800x=0$$

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.

$$x=\frac{-1800±\sqrt{1800^{2}}}{2\times 120}$$

Take the square root of 1800^{2}.

$$x=\frac{-1800±1800}{2\times 120}$$

Multiply 2 times 120.

$$x=\frac{-1800±1800}{240}$$

Now solve the equation x=\frac{-1800±1800}{240} when ± is plus. Add -1800 to 1800.

$$x=\frac{0}{240}$$

Divide 0 by 240.

$$x=0$$

Now solve the equation x=\frac{-1800±1800}{240} when ± is minus. Subtract 1800 from -1800.

$$x=-\frac{3600}{240}$$

Divide -3600 by 240.

$$x=-15$$

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 -15 for x_{2}.

$$120x^{2}+1800x=120x\left(x-\left(-15\right)\right)$$

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

$$120x^{2}+1800x=120x\left(x+15\right)$$

Evaluate

$$120x\left(x+15\right)$$