Solve 5x^2+18x

$$5 { x }^{ 2 } +18x$$

Factor

$$x\left(5x+18\right)$$

Solution Steps

Factor out x.

$$x\left(5x+18\right)$$

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.

$$5x^{2}+18x=0$$

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.

$$x=\frac{-18±\sqrt{18^{2}}}{2\times 5}$$

Take the square root of 18^{2}.

$$x=\frac{-18±18}{2\times 5}$$

Multiply 2 times 5.

$$x=\frac{-18±18}{10}$$

Now solve the equation x=\frac{-18±18}{10} when ± is plus. Add -18 to 18.

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

Divide 0 by 10.

$$x=0$$

Now solve the equation x=\frac{-18±18}{10} when ± is minus. Subtract 18 from -18.

$$x=-\frac{36}{10}$$

Reduce the fraction \frac{-36}{10} to lowest terms by extracting and canceling out 2.

$$x=-\frac{18}{5}$$

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 -\frac{18}{5} for x_{2}.

$$5x^{2}+18x=5x\left(x-\left(-\frac{18}{5}\right)\right)$$

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

$$5x^{2}+18x=5x\left(x+\frac{18}{5}\right)$$

Add \frac{18}{5} to x by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$5x^{2}+18x=5x\times \frac{5x+18}{5}$$

Cancel out 5, the greatest common factor in 5 and 5.

$$5x^{2}+18x=x\left(5x+18\right)$$

Evaluate

$$x\left(5x+18\right)$$