Solve 5x^2-18x)-(8x^2+6x)

$$5 x ^ { 2 } - 18 x ) - ( 8 x ^ { 2 } + 6 x )$$

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.

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

Take the square root of \left(-18\right)^{2}.

$$x=\frac{-\left(-18\right)±18}{2\times 5}$$

The opposite of -18 is 18.

$$x=\frac{18±18}{2\times 5}$$

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.

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

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

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

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

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

Divide 0 by 10.

$$x=0$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute \frac{18}{5} for x_{1} and 0 for x_{2}.

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

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

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

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

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

Evaluate

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