Solve 20u^2-55u

$$20 u ^ { 2 } - 55 u$$

Factor

$$5u\left(4u-11\right)$$

Solution Steps

Factor out 5.

$$5\left(4u^{2}-11u\right)$$

Consider 4u^{2}-11u. Factor out u.

$$u\left(4u-11\right)$$

Rewrite the complete factored expression.

$$5u\left(4u-11\right)$$

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.

$$20u^{2}-55u=0$$

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.

$$u=\frac{-\left(-55\right)±\sqrt{\left(-55\right)^{2}}}{2\times 20}$$

Take the square root of \left(-55\right)^{2}.

$$u=\frac{-\left(-55\right)±55}{2\times 20}$$

The opposite of -55 is 55.

$$u=\frac{55±55}{2\times 20}$$

Multiply 2 times 20.

$$u=\frac{55±55}{40}$$

Now solve the equation u=\frac{55±55}{40} when ± is plus. Add 55 to 55.

$$u=\frac{110}{40}$$

Reduce the fraction \frac{110}{40} to lowest terms by extracting and canceling out 10.

$$u=\frac{11}{4}$$

Now solve the equation u=\frac{55±55}{40} when ± is minus. Subtract 55 from 55.

$$u=\frac{0}{40}$$

Divide 0 by 40.

$$u=0$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute \frac{11}{4} for x_{1} and 0 for x_{2}.

$$20u^{2}-55u=20\left(u-\frac{11}{4}\right)u$$

Subtract \frac{11}{4} from u by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.

$$20u^{2}-55u=20\times \frac{4u-11}{4}u$$

Cancel out 4, the greatest common factor in 20 and 4.

$$20u^{2}-55u=5\left(4u-11\right)u$$

Evaluate

$$5u\left(4u-11\right)$$