Solve (4u^2+7u+9)+(8u+3)

$$( 4 u ^ { 2 } + 7 u + 9 ) + ( 8 u + 3 )$$

Evaluate

$$4u^{2}+15u+12$$

Solution Steps

Combine 7u and 8u to get 15u.

$$4u^{2}+15u+9+3$$

Add 9 and 3 to get 12.

$$4u^{2}+15u+12$$

Factor

$$4\left(u-\frac{-\sqrt{33}-15}{8}\right)\left(u-\frac{\sqrt{33}-15}{8}\right)$$

Steps Using the Quadratic Formula

Combine 7u and 8u to get 15u.

$$factor(4u^{2}+15u+9+3)$$

Add 9 and 3 to get 12.

$$factor(4u^{2}+15u+12)$$

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.

$$4u^{2}+15u+12=0$$

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.

$$u=\frac{-15±\sqrt{15^{2}-4\times 4\times 12}}{2\times 4}$$

Square 15.

$$u=\frac{-15±\sqrt{225-4\times 4\times 12}}{2\times 4}$$

Multiply -4 times 4.

$$u=\frac{-15±\sqrt{225-16\times 12}}{2\times 4}$$

Multiply -16 times 12.

$$u=\frac{-15±\sqrt{225-192}}{2\times 4}$$

Add 225 to -192.

$$u=\frac{-15±\sqrt{33}}{2\times 4}$$

Multiply 2 times 4.

$$u=\frac{-15±\sqrt{33}}{8}$$

Now solve the equation u=\frac{-15±\sqrt{33}}{8} when ± is plus. Add -15 to \sqrt{33}.

$$u=\frac{\sqrt{33}-15}{8}$$

Now solve the equation u=\frac{-15±\sqrt{33}}{8} when ± is minus. Subtract \sqrt{33} from -15.

$$u=\frac{-\sqrt{33}-15}{8}$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute \frac{-15+\sqrt{33}}{8} for x_{1} and \frac{-15-\sqrt{33}}{8} for x_{2}.

$$4u^{2}+15u+12=4\left(u-\frac{\sqrt{33}-15}{8}\right)\left(u-\frac{-\sqrt{33}-15}{8}\right)$$