Solve (6u+6)+(u^2+1)

$$( 6 u + 6 ) + ( u ^ { 2 } + 1 )$$

Factor

$$\left(u-\left(-\sqrt{2}-3\right)\right)\left(u-\left(\sqrt{2}-3\right)\right)$$

Steps Using the Quadratic Formula

Add 6 and 1 to get 7.

$$factor(6u+7+u^{2})$$

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.

$$u^{2}+6u+7=0$$

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.

$$u=\frac{-6±\sqrt{6^{2}-4\times 7}}{2}$$

Square 6.

$$u=\frac{-6±\sqrt{36-4\times 7}}{2}$$

Multiply -4 times 7.

$$u=\frac{-6±\sqrt{36-28}}{2}$$

Add 36 to -28.

$$u=\frac{-6±\sqrt{8}}{2}$$

Take the square root of 8.

$$u=\frac{-6±2\sqrt{2}}{2}$$

Now solve the equation u=\frac{-6±2\sqrt{2}}{2} when ± is plus. Add -6 to 2\sqrt{2}.

$$u=\frac{2\sqrt{2}-6}{2}$$

Divide -6+2\sqrt{2} by 2.

$$u=\sqrt{2}-3$$

Now solve the equation u=\frac{-6±2\sqrt{2}}{2} when ± is minus. Subtract 2\sqrt{2} from -6.

$$u=\frac{-2\sqrt{2}-6}{2}$$

Divide -6-2\sqrt{2} by 2.

$$u=-\sqrt{2}-3$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute -3+\sqrt{2} for x_{1} and -3-\sqrt{2} for x_{2}.

$$u^{2}+6u+7=\left(u-\left(\sqrt{2}-3\right)\right)\left(u-\left(-\sqrt{2}-3\right)\right)$$

Evaluate

$$u^{2}+6u+7$$

Solution Steps

Add 6 and 1 to get 7.

$$6u+7+u^{2}$$