Solve (6g+9)+(g^2+4g+1)

$$( 6 g + 9 ) + ( g ^ { 2 } + 4 g + 1 )$$

Evaluate

$$g^{2}+10g+10$$

Solution Steps

Combine 6g and 4g to get 10g.

$$10g+9+g^{2}+1$$

Add 9 and 1 to get 10.

$$10g+10+g^{2}$$

Factor

$$\left(g-\left(-\sqrt{15}-5\right)\right)\left(g-\left(\sqrt{15}-5\right)\right)$$

Steps Using the Quadratic Formula

Combine 6g and 4g to get 10g.

$$factor(10g+9+g^{2}+1)$$

Add 9 and 1 to get 10.

$$factor(10g+10+g^{2})$$

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.

$$g^{2}+10g+10=0$$

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.

$$g=\frac{-10±\sqrt{10^{2}-4\times 10}}{2}$$

Square 10.

$$g=\frac{-10±\sqrt{100-4\times 10}}{2}$$

Multiply -4 times 10.

$$g=\frac{-10±\sqrt{100-40}}{2}$$

Add 100 to -40.

$$g=\frac{-10±\sqrt{60}}{2}$$

Take the square root of 60.

$$g=\frac{-10±2\sqrt{15}}{2}$$

Now solve the equation g=\frac{-10±2\sqrt{15}}{2} when ± is plus. Add -10 to 2\sqrt{15}.

$$g=\frac{2\sqrt{15}-10}{2}$$

Divide -10+2\sqrt{15} by 2.

$$g=\sqrt{15}-5$$

Now solve the equation g=\frac{-10±2\sqrt{15}}{2} when ± is minus. Subtract 2\sqrt{15} from -10.

$$g=\frac{-2\sqrt{15}-10}{2}$$

Divide -10-2\sqrt{15} by 2.

$$g=-\sqrt{15}-5$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute -5+\sqrt{15} for x_{1} and -5-\sqrt{15} for x_{2}.

$$g^{2}+10g+10=\left(g-\left(\sqrt{15}-5\right)\right)\left(g-\left(-\sqrt{15}-5\right)\right)$$