# How do you complete the square for 3x^2 +18x + 5?

Apr 5, 2018

$3 {\left(x + 3\right)}^{2} - 22$

#### Explanation:

$3 {x}^{2} + 18 x + 5$

take out common factor to make coefficient ${x}^{2} = 1$

$= 3 \left({x}^{2} + 6 x\right) + 5$

now CTS as normal inside the bracket as follows

$= 3 \left({x}^{2} + 6 x + {3}^{2} - {3}^{2}\right) + 5$

$= 3 \left({\left(x + 3\right)}^{2} - 9\right) + 5$

$= 3 {\left(x + 3\right)}^{2} - 27 + 5$

$= 3 {\left(x + 3\right)}^{2} - 22$

Apr 5, 2018

$3 {\left(x + 3\right)}^{2} - 22$

#### Explanation:

Before completing the square a must equal 1, so first we divide all terms by 3.

= $\frac{3 {x}^{2}}{3} + \frac{18 x}{3} + \frac{5}{3}$

= $3 \left({x}^{2} + 6 x + \frac{5}{3}\right)$

= $3 \left({x}^{2} + 6 x + {\left(\frac{6}{2}\right)}^{2} - {\left(\frac{6}{2}\right)}^{2} + \frac{5}{3}\right)$

= $3 \left(\left({x}^{2} + 6 x + 9\right) - 9 + \frac{5}{3}\right)$

= $3 \left(\left({x}^{2} + 6 x + 9\right) - \frac{27}{3} + \frac{5}{3}\right)$

= $3 \left(\left({x}^{2} + 6 x + 9\right) - \frac{22}{3}\right)$

$\left({x}^{2} + 2 x y + {y}^{2}\right) = {\left(x + y\right)}^{2}$, so...

= $3 \left({\left(x + 3\right)}^{2} - \frac{22}{3}\right)$

= $3 {\left(x + 3\right)}^{2} - 22$