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

2 Answers
Apr 5, 2018

3(x+3)^2-22

Explanation:

3x^2+18x+5

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

=3(x^2+6x)+5

now CTS as normal inside the bracket as follows

=3(x^2+6x+3^2-3^2)+5

=3((x+3)^2-9)+5

=3(x+3)^2-27+5

=3(x+3)^2-22

Apr 5, 2018

3(x + 3)^2 - 22

Explanation:

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

= (3x^2)/3 + (18x)/3 + (5)/3

= 3(x^2 + 6x + 5/3)

= 3(x^2 + 6x +(6/2)^2 - (6/2)^2 + 5/3)

= 3((x^2 + 6x + 9) - 9 + 5/3)

= 3((x^2 + 6x + 9) - 27/3 + 5/3)

= 3((x^2 + 6x + 9) - 22/3)

(x^2 + 2xy + y^2) = (x + y)^2, so...

= 3((x + 3)^2 - 22/3)

= 3(x + 3)^2 - 22