How do you factor #9x^2 +18x + 8#?

1 Answer
Jul 15, 2017

Answer:

#9x^2+18x+8 = (3x+2)(3x+4)#

Explanation:

The quadratic:

#9x^2+18x+8#

is in the form:

#ax^2+bx+c#

with #a=9#, #b=18# and #c=8#.

This has discriminant #Delta# given by the formula:

#Delta = b^2-4ac = color(blue)(18)^2-4(color(blue)(9))(color(blue)(8)) = 324 - 288 = 36 = 6^2#

Since this is positive and a perfect square, this quadratic will factor with integer coefficients.

Let's use an AC method:

Find a pair of factors of #AC = 9*8 = 72# with sum #B=18#. (We look for a sum rather than a difference since the sign of #c# is positive).

The pair #12, 6# works in that #12*6=72# and #12+6=18#.

Use this pair to split the middle term and factor by grouping:

#9x^2+18x+8 = (9x^2+12x)+(6x+8)#

#color(white)(9x^2+18x+8) = 3x(3x+4)+2(3x+4)#

#color(white)(9x^2+18x+8) = (3x+2)(3x+4)#

#color(white)()#
We can find the same factorisation by completing the square:

Note that #9x^2 = (3x)^2# and #18x = 2(3x)(3)#.

Hence we find:

#9x^2+18x+8 = (3x)^2+2(3x)(3)+3^2-1#

#color(white)(9x^2+18x+8) = (3x+3)^2-1^2#

#color(white)(9x^2+18x+8) = ((3x+3)-1)((3x+3)+1)#

#color(white)(9x^2+18x+8) = (3x+2)(3x+4)#