Question

Which of the binomials below is a factor of this trinomial?

9x2 + 12x + 4

Answer 1

Okay, are there any common factors in all three terms here?

Answer 2

yes

Answer 3

Oh, what are they?

Answer 4

2 & 3 ?

Answer 5

Can you show me the result of factoring out those factors?

Answer 6

3x + 2?

Answer 7

well, no. We can't divide the terms by either of those numbers and get integer coefficients:
9/3 = 3
12/3=4
4/3= 4/3

9/2=9/2
12/2=6
4/2=2

There aren't any common factors to take out here. If the last term happened to be 3 instead of 4, or 6 instead of 4, then we would be able to factor out a 3, but that's not the case.

\[9x^2+12x+4\]We'll use factoring by grouping. We multiply the coefficients of the first and third terms:
\(9*4=36\)
Now we factor that result, and look for a pair of factors which will add up to the coefficient of the middle term, \(12\).
\[36*1\]\[18*2\]\[9*4\]\[6*6\]\[4*9\]\[2*18\]\[1*36\]
\[6+6=12\], so we'll choose that.
Now we rewrite your polynomial, splitting the middle term to use the two coefficients:
\[9x^2+6x + 6x +4\]Now we group that into two groups:\[(9x^2+6x)+(6x+4)\]Now we factor each group:\[3x(3x+2)+2(3x+2)\]Notice that \((3x+2)\) is now a factor of each product term?

Answer 8

We factor that again:\[3x(3x+2) + 2(3x+2) = (3x+2)(3x+2)\]

Answer 9

3x + 1?

Answer 10

No, \[(3x+2)(3x+2)\]is our factoring. You could also write that as\[(3x+2)^2\]

Answer 11

Checking our work:\[(3x+2)(3x+2) = 3x*3x + 3x*2 + 2*3x + 2*2 =\]\[9x^2+6x+6x+4=9x^2+12x+4\checkmark\]