Question

Factor 16j2 + 24j + 9.

Answer 1

j2 is j to the second power

Answer 2

yummydum u replying?

Answer 3

Do you recognize that this is a perfect square trinomial?
\[(a + b)^2 = a^2 + 2ab + b^2\]

Answer 4

\[16j^2+24j+9\]
multiply the first and last coefficient and find factors of the product that add up to the coefficient of the second term
\[9\times 16=144\]\[12\times12=144\]\[12+12=24\]
now split the middle term and write it using the two factors\[16j^2+12j+12j+9\]\[(16j^2+12j)+(12j+9)\]now find GCFs for the parenthesis
\[4j(4j+3)+3(4j+3)\]^this can also be written as\[4j+3(4j+3)\]
so the factored form of \[16j^2+24j+9\]is\[(4j+3)(4j+3)\]or\[(4j+3)^2\]

Answer 5

@alittlehaphazard do you get it? :)

Answer 6

Multiply 16 and 9 to get 144

Now list all the ways to multiply to 144


1*144 = 144
2*72 = 144
3*48 = 144
4*36 = 144
6*24 = 144
8*18 = 144
9*16 = 144
12*12 = 144
(-1)*(-144) = 144
(-2)*(-72) = 144
(-3)*(-48) = 144
(-4)*(-36) = 144
(-6)*(-24) = 144
(-8)*(-18) = 144
(-9)*(-16) = 144
(-12)*(-12) = 144


Of this somewhat large list, notice how 12 and 12 add to 24 AND multiply to 144


So we can break up 24j into 12j + 12j and factor by grouping


16j^2 + 24j + 9

16j^2 + 12j + 12j + 9

(16j^2 + 12j) + (12j + 9)

4j(4j+3) + (12j + 9)

4j(4j+3) + 3(4j + 3)

(4j+3)(4j+3)

(4j+3)^2


So 16j^2 + 24j + 9 completely factors to (4j+3)^2

Answer 7

i wrote all that too and then realized that alittlehaphazard closed his question...

Answer 8

haha thank u (:

Answer 9

your welcome :)