Factor:

Question

A.). 4(3x+1)^2 -5(3x+1) +1

B.) 5(2-3x)^2 -28(2-3x)+15

C.) 2(3a-4)^2 - (3a-4)(a+2)-6(a+2)^2

Answer 1

For the first two, do you notice that the binomial appears both as a square and as a first degree expression?
When you first learn to do these types, it might be a good idea to replace the binomial with a single variable.
e.g. in the first:
let 3x+1 = a , then your equation becomes
4a^2 - 5a + 1
which factors quite nicely to:
(4a+1)(a+1)
now put your value of a back in ...
= (4(3x+1) + 1)(3x+1 + 1)
= (12x + 5)(3x+2)

try the second in the same way.

for the third I would do:
let x = 3a-4 and y = a+2 to get
2x^2 - xy - 6y^2
= (x-2y)(2x+3y)
now put our replacements back in:
= (3a-4 - 2(a+2))(2(3a-4) + 3(a+2))
= (a - 8)(9a -2)

of course you could also expand each one, simplify them, and then factor them
You MUST of course get the same answer

Answer 2

Oh ok thanks:)