Asked by Rain
Factor:
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
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
Answers
Answered by
Reiny
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
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
Answered by
Rain
Oh ok thanks:)
There are no AI answers yet. The ability to request AI answers is coming soon!
Submit Your Answer
We prioritize human answers over AI answers.
If you are human, and you can answer this question, please submit your answer.