SOLUTION: Factor completely: 3(2y+3)^2+23(2y+3)-8 I started by using the substitution method, let u = 2y+3 3u^2+23u-8 (3u+24)(u-1) {3(2y+3)+24}{(2y+3)-1} (6y+6+24)(2y-2) (6y+30)(2y-

Question 102371: Factor completely:
3(2y+3)^2+23(2y+3)-8
I started by using the substitution method, let u = 2y+3
3u^2+23u-8
(3u+24)(u-1)
{3(2y+3)+24}{(2y+3)-1}
(6y+6+24)(2y-2)
(6y+30)(2y-2)
{6(y+5)}(2(y)}
Now I'm stuck. Please help.
Answer by bucky(2189) (Show Source): You can put this solution on YOUR website!
Given to factor completely:
.

.
You said:
.
I started by using the substitution method, let <===OK
<=== OK
<=== mistake. Notice that +24 times -1 does not equal -8. The factors are

.
. continue from the point where you now have the factors of
.
<=== in this line 2y + 3 is substituted for u in the factors
.
<=== in this line the 3 is multiplied times (2y+3)
.
<=== in this line the numbers in each set of parentheses are combined
.
<=== in this line the common factor of 2 is taken out of the first
set of parentheses
.
That's the answer .... is the complete factorization of the given
problem.
.
Hope this helps to straighten things out for you.
.
RELATED QUESTIONS
-3/2 + 1/3u = -4/5... (answered by tommyt3rd)

27+u=3-3u (answered by jim_thompson5910,actuary)

I am not good with inequalities so i need help with the following... 21-y=(-87)+2y... (answered by Bb)

Solve for u 3u^2+10u=-3 (answered by jim_thompson5910)

completely factor the expression 48u^4v^4-18u^2v^2-3u^8v^5 a. nonfactorable b.... (answered by ewatrrr)

step by step how would u do a problem like... (answered by checkley71)

Factor using the ac method 21x^2 + 2y -... (answered by jim_thompson5910)

Solve the system using the substitution method. x + 2y = 6 3x - 2y = 2 This is... (answered by jerryguo41)

-3x + 2y = 3 substitution Method (answered by Alan3354)