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-

Algebra ->  Expressions-with-variables -> 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-      Log On


   

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) About Me  (Show Source):
You can put this solution on YOUR website!
Given to factor completely:
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3%282y%2B3%29%5E2%2B23%282y%2B3%29-8
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You said:
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I started by using the substitution method, let u+=+2y%2B3 <===OK
3u%5E2%2B23u-8 <=== OK
%283u%2B24%29%28u-1%29 <=== mistake. Notice that +24 times -1 does not equal -8. The factors are
%283u+-+1%29%28u+%2B+8%29
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. continue from the point where you now have the factors of %283u-1%29%28u%2B8%29
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%283%282y%2B3%29-1%29%28%282y%2B3%29%2B8%29 <=== in this line 2y + 3 is substituted for u in the factors
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%286y%2B9-1%29%282y%2B3+%2B+8%29 <=== in this line the 3 is multiplied times (2y+3)
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%286y%2B8%29%282y%2B11%29 <=== in this line the numbers in each set of parentheses are combined
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2%283y%2B4%29%282y%2B11%29 <=== in this line the common factor of 2 is taken out of the first
set of parentheses
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That's the answer .... 2%283y%2B4%29%282y%2B11%29 is the complete factorization of the given
problem.
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Hope this helps to straighten things out for you.
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