Question 102371
Given to factor completely:
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{{{3(2y+3)^2+23(2y+3)-8}}}
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You said:
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I started by using the substitution method, let {{{u = 2y+3}}} <===OK
{{{3u^2+23u-8}}} <=== OK
{{{(3u+24)(u-1)}}} <=== mistake. Notice that +24 times -1 does not equal -8. The factors are 
{{{(3u - 1)(u + 8)}}}
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. continue from the point where you now have the factors of {{{(3u-1)(u+8)}}}
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{{{(3(2y+3)-1)((2y+3)+8)}}} <=== in this line  2y + 3 is substituted for u in the factors
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{{{(6y+9-1)(2y+3 + 8)}}} <=== in this line the 3 is multiplied times (2y+3)
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{{{(6y+8)(2y+11)}}} <=== in this line the numbers in each set of parentheses are combined
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{{{2(3y+4)(2y+11)}}} <=== 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(3y+4)(2y+11)}}} 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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