SOLUTION: Factor the polynomial by grouping
9x^2+90xy+225y^2
9*(x^2+10*x*y+25*y)
9*(x+5y)^2
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Question 191282: Factor the polynomial by grouping
9x^2+90xy+225y^2
9*(x^2+10*x*y+25*y)
9*(x+5y)^2
Found 3 solutions by Mathtut, kara!, solver91311:
Answer by Mathtut(3670) (Show Source): You can put this solution on YOUR website!
:
:
Answer by kara!(17) (Show Source): You can put this solution on YOUR website!
9x^2+90xy+225y^2
9*(x^2+10*x*y+25*y)
9*(x+5y)^2
Answer by solver91311(24713) (Show Source): You can put this solution on YOUR website!
You show the correct factorization, but you didn't factor by grouping. Pulling the 9 out was a good first step, but not absolutely essential. Let's go from there anyway.
 )
Now we want to factor
by grouping.
Step 1: Multiply the lead coefficient by the constant term, 1 X 25 = 25.
Step 2: Find two factors of the results of step 1 that sum to the coefficient on the center term. Here 5 X 5 = 25, and 5 + 5 = 10.
Step 3: Split the center term into two terms, using appropriate signs so that the coefficients are the two factors you discovered in step 2. Here, 5 and 5.
Step 4: Take the first group of two terms and factor out any common factors between the two terms. Here, the only common factor is x, so:
\ \ +5xy +25y )
Step 5: Do the same thing to the other two terms:
\ \ +5y(x + 5y) )
Step 6: Now we have two terms, each with a common factor of )
. Factor that out.
(x + 5y) = (x + 5y)^2)
And remember the factor of 9 that we left back at the beginning:
^2)
Since 9 is 3 squared, you could say:
^2 = 3(x + 5y)*3(x + 5y) = (3x + 15y)^2)
John
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