SOLUTION: I'm doing my homework and I'm not sure with some of my answers. Can you help me? Here it is: Find the product of [(d+e)-9]^2 Is that equal to d^2 + 2de + e^2 + 49 ?

Algebra ->  Polynomials-and-rational-expressions -> SOLUTION: I'm doing my homework and I'm not sure with some of my answers. Can you help me? Here it is: Find the product of [(d+e)-9]^2 Is that equal to d^2 + 2de + e^2 + 49 ?      Log On


   

Question 818532: I'm doing my homework and I'm not sure with some of my answers. Can you help me? Here it is:
Find the product of [(d+e)-9]^2
Is that equal to d^2 + 2de + e^2 + 49 ?
Found 2 solutions by jhunjiro, jsmallt9:
Answer by jhunjiro(67) About Me  (Show Source):
You can put this solution on YOUR website!

Solution:
[(d+e)-9]^2
(d+e)^2 -18(d+e)+81
-
(d^2 +2de +e^2)-18d -18e +81
-
d^2 +e^2 +2de -18d -18e +81
Final Answer: d^2 +e^2 +2de -18d -18e +81

Answer by jsmallt9(3759) About Me  (Show Source):
You can put this solution on YOUR website!
Short answer: No

%28%28d%2Be%29-9%29%5E2++
There are a couple of ways to figure this out:
  • Multiply it out:
    1. Rewrite as %28%28d%2Be%29-9%29%28%28d%2Be%29-9%29
    2. Multiply each term of the first (d+e)-9 times each term of the second (d+e)-9. This will be 9 multiplications!
    3. Add like terms, if any.
  • Use patterns:
    • %28a%2Bb%29%5E2=+a%5E2%2B2ab%2Bb%5E2
    • %28a-b%29%5E2=+a%5E2-2ab%2Bb%5E2
    • I am going to show you how to use the patterns.

    At first you might think: The patterns are for squaring two-term expressions but I'm trying to square a three-term expression. How is that possible?

    Well, the problem already has the d+e grouped. We can treat it as the "a" in %28a-b%29%5E2. Using (d+e) as the "a" and the "9" as the "b", the pattern shows us how to square your expression:
    %28d%2Be%29%5E2-2%28d%2Be%29%289%29%2B%289%29%5E2++
    If you can't see the pattern in this, then check out the P.S. at the end. Simplifying the last parts:
    %28d%2Be%29%5E2-18%28d%2Be%29%2B81+
    %28d%2Be%29%5E2-18d-18e%2B81+
    To simplify the beginning we can use the %28a%2Bb%29%5E2 pattern (in a more obvious way than above):
    %28d%29%5E2%2B2%28d%29%28e%29%2B%28e%29%5E2-18d-18e%2B81+
    which simplifies:
    d%5E2%2B2de%2Be%5E2-18d-18e%2B81+
    There are no like terms. So the expression is fully simplified.

    P.S. If you're having trouble with how we used the %28a-b%29%5E2 pattern on %28%28d%2Be%29-9%29%5E2++ then using a temporary variable can help:
    Let q = (d+e). Substituting this in for (d+e) we get:
    %28q-9%29%5E2++
    Use of the %28a-b%29%5E2 pattern should be clear now:
    %28q%29%5E2-2%28q%29%289%29%2B%289%29%5E2
    which simplifies:
    q%5E2-18q%2B81
    Now we need to replace the "q" with "(d+e)":
    %28d%2Be%29%5E2-18%28d%2Be%29%2B81
    which is the same as we had earlier. It simplifies to the same answer we got above: d%5E2%2B2de%2Be%5E2-18d-18e%2B81+

    P.P.S. If you insist on multiplying %28%28d%2Be%29-9%29%5E2 out instead of using the patterns, you should end up with the same result.