Question 139591This question is from textbook Algebra Structure and Method
: I have been struggling with this problem and I was wondering if anyone could help me? I would deeply appreciate it! Please and Thank you!
Factoring Patterns for ax^2+bx+c
Factor. Check by multiplying the factors. If the polynomial is not factorable, write prime.
2(x-y)^2-9(x-y)z-5z^2
This question is from textbook Algebra Structure and Method
Found 2 solutions by jim_thompson5910, solver91311:
Answer by jim_thompson5910(35256)   (Show Source):
You can put this solution on YOUR website!  Start with the given equation
Let 
 Plug in
 Sort the terms in descending order
 Factor out a negative one
Now let's factor the inner polynomial 
Looking at we can see that the first term is  and the last term is  where the coefficients are 5 and -2 respectively.
Now multiply the first coefficient 5 and the last coefficient -2 to get -10. Now what two numbers multiply to -10 and add to the middle coefficient 9? Let's list all of the factors of -10:
Factors of -10:
1,2,5,10
-1,-2,-5,-10 ...List the negative factors as well. This will allow us to find all possible combinations
These factors pair up and multiply to -10
(1)*(-10)
(2)*(-5)
(-1)*(10)
(-2)*(5)
note: remember, the product of a negative and a positive number is a negative number
Now which of these pairs add to 9? Lets make a table of all of the pairs of factors we multiplied and see which two numbers add to 9
| First Number | Second Number | Sum | | 1 | -10 | 1+(-10)=-9 | | 2 | -5 | 2+(-5)=-3 | | -1 | 10 | -1+10=9 | | -2 | 5 | -2+5=3 |
From this list we can see that -1 and 10 add up to 9 and multiply to -10
Now looking at the expression , replace  with  (notice adds up to . So it is equivalent to )
Now let's factor  by grouping:
 Group like terms
 Factor out the GCF of  out of the first group. Factor out the GCF of  out of the second group
 Since we have a common term of  , we can combine like terms
So factors to
So factors to
This means that factors to  (remember, we pulled out a negative one previously)
 Now replace "w" with 
 Distribute
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Answer:
So factors to
Answer by solver91311(24713)  (Show Source):
You can put this solution on YOUR website! There is a slight flaw in your terminology. ALL 2nd degree trinomials can be expressed as the product of two factors,  and  . The thing is, r and s aren't always integers or even rational numbers. I suspect your instructor meant to say "If the polynomial is not factorable over the integers (or rationals), write prime."
With this ungodly horror, , recognize that the independent variable is , so let's clean it up a bit by saying  . That let's us rewrite the expression:  . Now we can see that the pattern  means that  ,  , and 
Now we can use the fact that  and  are roots of  if and only if and are factors of  .
So let's find the roots of  using the quadratic formula:
 or 
 or 
 or 
That means that either  or  (which is equivalent to  .
Hence our factors so far are  , but remember that , so:
 are your two factors.
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