SOLUTION: 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 F

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

-------------------------------
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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