SOLUTION: please help me factor completely with respect to the integers x^4-13x^2+36

Question 117545: please help me
factor completely with respect to the integers
x^4-13x^2+36
Answer by jim_thompson5910(35256) (Show Source): You can put this solution on YOUR website!

Looking at we can see that the first term is and the last term is where the coefficients are 1 and 36 respectively.

Now multiply the first coefficient 1 and the last coefficient 36 to get 36. Now what two numbers multiply to 36 and add to the middle coefficient -13? Let's list all of the factors of 36:

Factors of 36:
1,2,3,4,6,9,12,18

-1,-2,-3,-4,-6,-9,-12,-18 ...List the negative factors as well. This will allow us to find all possible combinations

These factors pair up and multiply to 36
1*36
2*18
3*12
4*9
6*6
(-1)*(-36)
(-2)*(-18)
(-3)*(-12)
(-4)*(-9)
(-6)*(-6)

note: remember two negative numbers multiplied together make a positive number

Now which of these pairs add to -13? Lets make a table of all of the pairs of factors we multiplied and see which two numbers add to -13

First Number	Second Number	Sum
1	36	1+36=37
2	18	2+18=20
3	12	3+12=15
4	9	4+9=13
6	6	6+6=12
-1	-36	-1+(-36)=-37
-2	-18	-2+(-18)=-20
-3	-12	-3+(-12)=-15
-4	-9	-4+(-9)=-13
-6	-6	-6+(-6)=-12

From this list we can see that -4 and -9 add up to -13 and multiply to 36

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

Now factor and using the difference of squares

-------------------------------
Answer:

So completely factors to
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