SOLUTION: x^3+13x^2+32x+20 please help me factorizing above equation =(x)(x^2+12x+x+12)+20(x+1) =(x)(x+1)(x+12)+20(x+1) =(x+1)[(x)(x+12)+20] =(x+1)(x^2+12x+20) =(x+1)(x+10)(x+2)

Algebra ->  Polynomials-and-rational-expressions -> SOLUTION: x^3+13x^2+32x+20 please help me factorizing above equation =(x)(x^2+12x+x+12)+20(x+1) =(x)(x+1)(x+12)+20(x+1) =(x+1)[(x)(x+12)+20] =(x+1)(x^2+12x+20) =(x+1)(x+10)(x+2)      Log On


   

Question 637281: x^3+13x^2+32x+20
please help me factorizing above equation
=(x)(x^2+12x+x+12)+20(x+1)
=(x)(x+1)(x+12)+20(x+1)
=(x+1)[(x)(x+12)+20]
=(x+1)(x^2+12x+20)
=(x+1)(x+10)(x+2)
Answer by jsmallt9(3758) About Me  (Show Source):
You can put this solution on YOUR website!
It doesn't look like you need any help. Everything you've done is correct.

Another way to do this is to try the rational roots. The possible rational roots of your polynomial are all the possible ratios, positive and negative, that can be formed using a factor of the constant term (at the end) over a factor of the leading coefficient. For your polynomial the possible rational roots are:
+1/1 or +1
+2/1 or +2
+4/1 or +4
+5/1 or +5
+10/1 or +10
+20/1 or +20

We can use synthetic division to see if x-r (where r is the root) is a factor of x%5E3%2B13x%5E2%2B32x%2B20. For reasons you may already know, I already know that none of the possible positive roots can work (because all the coefficients are positive). So I will only try negative ones:
-1  |   1   13   32   20
            -1  -12  -20
       ------------------
        1   12   20    0

The remainder is in the lower right corner. The zero remainder tells us that x-(-1) divided evenly. IOW, x-(-1) (or x+1) is a factor. Not only that, but the rest of the bottom row tells us the other factor. The "1 12 20" translates into x%5E2%2B12x%2B20. So
x%5E3%2B13x%5E2%2B32x%2B20+=+%28x%2B1%29%28x%5E2%2B12x%2B20%29
The second factor can itself be factored:


As you can see, we get the same answer either way.