SOLUTION: Factoring trinomials:
-60z^5 + 65z^4 + 20z^3
I have worked this far...
-5z^3(12z^2-13z-4)
12*4= 48 with a sum of -13
Just not quite sure how to always pick the two i
Algebra ->
Polynomials-and-rational-expressions
-> SOLUTION: Factoring trinomials:
-60z^5 + 65z^4 + 20z^3
I have worked this far...
-5z^3(12z^2-13z-4)
12*4= 48 with a sum of -13
Just not quite sure how to always pick the two i
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Question 872713: Factoring trinomials:
-60z^5 + 65z^4 + 20z^3
I have worked this far...
-5z^3(12z^2-13z-4)
12*4= 48 with a sum of -13
Just not quite sure how to always pick the two integers to get the product of 48 and the sum of -13. This is not the only question. Is there a faster simpler way to find the two numbers? Thank you for your help.
James Answer by rothauserc(4718) (Show Source):
You can put this solution on YOUR website! -60z^5 + 65z^4 + 20z^3
-5z^3(12z^2 -13z -4)
now factor the expression within parenthesis, start by finding the prime factors of 4 and 12
4 = 2^2 * 1 and 12 = 3*2^2
for 12, I will try 3 and 4 and for 4, I will try 4 and 1
(3z - 4)(4z + 1) = 12z^2 -13z -4
therefore
-60z^5 + 65z^4 + 20z^3 = -5z^3(3z - 4)(4z + 1)
