n^3 - 9n^2 - n + 9 = 0
(4x + 9y)^2 = (squaring binomials)
-25x^2 - 115x + 210 = -5(5x-7)(x+6)
check my answer to my other problems please!!!
-4m^2n^3 - 36m^2n^2 = 4m^2n^2 (n + 9)
9x (-x + 2) = -9x + 18x (is my answer right)
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n^3 - 9n^2 - n + 9 = 0
When you factorize,the final answer is:
(n-1)(n+1)(n-9)=0
(4x + 9y)^2 =16x^2+72xy+81y^2
-4m^2n^3 - 36m^2n^2=-4m^2n^2(n+9)
9x (-x + 2)=-9x^2+18x
-25x^2 - 115x + 210 = -5(5x-7)(x+6) is correct
notthejake
n^3 - 9n^2 - n + 9 = 0
factor by grouping
(n^3 - 9n^2) - (n - 9) = 0
n^2 (n - 9) - 1(n - 9) = 0
(n^2 - 1)(n - 9) = 0
(n + 1)(n - 1)(n - 9) = 0
n + 1 = 0 ==> n = -1
n - 1 = 0 ==> n = 1
n - 9 = 0 ==> n = 9
so either n = 1 , -1, or 9
*****
(4x + 9y)^2 = (squaring binomials)
16x^2 + 36xy + 36xy + 81y^2 =
16x^2 + 72xy + 81y^2
*****
-25x^2 - 115x + 210 = -5(5x-7)(x+6)
-25x^2 - 115x + 210 = -5(5x^2 +23x - 42)
-25x^2 - 115x + 210 = -25x^2 - 115x + 210 yep, that's right
*****
GCF: -4m^2n^2, leaving n + 9
you left out a negative sign
-4m^2 n^3 - 36m^2 n^2 = -4m^2 n^2(n + 9)
***
left out an x^2:
9x(-x + 2) = -9x^2 + 18x
docktet
What are the instructions for the first problem? Is there some lead in material that might help us out. We could use a graphing calulator; but, I don't think that is what you are supposed to do.
If it is the solutions are n = -1, +1, 9.
Squaring the binomial has a rule of sq the first term of the binomial, sq the last term of the binomial, multiply the two pieces of the binomial then multiply by 2.
(4x)^2 +2(4x)(9y) +(9y)^2
16x^2 +72xy +81y^2
It looks like the third problem has already been factored. Good job.
The other two items are also correct. Congratulations.
GKumar
What is problem