Question 596010: I Factored these questions and here are my answers
1)15^3-60x
15x(x^2-4)
2)5x^3-2x^2+5x-2
x^2(5x-2) (5x-2)
(x^2+1)(5x-2)
3)3x^2+x-2
ac=3(-2)=-6 b=1
3x^2+3x-2x-2
3x(x+1)-2(x+1)
(3x-2)(x+1)
Simplify a)(x+y)(x^2-5y^4)
answer:(x^3-xy^4)+(x^2 -5y^5)
4)write the set {...,-3,-1,1,3,5,7,...}using set builder notation.
Found 3 solutions by jim_thompson5910, Alan3354, solver91311:
Answer by jim_thompson5910(35256)   (Show Source):
You can put this solution on YOUR website! # 1
you can keep going and factor x^2-4 into (x-1)(x+1)
So the final answer is 15x(x-1)(x+1)
=======================================================
# 2
Correct
=======================================================
# 3
Correct
=======================================================
# 4
Answer by Alan3354(69443)  (Show Source):
You can put this solution on YOUR website! I Factored these questions and here are my answers
1)15^3-60x
15x(x^2-4)
You can factor further:
= 15x(x-2)(x+2)
-------------
2)5x^3-2x^2+5x-2
x^2(5x-2) (5x-2)
(x^2+1)(5x-2)
----------
3)3x^2+x-2
ac=3(-2)=-6 b=1
3x^2+3x-2x-2
3x(x+1)-2(x+1)
(3x-2)(x+1)
----------
Simplify a)(x+y)(x^2-5y^4)
answer:(x^3-xy^4)+(x^2 -5y^5)
"Simplified" can be a matter of opinion.
----------------
Answer by solver91311(24713)  (Show Source):
You can put this solution on YOUR website!
All of your factoring problems have correct answers.
The simplify problem is wrong. You are multiplying two binomials, so use the FOIL process. Start over and do each multiplication carefully.
The set is just the set of all odd integers.
John
My calculator said it, I believe it, that settles it
|
|
|
