I need some help.....I've been trying to do these for a while and just can't get it!
1) (x-(7/6))(x+(6/7))
2)(x^6 - 5x^3 + 6) / (x^3 - 3)
Show work if possible Thanks![]()
I need some help.....I've been trying to do these for a while and just can't get it!
1) (x-(7/6))(x+(6/7))
2)(x^6 - 5x^3 + 6) / (x^3 - 3)
Show work if possible Thanks![]()
Last edited by JPHamlett; 08-19-2011 at 04:10 PM.
x^2-49/36
the 7/6 and the -7/6 cancel out
Thats just doing it off the top of my head.
What you have to do is the foil method.
First, Outer, Inner, Last
What it means, is everyone on the left side gets multiplied on the right side.
FIRST OUTER INNER LAST
x with x, x with 7/6, then x with -7/6 and last is 7/6 with 7/6
EDIT: didnt see the negative in front of 7/6 bt you get the idea
Last edited by Overtime; 08-19-2011 at 04:19 PM.
Are you just trying to find the roots?
The first one is already done, for the second one make a simple substitution and it will look very easy!
Let x^3 = y;
So 2) becomes:
=(y^2-5y+6)/(y-3)
=(y-3)*(y-2)/(y-3)
=y-2
=x^3-2
x=2^(1/3)
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I thought these were polynomials...
Ya I need to factor them out
Well Overtime did the math wrong but he gave you the steps right.
Just for more detail..
(x-(7/6))(x+(6/7))
x^2
(x-(7/6))(x+(6/7))
(6/7)x
(x-(7/6))(x+(6/7)) (notice the negative)
-(7/6)x
(x-(7/6))(x+(6/7))
-(7/6) * (6/7)
x^2 + (6/7)x - (7/6)x - (7/6) * (6/7)
Then at this point, do - (7/6 * 6/7) (-1), and that brings us to
x^2 + (6/7)x - (7/6)x - 1
and the last step is just (6/7)x - (7/6)x.
x^2 - (13x)/42 - 1 (take note of the signs)
Hope that helps a bit more.
Last edited by i luffs yeww; 08-19-2011 at 06:22 PM.
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