SRL Archive - Mathematics

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 :)

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

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)

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.