Question

t^6-1= help pkease

Answer 1

Trying to factor it?

Answer 2

yes @whpalmer4

Answer 3

Well, do recall that when we multiply polynomials, the final term always is just the product of the final terms in the polynomials?

For example:
\[(x+a)(x+b) = x^2+ ax + bx + ab\]
That means our factors must have final terms that are factors of -1, right?

Answer 4

Also, you should be looking at that and thinking "difference of squares!" because you can easily take the square root of both \(t^6\) and \(1)...

Answer 5

sorry, \(t^6\) and \(1\)...

Answer 6

So, two routes that we can go here (at least):

1) difference of squares
2) unraveling the factors by the rational root theorem

Which do you want to do? Both is a perfectly acceptable response!

Answer 7

difference of squares

Answer 8

Okay, what squared gives us \(t^6\)?

Answer 9

\[\sqrt{t^6} =\]

Answer 10

t^3

Answer 11

got it

Answer 12

Right, so if we express \(t^6-1 = (t^3)^2-1\) as a product of two binomials, what are they?

Answer 13

Now you can factor each of those again as a sum or difference of cubes...

Answer 14

\[(t+1)(t ^2-t+1)(t-1)(t^2+t+1)\]

Answer 15

You got it!

Answer 16

thank you i got 3 more problems if u cxan help me

Answer 17

by the RRT, we know that any factors must have +1 or -1, so we test them out to see if they give us 0:

\[(1)^6-1 = 0\]\[(-1)^6-1=0\]So both of those are roots, which means we have factors of \((t-1)(t+1)\) which we can divide out and then hopefully recognize that we can factor the remaining part...

Answer 18

let's see the first one

Answer 19

c^2-4c+4

Answer 20

wait wrong one
t^2-24+144

Answer 21

\[t^2-24t+144\]First quick check to make when factoring a trinomial like this: is it a perfect square? Take the coefficient of \(t\), divide by 2, square the result. If that equals the 3rd term, it's a perfect square.