Question

Factor the polynomial:
x^(2) - 121

Answer 1

Also:
t^(3) + 2t^(2) - 9t - 18

Answer 2

121 is nothing but \(11^2\)..

So now you can write it as:

\(x^2 - 11^2\)

Now use the formula:

\[a^2 - b^2 =(a+b)(a-b)\]

Here a = x and b = 11..

Answer 3

What did you get by using that formula ??

Answer 4

Ok, I get it now! I got:
(x + 11)(x - 11)
x^(2) - 11x + 11x - 121
x^(2) - 121

Answer 5

Yep..

Answer 6

Thanks :D

Answer 7

\[\large (t^3 + 2t^2) - (9t - 18)\]

factor out \(t^2\) from first grouping and -9 from second grouping:

\[\implies t^2(t + 2) - 9(t + 2)\]

\[(t+2)(t^2-9)\]

Answer 8

Now do the same for \(t^2 - 9\)

Factorize it..

Answer 9

t^(2) - 9 = ...
t^(2) - 3^(2)?

Answer 10

Yes go ahead and have faith you are doing right..

Answer 11

Thanks again :D

Answer 12

Welcome..