Question

factor completely:
11x^2-99

Answer 1

(x+9)(x+11) ?

Answer 2

First take out the common factor 11. Can you do that?

Answer 3

if you foil (x+9)(x+11) you will not get the original, im only sayng this to let you know you can check your answers on these

Answer 4

k

Answer 5

what is 11/11 and 99/11?

Answer 6

idk than

Answer 7

1 an 9

Answer 8

11(x^2 - 9)
Can you recognise what the (x^2 - 9) is?

Answer 9

yes

Answer 10

so (x-1)(x+11)

Answer 11

so you said 11(11/11x^2+99/11) what does this simplify too?

Answer 12

now i'm getting confused

Answer 13

isn't the anser (x-1)(x-11)

Answer 14

simplify what I wrote, when you factor out a number you are deviding the inside terms by that number , and then put that number on the outside. 11(x^2+9)

Answer 15

(x^2 - 9) is the difference of two squares and factorise to (x + 3)(x - 3)
The final answer is:
11(x + 3)(x - 3)

Answer 16

\[11x^2-99\]\[11(x^2)-(11)9\]
By using the fact that
\[a(b) - c(b) = b(a-c)\]
We can do the same thing with this
11 is the common fact, so c
\[11(x^2-9)\]

Answer 17

no foil your answer out and it wont match

Answer 18

so is the polynomial prime?

Answer 19

\[11x^2 -99 = 11(x^2 - 9) = 11(x-3)(x+3)\]

Answer 20

And by using the fact that
\[(a-b)(a+b) = a^2 - b^2\]
We can factor
\[x^2−9 = (x-3)(x+3)\]

Answer 21

So the final answer would be
\[11(x-3)(x+3)\]

Answer 22

is the 11 suppose to stil lb the front to make it 11(x-3)(x+3) ?

Answer 23

yes

Answer 24

Yep

Answer 25

thanks

Answer 26

factor the 11 into the first term and then foil the two binomials and you will get back to where you started

Answer 27

err distribute the 11 into the first factor and then foil the two binomails to get the original