Question

a²-b²+6b-9
help my factor this, please

Answer 1

group them like this
a^2 - 9 + b^2 + 6b
now factor first 2 terms and last 2 - this will help

Answer 2

thats
a^2 - 9 - b^2 + 6b

Answer 3

(a+3)(a-3)-b(b-6)

Answer 4

thats right

Answer 5

thank you ! i've been trying to figure this out since last night :)

Answer 6

you are welcome but that hasn't factored it completely - perhaps that the best we can do - ill check it out

Answer 7

okay, thanks:)

Answer 8

no - hold on this should factor into 2 trinomials

=a^2 - b^2 + 6b - 9
= (a + b)(a - b) + 6b - 9
so last term in each parentheses will b + 3 and -3 to give -9

i think its ( a + b - 3)(a - b + 3)

Answer 9

lets expand that to check
= a^2 -ab + 3a + ab - b^2 + 3b -3a + 3b - 9
= a^2 - b^2 + 6b - 9

yes

Answer 10

these are tricky ones

Answer 11

cant it be (a+b)(a-b)+[3(2b-3)
3(a+b)(a-b)(2b-3)

Answer 12

no - that final part does not equal previous line :- (a+b)(a-b) is multiplied by in last but not in previous

Answer 13

* multiplied by 3

Answer 14

before you can take out part of each term there must be common terms
eg

3(x + 2) - y(x+ 2)

= (x+2)(3-y)

Answer 15

there are no common terms in
(a+b)(a-b)+[3(2b-3)

Answer 16

i think we're allowed to have the final answer in 3 brackets

Answer 17

i'm very confused, i still dont get why we cant factor out 3 in 6b-9

Answer 18

yes you can that = 3(2b - 3)
and
(a+b)(a - b) + 3(2b - 3)
is another way of writing the original function but its not factoring it completely

Answer 19

and there are no common factors in the above 2 terms so thats as far as you cab take it

Answer 20

yeah okay i understand now, thanks :)

Answer 21

yw

Answer 22

@burhan101
This expression factors as shown below:

a²-b²+6b-9 =

a^2 - ( b^2 -6b + 9) =

a^2 - (b - 3) ^2 =

[a + (b - 3) ] [a - (b - 3)] =

(a + b - 3) (a - b + 3)