Question

Factor completely: −3x2 + 6x − 9

Answer 1

-3x^2+6x-9

Answer 2

\[\large{-3x^2+6x-9}\] do u know about middle term splitting method ?

Answer 3

No can you explain?

Answer 4

this cannot be factorise it is irreducible.

Answer 5

it is either −3(x2 − 2x + 3) or

or −1(x2 − 6x + 9)

or −3x(x2 − 2x + 3)

or −3(x2 + 2x − 3)

Answer 6

yes you are taking common cannot be factorise in the real no although with iota they can be factorised

Answer 7

iota?

Answer 8

=-3x^2+6x-9
you have to break the middle term i.e 6x in such a way that their product will be 27(since -3x^2*-9=27)
and sum will be = 6 (from 6x)
in this case the middle term will be broken into 3x and 9x becoz 3 n 9 will gave product 27 and sum 6
then place 3 n 9 in place of middle term n solve it
=-3x^2+6x-9
=-3x^2+3x+9x-9
take common out them
=-3x(x-1)+9(x-1)
=(-3x+9) (x-1) this should be the answer =)

Answer 9

yes it is complex no representation and \[\iota=\sqrt{-1}\]

Answer 10

@Qibtiya 3x+9x =12x not +6x the middle which it should be

Answer 11

Since 3,6, and 9 are all multiples of 3, you can factorize 3 from the equation =)

If you divide each term by 3, you get: 3(-x^2 + 2x -3)
Since you almost always want the highest term to be positive, you can factor the negative sign out, too!

So the answer would be -3(x^2 - 2x + 3)

Answer 12

@sami-21 no for sum we take the middle term as it is
the two terms which we select or use must give the sum equal to the middle term i.e 6
hope you understand it =)

Answer 13

@Qibtiya you wrote
=-3x^2+3x+9x-9
if i combine them i get -3x^2+12x-9 which is not the origional polynomial you have added 6x from your own maths rule is if you add spmething subtract it as well
also if you multiply the factors u got u do not get the origional question it means something wrong !

Answer 14

the two terms you selected are 3x and 9x which adds up to 12 x not 6x i hope you got ur typo error !

Answer 15

yeah thankx sami for the correction we can solve it by taking out the common i.e from whole eq
=−3(x2 + 2x − 3)
now break the middle term product= -3 and sum =+2 so the two possible values are 3 and -1
=−3(x2 + 2x − 3)
=-3[x^2-x+3x-3]
=-3[x(x-1)+3(x-1)]
=-3[(x+3) (x-1)] this is right answer