How do you factorise : 2x^2 - 3x-9?

Question

Im kinda confused
:/

Answer 1

2x^2-3x-9
=2x^2-6x+3x-9
=2x(x-3)+3(x-3)
=(2x+3)(x-3)

Answer 2

2x² - 3x - 9 = 0

( 2x + 3) (x - 3) = 0 ( factorised equation )

if you want to find x,

from ( 2x + 3 ),

2x + 3 = 0

2x = -3

x = -3/2 or -1.5


from ( x -3 ),

x - 3 = 0

x = 3



therefore, x = 3 or -3/2

Answer 3

9*2=18
6*3=18 & 6-3=3 this is all u hav 2 do

then write
2x^2-3x-9
=2x^2-6x+3x-9
=2x(x-3)+3(x-3)
=(x-3)(2x+3)

Answer 4

2x² - 3x -9
it will be

2x²-6x+3x-9
2x(x-3)+3(x-3)
or (2x+3)(x-3) <<<<<< solution

For further reference:
For questions of this sort you need to split the middle term in such a way so that you can get

ax²+bx+cx+d
so a*d = b*c

Like above I split -3x into +6x and -3x
+6x * -3x = -18x²
which is equal to -9*2x² = -18x².

Use this simple rule to factorise .

Good luck
Have a nice day
:)

Answer 5

(2x+3)(x-3)

To get 2x^2, toy need x and 2x....to get -9, you need two numbers that will multiply for negative 9 and add to -3......once you multiply the 2x by -3 and get -6x and then add 3 x, it works out

Answer 6

(2x+3) (x-3)

To check:
(2x+3)(x-3)
2x^2-6x+3x-9
2x^2-3x-9

Answer 7

Try this site. It will answer this question and further math problems.