Question

2x^2-5x-3=0
please factor

Answer 1

Do u know about Splitting the middle term?

Answer 2

what do you mean?

Answer 3

\[2x^2 - 5x - 3 = 0\] that's better..

DDo you know about groups factoring?

Answer 4

@user123
http://www.purplemath.com/modules/factquad.htm
have a look :)

Answer 5

2 times 3 is 6

Answer 6

clearly no so what you need to do is to find 2 number with sum equal to -5 and when multiplied they must equal to -6
Any ideas about those 2 numbers?

Answer 7

just a HINT
- 3 - 2 = -5
-3 * (-2) = 6

Answer 8

Correct rishavraj well team work
So those 2 numbers - 3 and -2 you need to add them instead of -5x so we would have the following
\[(2x^2 - 2x )(- 3x -3 )= 0\]

Answer 9

actually @rishavraj is wrong he picked the wrong numbers
when multiplied they must equal -6 not 6 and the sum must equal to -5
So those numbers would be -6 and 1
so therefor
\[(2x^2 + x) (-6x - 3) = 0\]

Answer 10

@Will.H oopps my bad...i didnt notice tht tiny "-"

Answer 11

the combination would be
-6 + 1 :P lol

Answer 12

and now we have to find the GCF for the 1st binomial which is x so therefor
\[x(2x + 1)\]
and now find the GCF of the 2nd binomial which is -3 so therefor
\[-3(2x + 1)\]
And now we have the new expression as \[(2x + 1)( x - 3)\]

Answer 13

yeah @rishavraj lol it is fine.. So yeah if you have any questions about this tag me

Answer 14

@user123
\[2x^2 - 5x - 3 = 0\]
\[2x^2 + x - 6x - 3 = 0\]
\[x(2x + 1) - 3 (2x + 1) = 0\]
\[(x - 3)(2x + 1) = 0\]

Answer 15

thanks. how should you solve this with the zero product property?

Answer 16

@Will.H

Answer 17

you mean you wanna find the zeros?

Answer 18

yea it says to solve it