Question

Factorize \(x^2+3xy+2y^2+4x+5y+3\) Into product of two linear factors.

Answer 1

this is not a common way but u can make a quadratic in terms of x for example and solve for its roots

Answer 2

Can you explain a little?

Answer 3

ok when we have a quadratic like this :\[x^2+4x+3\]and we want to factor it...actually we're going to find its roots...after factoring we write it like this \[(x+1)(x+3)\]-1 and -3 are the roots of quadratic

now we have\[ x^2+3xy+2y^2+4x+5y+3=0\]find the roots of this quadratic for x

Answer 4

Wow! That looks complicated! But, 3xy, 2y^2, 5y ... What to do with them?

Answer 5

I just edited the question. Please see if it is easier.

Answer 6

\[x^2+(3y+4)x+2y^2+5y+3=0\]first step is to evaluate discriminant...

Answer 7

its same

Answer 8

it will be a pretty expression...i promise

Answer 9

Δ = (3y+4)^2 - 4(2y^2 + 5y + 3) = y^2 + 4y + 4 = (y+2)^2 ?!

Answer 10

yep

Answer 11

so what are the roots?

Answer 12

\[x =\frac{ -(3y+4) \pm(y+2)}{2}\]\[ = -y -1 \ or \ -2y-3\]?

Answer 13

done

Answer 14

\[x^2+3xy+2y^2+4x+5y+3=?\]

Answer 15

Wow! That's a faster method!

Answer 16

(x+2y+3)(x+y+1)

Answer 17

Here's the slow one :(
x^2 + 3xy + 2y^2 = (x+2y)(x+y)

x^2 + 3xy + 2y^2 + 4x + 5y + 3
= (x+2y+l)(x+y+m)
= x^2 + 3xy + 2y^2 + (l+m)x + (l+2m)y + lm

l+m = 4
l+2m = 5
lm = 3

=> l = 3 ; m = 1

So,
x^2 + 3xy + 2y^2 + 4x + 5y + 3
= (x+2y+3)(x+y+1)

Thanks! I learnt a better method now!

Answer 18

welcome :)