Question

If the polynomial \(kx^2 - 2xy - 3y^2 + 3x - 5y +2 \) can be factorized into product of two linear factors, what is k?

Answer 1

(x-3y+1)(x+y+2) where k=1

Answer 2

Looking at the terms, I decided that they had to be two factors each with (ax+by+c).
I started by thinking that the trailing term of 2 must come from 1 and 2.
I looked at the -3y^2 term and tried to come up with different ways to get it if the result gave me -5y down the road. Same with -2xy. I used trial and error with 3 different sets of numbers, each time tweaking a minus sign or two coefficients until I say how the pattern was coming out. I wish I had a better analytical method.

Answer 3

Trial and error :(

First, factorize -3y^2 - 5y + 2
-3y^2 - 5y + 2 = (y+2)(-3y+1)

Let
kx^2 - 2xy -3y^2 + 3x - 5y +2
= (lx + y+2)(mx - 3y + 1)
= lm x^2 + (m-3l)xy - 3y^2 + (l+2m)x -5y + 2

By equating the coefficients,
lm = k
m - 3l = -2
2m + l = 3

=> l = 1, m = 1 , k = lm = 1

Any better method?

Answer 4

I like that much better. I was thinking that with my method, if k was something other than 1 I would be in trouble. Your method seems to cut through that.

Answer 5

one can let\[kx^2−2xy−3y^2+3x−5y+2=(kx+ay+1)(x+by+2)\]and solve for a,b,k...without trial and error

but this is not easier, is it?

Answer 6

allow me to do some effort

Answer 7

Another way I've tried, but I don't know what I am actually doing:
kx^2 - 2xy - 3y^2 + 3x - 5y + 2 = 0
-3y^2 - (5+2x)y + kx^2 + 3x + 2 = 0
Δ = [-(5+2x)]^2 - 4 (-3)(kx^2 + 3x + 2)
= 25 + 20x + 4x^2 + 12kx^2 + 36x + 24
= (4+12k)x^2 + 56x + 49

Since it can be factored, it should be in (ax+b)^2, and b is 7, a = 56/(7x2) = 4
So,

\(\sqrt{4+12k} = 4\)
4 + 12k = 16
k = 1

Answer 8

after simplifying\[kx^2−2xy−3y^2+3x−5y+2\\=kx^2+(a+bk)xy+aby^2+(2k+1)x+(2a+b)y+2\]2k+1=3--->k=1
a+b=-2
2a+b=-5---->a=-3 , b=1

Answer 9

Allow me to type something here.

\[(A_1x + B_1y+C_1)(A_2x + B_2y+C_2)\]\[=(A_1x + B_1y)(A_2x + B_2y)+C_1(A_2x + B_2y)+C_2(A_1x + B_1y) +C_1C_2\]
\((A_1x + B_1y)(A_2x + B_2y)\) contains \(x^2, xy, y^2\) terms;
\(C_1(A_2x + B_2y)+C_2(A_1x + B_1y)\) contains x and y terms;
\(C_1C_2\) is a constant.

So, we can factorize the x^2, xy and y^2 terms first,

Answer 10

By the way, does my second method make sense?

Answer 11

some trial and error used for last step but makes sense :)

Answer 12

Trial and error? Where? Why?

Answer 13

lol...wait..its accurate

Answer 14

Wait.
First method = first comment
Second method = Δ one
''Something'' that I typed = background for the first method.

Answer 15

Hmm, I guess I mixed that ''something'' up with the previous question. In this question, the term y^2, y and constant are factorized first as k, which is the coefficient of x^2, is the unknown.

Answer 16

Look like I don't really understand the first method :(

Answer 17

I used the first method again, but with a few more variables (allowing k to be a product other than (k)(1). It was a total mess with a couple mistakes that I had to find, but I got the original answer along with (2x-6y+2)(0.5x+0.5y+1) where k=1. No trial and error.
Thank you for teaching me a new skill.

Answer 18

You're welcome! Actually, I just learnt it today :P