Factorization of $ x^2 +xy+5x+m+5 $

Question

I want to factorize $ x^2 +xy+5x+m+5 $ . For what value of m , $ x^2 +xy+5x+m+5 $ can be resolved into linear factors ?

My try : $ x^2 +xy+5x+m+5 $ = $ x^2 +(5+y)x+(m+5) $

To get the linear factors , we must have the determinant of this eqn is >= 0 .

D = $ (5+y)^2-4(m+5)$ . Then I can not proceed .

You want to factor over the rational integers, $\mathbb{Z}$? — TBrendle, Dec 06 '13 at 04:43
http://www.askiitians.com/iit-jee-straight-line/pair-of-straight-lines/ — lab bhattacharjee, Dec 06 '13 at 04:44
If it factors, it factors as $(ax+by+c)(dx+ey+f)$ for some real $a,b,c,d,e,f$. $b$ and $e$ can't both be zero; say $b\ne0$. Then $e=f=0$. So we are talking $dx(ax+by+c)$, so we need $m=-5$. — Gerry Myerson, Dec 06 '13 at 04:48
@GerryMyerson Why did you put this as a comment when it is clearly an answer? — TBrendle, Dec 06 '13 at 04:48
@TBrendle, I don't know. I'm sure I had a good reason at the time. — Gerry Myerson, Dec 06 '13 at 04:51
Yes . @labbhattacharjee , Many many thanks . I have done it . — Way to infinity, Dec 06 '13 at 04:51

score 3 · Accepted Answer · answered Dec 06 '13 at 04:51

Note that factoring we will expect the factors to be of the form: $$(x+\alpha y + \beta)(x+\gamma y + \delta)$$ Comparing the coefficient of $y^2$ gives us $\alpha \gamma = 0$. Due to symmetry, let us set $\gamma =0$. Hence, we will expect the factors to be of the form: $$(x+\alpha y + \beta)(x + \delta)$$ Now comparing the coefficient of $y$, we get that $$\alpha \delta = 0$$ If $\alpha = 0$, there will be no $y$ term. Hence, $\delta = 0$. Hence, the factorization has to be of the form $$(x+\alpha y + \beta)x$$ Hence, $m+5 = 0 \implies m=-5$.

score 1 · Answer 2 · edited Dec 06 '13 at 04:57

1

Since now you want to have the discriminant greater than $0$ which in itself is a quadratic polynomial ie $y^2+10y+(5-4m)$ . We see that this will be always greater than $0$ . If its discriminant is always less than $0$ . Because if a quadratic has real roots than it will be positive and negative .

The determinant of $y^2+10y+(5-4m)$ is $ D = 100 -4(5-4m) $ = $80+16m$

Now $D=0$ => $ 80+16m = 0 $ =>$ m = -5 $

edited Dec 06 '13 at 04:57

Way to infinity

1,884

answered Dec 06 '13 at 04:47

abkds

2,210

I have understood your method . It is more charming . – Way to infinity Dec 06 '13 at 04:53
1

Nice to hear that :) ! – abkds Dec 06 '13 at 04:54

Factorization of $ x^2 +xy+5x+m+5 $

2 Answers2