Prove that $a+b+1 = 0$

Question

The polynomials $x^2+ax+b$ and $x^2+bx+a$ have common factors.prove that $a+b+1=0$.

My attempt- I could do nothing other than dividing the polynomials to get $x^2+bx+a$=$x^2+ax+b+bx-ax+a-b$.Please help me what to do.

Let $p$ and $q$ be common factors of the polynomials, then substitute to get 2 equations. — Hassan Muhammad, Sep 16 '14 at 05:59
@HassanMuhammad currently i study in 9.Hope that one day i will also become like you! — SNEHIL SINHA, Sep 16 '14 at 06:19
The question should read "have a common factor", not "have common factors". — TonyK, Sep 28 '14 at 13:52

lab bhattacharjee · Answer 1 · 2014-09-16T06:04:33.143

3

If $x-c$ is the common factor,

$x-c$ will divide $(x^2+bx+a)-(x^2+ax+b)=(b-a)(x-1)\implies c=1$

If $x^2+bx+a=(x-1)(x-d),$

$\iff x^2+bx+a=x^2-x(d+1)+d$

Comparing the constants & the coefficients of $x,$

$ d=a,b=-(d+1)=-(a+1)$

edited Sep 16 '14 at 06:04

answered Sep 16 '14 at 05:58

lab bhattacharjee

274,582

upvote thank you too for help. – SNEHIL SINHA Sep 16 '14 at 06:19
The best answer would combine the first half of this one with the second half of NumThcurious's answer. – TonyK Sep 28 '14 at 13:54

score 0 · Accepted Answer · answered Sep 16 '14 at 06:08

0

If $p$ and $q$ are common factors of the polynomials, then $$p^2+pb+a=p^2+ap+b$$ and $$q^2+qb+a=q^2+aq+b$$ So, $(b-a)p=b-a$ and $(b-a)q=b-a$, this shows that $p=q=1$ Substituting $p=q=1$ in each of the two equations yeilds $a+b+1=0$

answered Sep 16 '14 at 06:08

Hassan Muhammad

4,282

1

$p=q=1$ are not the common factors. – Sep 16 '14 at 07:06
The polynomials have one common factor, which must be of the form $x-c$ for some $c$. What you have written here makes no sense. – TonyK Sep 28 '14 at 13:51

score 0 · Answer 3 · answered Sep 16 '14 at 07:04

let $f(x)=x^2+ax+b$ and $g(x)=x^2+bx+a$.

If the 2 polynomials have a common factor, then there exists an $x$ such that can write $$ f(x)=g(x)$$ Then we obtain, $ax+b=bx+a$, which implies $x=1$ thereby $x-1$ is the common factor of the $2$ polynomials for all $a-b \neq 0$.

Since $x-1$ is a common factor of $f(x)$ and $g(x)$, then $f(1)=g(1)=0$. By replacing $x=1$ in both functions, the result is simply identical: $$ 1+b+a=1+a+b=0$$

Prove that $a+b+1 = 0$

3 Answers3