Question

if h.c.f of $ax^2+bx+c and bx^2+ax+c /: is (x+1) then prove that c=0 and a=b$​

Answer 1

GIVEN

H.c.f = x+1=0

putting x=-1 in $ax^2+bc+c\\ a(-1)^2+b(-1)+c=0\\ a-b+c=0----eq(1)$

Again,

putting x= -1 in$bx^2+ax+c\\ b(-1)^2+a(-1)+c=0\\ b+a+c=0-----eq(2)$

Adding eq (1) and eq (2)

$2c=0\\c=0$

putting c=0 in equation (1)

a-b+0=0

a-b=0

a=b

Hence,

c=0,a=b proved√√

Answer 2

нε¥ ღ∀⊥ε ʊя ∀η﹩ẘεя..

GIVEN

H.c.f = x+1=0

putting x=-1 in \begin{lgathered}ax^2+bc+c\\ a(-1)^2+b(-1)+c=0\\ a-b+c=0----eq(1)\end{lgathered}

ax

2

+bc+c

a(−1)

2

+b(−1)+c=0

a−b+c=0−−−−eq(1)

Again,

putting x= -1 in\begin{lgathered}bx^2+ax+c\\ b(-1)^2+a(-1)+c=0\\ b+a+c=0-----eq(2)\end{lgathered}

bx

2

+ax+c

b(−1)

2

+a(−1)+c=0

b+a+c=0−−−−−eq(2)

Adding eq (1) and eq (2)

\begin{lgathered}2c=0\\c=0\end{lgathered}

2c=0

c=0

putting c=0 in equation (1)

a-b+0=0

a-b=0

a=b

Hence,

c=0,a=b proved√√℘яøṽε∂ ṽεяḯḟḯε∂....

﹩øяя¥.....

нø℘ε ḯ⊥ нεʟ℘﹩....