Question

If ax² + bx+c and bx²+ax+c have a
common factor x+1 then show that
c=0 and a=b.​

Answer 1

Given :

• ax²+bx+c .........(1)

• bx²+ax+c...........(2)

Both equations have a common factor x+1

To find :

➡ Show that : c=0 and a=b

Solution :

➡ The Given polynomial are A(x)=ax²+bx+c and B(x)=bx²+ax+c .

It is also given that (x+1) is the common factor of A(x) and B(x) which means that

➡ x+1=0 ➡ x=-1

A(-1)=0 and B(-1)=0

let us first substitute A(-1)=0 in A(x) =ax²+bx+c as shown below :

$\implies \bf \: a(x) = {ax}^{2} + bx + c \\ \\ \implies \bf \: a( - 1) = {a( - 1)}^{2} + b( - 1) + c \\ \\ \implies \bf \: 0 = a \times 1 - b \times c \\ \\ \implies \boxed{ \bf{a - b + c = 0 \: \: }}......(1)$

Now substitute B(-1)=0 in B(x)=bx²+ax+c as shown below :

$\\ \implies \bf \: b(x) = {bx}^{2} + ax + c \\ \\ \implies \bf \: b( - 1) = {b( - 1)}^{2} + (a \times 1) + c \\ \\ \implies \bf \: 0 = b \times 1 - a + c \\ \\ \implies \boxed{ \bf \: - a + b + c = 0} \: .....(2)$

Now subtracting both equations (1) and (2) we get

$\\ \implies \bf \: (a - ( - a)) - b - b + c - c = 0 \\ \\ \implies \bf \: a + a - 2b = 0 \\ \\ \implies \bf \: 2a - 2b = 0 \\ \\ \implies \bf \: 2a = 2b \\ \\ \implies \boxed{ \bf{a = b}}$

Now substitute a=b in equation (1) we get

$\\ \implies \bf \: a - a + c = 0 \\ \\ \implies \bf \cancel{a - a} + c = 0 \\ \\ \implies \boxed{\bf \: c = 0}$

➡ Hence a=b and c=0

Hence proved !