If the HCF of Polynomial ( ax2 + bx + c ) and ( bx2 + ax +c ) is ( x + 1 ) Then Prove c = 0 or a = b

Que. If the HCF of Polynomial ( ax2 + bx + c ) and ( bx2 + ax + c ) is ( x + 1 ) Then Prove c = 0 or a = b

Solution: We have two Polynomials (ax² + bx + c) and (bx² + ax + c)

here we have HCF = (x+1)
So, x+1 = 0
x = -1

Let’s put the value of x = -1 in both polynomial one by one

(ax² + bx + c) = 0
a(-1)² + b(-1) + c =0
a – b + c = 0

Let’s suppose this is equation 1
a -b + c = 0 …..equ. 1

Now we will put the value of x = -1 in second polynomial

(bx² + ax + c) = 0
b(-1)² + a(-1) + c = 0
b – a + c = 0

Let’s suppose this is equation 2.
b – a + c = 0 ……equ. 2

Now we will add both the equations ( equ. 1 + equ. 2 )

[ a -b + c = 0 ] …..equ. 1
+ [ b – a + c = 0 ]……equ. 2

on addition we get

a – b +c + b – a + c = 0
Here + a and – a will cancel each other and + b & – b will cancel out each other
c + c = 0
2c = 0
c = 0

So, when we add both equations we get c = 0. We prove the relation.

Now we will subtract these equations

[ a -b + c = 0 ] …..equ. 1
– [ b – a + c = 0 ]……equ. 2

on subtraction we get

a – b +c – b + a – c = 0
Here + c and – c will cancel each other
we will get the following:

a – b – b + a = 0
2a – 2b = 0
a – b = 0
a = b

So, we have solved the relation a = c.

In the solution to this problem we will use the following steps:

First we will find the value of x from HCF = 0.

X+1 = 0
x = -1

Now we will make two separate equations by putting the value of x in both polynomials one by one.

[ a -b + c = 0 ] …..equ. 1
[ b – a + c = 0 ]……equ. 2

For proving the relationship we will add and subtract these both equations.

When we add both the equations we get
c = 0

When we subtract both the equations we get

a = b

So, these were the condition and we have solved them. This is our answer.

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