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