**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 (ax^{2} + bx + c) and (bx^{2} + 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

(ax^{2} + 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

(bx^{2} + 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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