Question

The H.C.F. of ax2 - ay2 and ax + ay is . . . . .

Answer 1

The given polynomials are P(x)=ax

2

+bx+c and Q(x)=bx

2

+ax+c

It is also given that (x+1) is the common factor of P(x) andQ(x) which means that P(−1)=0 and Q(−1)=0.

Let us first substitute P(−1)=0 in P(x)=ax

2

+bx+c as shown below:

P(x)=ax

2

+bx+c

⇒P(−1)=a(−1)

2

+(b×−1)+c

⇒0=(a×1)−b+c

⇒a−b+c=0.........(1)

Now, substitute Q(−1)=0 in Q(x)=bx

2

+ax+c as shown below:

Q(x)=bx

2

+ax+c

⇒Q(−1)=b(−1)

2

+(a×−1)+c

⇒0=(b×1)−a+c

⇒−a+b+c=0.........(2)

Now subtracting the equations 1 and 2, we get

(a−(−a))−b−b+c−c=0

⇒a+a−2b=0

⇒2a−2b=0

⇒2a=2b

⇒a=b

Now substitute a=b in equation 1:

a−a+c=0

⇒c=0

Hence a=b and c=0.

Answer 2

Answer:

ax+ay

Step-by-step explanation:

since

ax²-ay²=a(x²-y²)

=a(x+y)(x-y)

=ax+ay(x-y)

So hcf= ax+ay

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