Question

Please answer it ASAP. ......... ( class 10 )

Answer 1

(-a)² + p·(-a) + q = 0
->a² -ap+ q = 0 (-a)² + m·(-a) + n = 0
-> a² -am + n = 0
Since they both equal 0, you can set them equal each other: a² -ap + q = a² -am + n -ap + q + -am + n am - ap = n-q a(m-p) = n-q
I hope it helps u

Answer 2

hi friend,

given, x+a is a factor of the polynomial x²+px+q and x²+mx+n

<>then -a is the zero of x²+px+q and x²+mx+n

→(-a)²+p(-a)+q=0 and (-a)²+m(-a)+n

→a²-pa+q =0 and a²-ma+n=0

→a²=pa-q. and a²=ma-n

by equating both of them .

pa-q=ma-n

pa-ma=q-n

a(p-m)=q-n

a=(q-n)/(p-m) [here q-n can written as -(n-q) and (p-m) can be written as -(m-p)]

however,later by dividing '-' '-' get cancel each other

a=(n-q)/(m-p)

hence proved

I hope this will help u ;)