# If (x+a) is a factor of two polynomials x2+px+q and x2+mx+n, then prove that :a=n- q/m-p

p(x)= x2+px+q

g(x)= x+a

x = -a

p(-a)= a2-ap+q=0 1st equatioon

p1(x)=x2+mx+n

p1(-a)=a2-am+n=0 2nd equation

on equating 1 and 2

a2-am+n=a2-ap+q

-am+ap=q-n

-a(m-p)=q-n

-a= q-n/m-p

a= n-q/m-p

hence proved

• 149

alpha and beta are the zeroes of x2-6x+a. find a if 3alpha+2beta=20

• -3
from which book did u get this question ?
• -13
If is a factor in and in , then and .
That would give as a system of equations that looks confusing, but we only need to solve it partially. --> --> --> --> --> --> --> • 14
If is a factor in and in , then and .
That would give as a system of equations that looks confusing, but we only need to solve it partially --> --> --> --> --> --> --> • 63
p(x)=x2+px+q
g(x)=x+a
x= -a
• -12
• -1
Thank you Vishes Rao.
• -8
x+a is a factor of polynomial

• 0
Thnx
• -3 • 8
p(x)= x2+px+q g(x)= x+a x = -a p(-a)= a2-ap+q=0 1st equatioon p1(x)=x2+mx+n p1(-a)=a2-am+n=0 2nd equation on equating 1 and 2 a2-am+n=a2-ap+q -am+ap=q-n -a(m-p)=q-n -a= q-n/m-p a= n-q/m-p hence proved
• 1
1. (x-2) if two equation x2+ px+ tatha x2+mx + n samanya factor hai .to v-a/m-v value bataye
• 0
hiiiiiiiiiiiiii
• 0
By remainder theorem
let p(x) be x2 + px+q
let f(x) be x2+mx+n
to find the value of x:
x+a=0
x=-a
substituting
p(-a)=(-a)2+p(-a)+q
=a2-pa+q
f(-a)=(-a)2+m(-a)+n
=a2-ma+n
According to question, x+a is a factor of p(x) and f(x). ( by factor theorem)
therefore, p(-a)=f(-a)=0
evaluating
a2-pa+q = a2-ma+n
-pa+q = -ma+n    (a2 gets cancelled, resulting in 0)
-(pa-q) = -(ma-n)  ('-' is common)
pa-q = ma-n
n-q = ma-pa   (transposing values)
n-q/a = ma-pa/a   (dividing both the sides by a)
n-q/a = m-p
a/n-q = 1/m-p         (inverting both the sides)
a = n-q/m-p             (bringing 'n-q' to the R.H.S)
Hence, proven

• 1
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