If (x+a) is a factor of two polynomials x2+px+q and x2+mx+n, then prove that :

a=n- q/m-p

from which book did u get this question ?

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.
 -->  -->  -->  -->  -->  -->  --> 

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
 -->  -->  -->  -->  -->  -->  --> 

p(x)=x²+px+q
g(x)=x+a
x= -a

thank you for the answer

Thank you Vishes Rao.

x+a is a factor of polynomial
 

Thnx

Refer to the pic for your answer Hope it helps...😊😊

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

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 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