Question

if (n-k) is a factor of the polynomials x²+px+q&x²+mx+n .Then the value of k is
please tell the answer ​

Answer 1

\bold{k=\frac{n-q}{p-m}}k=

p−m

n−q

is proven when n-k is factor of polynomials \bold{x^{2}+p x+q\ and\ x^{2}+m x+n}x

2

+px+q and x

2

+mx+n

Solution:

Let f(x)=x^{2}+p x+qf(x)=x

2

+px+q and g(x)=x^{2}+m x+ng(x)=x

2

+mx+n

Given: f(x) and g(x) has factor (n-k).

If (x-a) is factor of f(x), then f (a) = 0.

So, f(k) = 0 ⇒ k^{2}+p k+q=0k

2

+pk+q=0

And g(k) = 0⟹k^{2}+m k+n=0k

2

+mk+n=0

Equating both the equations.

\begin{gathered}\begin{array}{l}{k^{2}+p k+q=k^{2}+m k+n} \\ \\{p k-m k=n-q} \\ \\{k(p-m)=(n-q)} \\ \\{k=\frac{n-q}{p-m}}\end{array}\end{gathered}

k

2

+pk+q=k

2

+mk+n

pk−mk=n−q

k(p−m)=(n−q)

k=

p−m

n−q

hope it helps you