Question

if (n-k) is a factor of polynomial (x²+px+q+m) show: [k=n+n+q/m-p]​

Answer 1

Correct Given :- (n - k) is a factor of the polynomials (x^2+px+q) and (x² + mx + n)

To Show :- k = n + {(n + q)/m - p)} .

Solution :-

we know that, if (x - a) is a factor of a polynomial f(x) , then,

f(a) = 0 .

so, (n - k) is factor of both given polynomials .

then,

→ (x - a) = {x - (n - k)}

Let us assume that, a = (n - k) .

so,

→ f(x) = x² + px + q

→ f(a) = a² + pa + q -------- Eqn.(1)

and,

→ f(x) = x² + mx + n

→ f(a) = a² + ma + n ------ Eqn.(2)

comparing both Eqn.(1) and Eqn.(2),

→ a² + pa + q = a² + ma + n

→ q - n = ma - pa

→ (q - n) = a(m - p)

→ a = (q - n)/(m - p)

putting value of a now,

→ (n - k) = (q - n) / (m - p)

→ k = n - {(q - n) / (m - p)}

taking (-1) common from RHS numerator,

→ k = n + {(n - q) / (m - p)} (Proved.)

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