Question

if x2+px+q And x2+mx+n , x+a is a common factor than prove that
a= n-q
p-m​

Answer 1

Solution :-

I) When x + a is thate factor of x² + px + q

Let f(x) = x² + px + q

By factor theorem

f(-a) = 0

⇒ (-a) ² + p(-a) + q = 0

⇒ a² - ap + q = 0 --- eq(1)

II) When x + a is that Factor of x² + mx + n

Let p(x) = x² + mx + n

By factor theorem

p(-a) = 0

⇒ (-a) ² + m(-a) + n = 0

⇒ a² - am + n = 0 --- eq(2)

From (1) and (2)

⇒ a² - ap + q = a² - am + n

⇒ - ap + q = - am + n

⇒ am - ap = n - q

⇒ a(m - p) = n - q

⇒ a = (n - q)/(m - p)

Hence proved.

Answer 2

Solution :

Given that (x + a) is a common factor of the polynomials x² + px + q and x² + mx + n

Thus,x = - a would be a solution of both the polynomials

Putting x = - a in x² + px + q,we write :

a² - ap + q ---------(1)

Putting x = - a in x² + mx + n,we write :

a² - am + n --------(2)

Equating equations (1) and (2),we write :

a² - ap + q = a² - am + n

→ am - ap = n - q

→ a = (n - q)/(m - p)

Hence, ProveD