Question

If x+a is a factor of x^2+ px + q and x^2+ mx+ n then prove that a= n-q/ m-p

Answer 1

x+a=0
=x=-a
=(x)^2+p(x)+q
=-a^2+p(-a)+q
=a^2-ap+q

=(x)^2+mx+n
=a^2-am+n
Since x+a is a factor of x^2+px+q and x^2+mx+n
=a^2-ap+q=a^2-am+n
=a^2-a^2-ap+q=-am+n
=-am+n=-ap+q
=q-n=am-ap
=n-q=a(m-p)
=(n-q)÷(m-p)=a

Answer 2

Answer:

$a = \frac{n-q}{m-p}$    

Step-by-step explanation:

$x+a=0$  [$x+a$ is a factor]

$x=-a$ -------------(i)

$x^{2} +px+q$  and  $x^{2} + mx+n$  are factor of  $x+a$.

Then we can write

$x^{2} +px+q = x^{2} +mx+n$

$(-a)^{2} +p(-a)+q = (-a)^{2} +m(-a)+n$  [∵ $x=-a$ {from $eq^{n}$(i)}]

$a^{2} -pa+q = a^{2} -ma+n$

$-pa+q = -ma+n$

$-pa+ma = n-q$

$a(-p+m) = n-q$

$a(m-p) = n-q$

$a = \frac{n-q}{m-p}$

Hence Proved

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