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

Question

asked Oct 8, 2023 165k views

1 Answer

Thijser · Answer 1 · 2023-10-14T14:12:04+0000

Given polynomials are,

$\rm{f(x)=x^2+px+q}$

$\rm{g(x)=x^2+mx+n}$

According to Remainder Theorem, if (x + a) is a factor of f(x), then,

$\longrightarrow\rm{0=f(-a)}$

$\longrightarrow\rm{0=(-a)^2+p(-a)+q}$

$\longrightarrow\rm{0=a^2-pa+q\quad\dots(1)}$

Similarly, if (x + a) is a factor of g(x), then,

$\longrightarrow\rm{0=g(-a)}$

$\longrightarrow\rm{0=(-a)^2+m(-a)+n}$

$\longrightarrow\rm{0=a^2-ma+n\quad\dots(2)}$

Subtracting (2) from (1),

$\longrightarrow\rm{0=(a^2-pa+q)-(a^2-ma+n)}$

$\longrightarrow\rm{0=a^2-pa+q-a^2+ma-n}$

$\longrightarrow\rm{0=(m-p)a+q-n}$

$\longrightarrow\rm{0=(m-p)a-(n-q)}$

$\longrightarrow\rm{(m-p)a=n-q}$

$\longrightarrow\rm{\underline{\underline{a=(n-q)/(m-p)}}}$

Hence Proved!