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

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!