question_answer

Obtain all other zeroes or the polynomial\[{{x}^{4}}+6{{x}^{3}}+{{x}^{2}}-24x-20\], if two of its zeroes are \[+2\] and \[-5\].

Answer:

Given, \[2,-5\] are the zeroes of polynomial

\[p(x)={{x}^{4}}+6{{x}^{3}}+{{x}^{2}}-24x-20\]

So \[(x-2)\]and \[(x+5)\] are factors of \[p(x)\]

\[\Rightarrow \]   \[(x-2)(x+5)\] is also a factor of \[p(x)\]

So        \[(x-2)(x+5)={{x}^{2}}+3x-10\]

So, by remainder theorem,

Dividend = Divisor \[\times \] Quotient + Remainder

\[{{x}^{4}}+6{{x}^{3}}+{{x}^{2}}-24x-20=({{x}^{2}}+3x-10)\times ({{x}^{2}}+3x+2)+0\]

\[=({{x}^{2}}+3x-10)\text{ (}{{x}^{2}}+2x+x+2)\]

\[=({{x}^{2}}+3x-10)[x(x+2)+1(x+2)]\]

\[=({{x}^{2}}+3x-10)\text{(}x+2)(x+1)\]

So other zeros are \[-2\] and \[-1\].