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 \[-1\].
You need to login to perform this action.
You will be redirected in
3 sec