Find the all the zeroes of the polynomial x4+x3-34x2-4x+120 if its two zeroes are 2 and -2
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since it's two zeroes are 2 and -2 are it's zeroes
then (x-2)(x+2) completely divides x^4+x^3-34x^2-4x+120
so by division algorithm
i.e. p(x)=g(x)*q(x)+r(x) [here r(x)=0 because g(x) completely divides p(x)
we get q(x)=x^2+x-30
factoring q(x) we get (x-5)(x+6)
SO ALL THE ZEROES OF THE POLYNOMIAL ARE (x-2)(x+2)(x-5)(x+6)
then (x-2)(x+2) completely divides x^4+x^3-34x^2-4x+120
so by division algorithm
i.e. p(x)=g(x)*q(x)+r(x) [here r(x)=0 because g(x) completely divides p(x)
we get q(x)=x^2+x-30
factoring q(x) we get (x-5)(x+6)
SO ALL THE ZEROES OF THE POLYNOMIAL ARE (x-2)(x+2)(x-5)(x+6)
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