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find the zeroes of polynomial p(x)=2x^3-5x^2-14x+8

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Solution:  We know that , if  \ x=alpha is Zero of a polynomial  then \ x-alpha is factor of \ f(x)

putting \x=-2 in \ p(x)=2x^3-5x^2-14x+8 

Now, \ p(x)=0

Therefore (x+2) is a factor of p(x) 

Now,we divide \ p(x)=2x^3-5x^2-14x+8 by \ (x+2)

Hence , p(x)=2x^3-5x^2-14x+8=(x+2)(2x^2-9x+4)

Zeros of \ p(x) is x=-2,4,-1/2

 

Posted by

Deependra Verma

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