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

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 $x=-2,4,-1/2$

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