Question

Find all other zeros of the polynomial p(x) = 2x 3 + 3x 2 - 11x - 6, if one of its zero is -3.

Answer 1

answer is  x= +2,-1/2,-3 and i soved it and solution is given in the pic look into it...for any query...if u have any doubt in solution contact me....

Answer 2

All other zeros of the polynomial $2 x^{3}+3 x^{2}-11 x-6$ is $\bold{2,-\frac{1}{2},-3}$.

We know that $A x^{n}+B x^{n}-1+C x^{n}-2+--+z X^{0}=0$, The sum of the roots $=\frac{B}{A}$.

Product of the roots $=\frac{z}{A}$ when n is even,  

$=-\frac{z}{A}$ when n is odd

For the given equation $2 x^{3}+3 x^{2}-11 x-6$.

Sum of the roots $=-\frac{3}{2}$

Product of the roots $=-\frac{-6}{2}=\frac{6}{2}$.

If a is the root of F(x), if that function is in numerator we called it as ZERO.

If the function is in denominator, we called it as POLE.  

As given information -3 is one of zero, for triple polynomial it has two other roots such as a, b

$-3+a+b=-\frac{3}{2}$

$\begin{array}{c}{A+B=\frac{3}{2}-----(1)} \\ {A b=-1 / 2----------(2)}\end{array}$

$\begin{array}{c}{x-\frac{1}{x}=3 / 2=1} \\ {x^{2}-\frac{1}{x}=3 / 2} \\ {2 x^{2}-4 x+x-2=0} \\ {X=-\frac{1}{2}, x=2}\end{array}$

Similarly will get the same values of $2,-\frac{1}{2}$

By solving above two equations, we will get value of zeros are $2,-\frac{1}{2},-3$.