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Obtain all zeroes of the polynomial $f(x)\ =\ 2x^4\ +\ x^3\ –\ 14x^2\ –\ 19x\ –\ 6$, if two of its zeroes are $-2$ and $-1$.
Given: $f(x)\ =\ 2x^4\ +\ x^3\ –\ 14x^2\ –\ 19x\ –\ 6$ and its two zeros are $-2$ and $-1$.  
To do: Obtain all zeroes of the polynomial.
Solution:
Given expression is $f(x)\ =\ 2x^4\ +\ x^3\ –\ 14x^2\ –\ 19x\ –\ 6$.
Zeros of the given expression are $-2$ and $-1$.
From Given Zeroes, we have following factors $(x\ +\ 2)$ and $(x\ +\ 1)$
⇒ $x^2\ +\ 3x\ +\ 2$
Now We divide given polynomial by $x^2\ +\ 3x\ +\ 2$ to get another quadratic polynomial.
$x^2\ +\ 3x\ +\ 2)\ 2x^4\ +\ x^3\ –\ 14x^2\ –\ 19x\ –\ 6\ (2x^2\ -\ 5x\ -\ 3$
$2x^4\ +\ 6x^3\ +\ 4x^2$
$-$ $-$ $-$
----------------------------------------------
$-5x^3\ -\ 18x^2\ -\ 19x\ -\ 6$
$-5x^3\ -\ 15x^2\ -\ 10x$
$+$ $+$ $+$
----------------------------------------------
$-\ \ 5x^2\ \ -\ 9x\ -\ 6$
$-\ \ 5x^2\ \ -\ 9x\ -\ 6$
$+$ $+$ $+$
----------------------------------------------
0
After division we get, Quotient $=\ 2x^2\ -\ 5x\ -\ 3$
Now split the middle term and get two values.
⇒ $2x^2\ -\ 2x\ -\ 3x\ -\ 3$
⇒ $2x(x\ -\ 1)\ -\ 3(x\ -\ 1)$
⇒ $(2x\ -\ 3)(x\ -\ 1)$
So, $x\ =\ -\frac{3}{2}$ and $x\ =\ 1$
Thus, the four Zeroes are $\mathbf{1,\ -1,\ -2}$ and $\mathbf{-\frac{3}{2}}$.
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