Question
Using factor theorem, factorize the following polynomials:
$y^3 - 7y + 6$

Answer

Let $f(y)=y^3-7 y+6$
The factors of constant term in $f(y)$ are $\pm 1, \pm, 2, \pm 3$ and $\pm 6$.
We have,
$f (1)=1-7+6=0$
$\Rightarrow( y -1)$ is a factor of $f ( y )$
$f (-1)=-1+7+6=12$
$\Rightarrow(y+1)$ is a factor of $f(y)$
$f(2)=8-14+6=0$
$\Rightarrow( y -2)$ is a factor of $f ( y )$
$f(-2)=-8+14+6=12$
$\Rightarrow(y+2)$ is not a factor of $f ( y )$
$f(3)=27-21+6=12$
$\Rightarrow(y-3)$ is not a factor of $f(y)$
$f(-3)=-27+21+6=0$
$\Rightarrow(y+3)$ is a factor of $f(y)$
Since $f(y)$ is a polynomial of degree 3 . So, it cannot have more than 3 linear factors.
Thus, factors of $f(y)$ are $(y-1)(y-2)$ and $(y+3)$.
Therefore,
$f(y)=k(y-1)(y-2)(y+3)$
$y^3-7 y+6=k(y-1)(y-2)(y+3) \ldots(1)$
Putting $y=0$ on both sides, we get,
$6=k(-1)(-2)(3)$
$6=6 k$
$k=1$
Substituting $k=1$ in (1), we get,
$y^3-7 y+6=(y-1)(y-2)(y+3)$

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