Question
Using factor theorem, factorize the following polynomials: $y^3 - 7y + 6$
✓
Answer
Let $f(y) = y^3 - 7y + 6$
The factors of constant term in f(y) are $\pm1,\pm,2,\pm3$ and $\pm6.$
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, $y^3-7 y+6=(y-1)(y-2)(y+3)$
The factors of constant term in f(y) are $\pm1,\pm,2,\pm3$ and $\pm6.$
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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