Question

Using factor theorem, factorize each of the following polynomial:
y³-7y+ 6

Answer 1

Answer : $(y-1)(y-2)(y+3)$

Factor Theorem

*$f(x)=a(x-\alpha )...$ is divisible by $x-\alpha$ if $f(\alpha )=0$.

Proof

$f(\alpha )=a(\alpha -\alpha )...=0$

Hence shown.

Let $P(x)=y^3-7y+6$.

$P(1)=1-7+6=0$

∴ $y-1$ is a factor of $P(x)$.

$P(2)=8-14+6=0$

∴ $y-2$ is a factor of $P(x)$.

$P(-3)=-27+21+6=0$

∴ $y+3$ is a factor of $P(x)$.

The answer is $(y-1)(y-2)(y+3)$.

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