Question
Using factor theorem, factorize the following polynomials: $3x^3 - x^2 - 3x + 1$
✓
Answer
Let $f(x)=3 x^3-x^2-3 x+1$
The factor of the coefficient of $x^3$ is $3 .$
So, the possible rational roots of $f(x)$ are $\pm 1$ and $\pm \frac{1}{3}$
We have, $f(1)=3-1-3+1=0$
$\Rightarrow(x-1)$ is a factor of $f(x) f(-1)=-3-1+3+1$
$\Rightarrow(x+1)$ is a factor of $f(x)$
So, $(x-1)$ and $(x+1)$ are factors of $f(x)$
$\Rightarrow(x-1)(x+1)$ is a also a factor of $f(x)$
$\Rightarrow x^2-1$ is a factor of $f(x)$.
Let us now divide $f(x)=3 x^3-x^2-3 x+1$ by $x^2-1$ to get
the other factors of $f(x)$. By long division, we have:
Therefore, $3x^3 - x^2 - 3x + 1 = (x^2 - 1)(3x - 1)$
Now, $(x^2 - 1) = (x - 1)(x + 1)$
Hence, $3x^3 - x^2 - 3x + 1 = (x - 1)(x + 1)(3x - 1)$
The factor of the coefficient of $x^3$ is $3 .$
So, the possible rational roots of $f(x)$ are $\pm 1$ and $\pm \frac{1}{3}$
We have, $f(1)=3-1-3+1=0$
$\Rightarrow(x-1)$ is a factor of $f(x) f(-1)=-3-1+3+1$
$\Rightarrow(x+1)$ is a factor of $f(x)$
So, $(x-1)$ and $(x+1)$ are factors of $f(x)$
$\Rightarrow(x-1)(x+1)$ is a also a factor of $f(x)$
$\Rightarrow x^2-1$ is a factor of $f(x)$.
Let us now divide $f(x)=3 x^3-x^2-3 x+1$ by $x^2-1$ to get
the other factors of $f(x)$. By long division, we have:
Therefore, $3x^3 - x^2 - 3x + 1 = (x^2 - 1)(3x - 1)$
Now, $(x^2 - 1) = (x - 1)(x + 1)$
Hence, $3x^3 - x^2 - 3x + 1 = (x - 1)(x + 1)(3x - 1)$
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