Question
Using factor theorem, factorize the following polynomials:
$3x^3 - x^2 - 3x + 1$
$3x^3 - x^2 - 3x + 1$
✓
Answer
Let $f(x) = 3x^3 - x^2 - 3x + 1$
The factor of the coefficient of $x^3$ is $3$.
So, the possible rational roots of $f(x)$ are $\pm1$ 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 $f(x) = 3x^3 - x^2 - 3x + 1$ 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 $\pm1$ 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 $f(x) = 3x^3 - x^2 - 3x + 1$ 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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