Factorize $2 x^{3} - 3 x^{2} - 3 x + 2$ into linear factors.

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Solution

Let $p (x) = 2 x^{3} - 3 x^{2} - 3 x + 2$
Now. $p (1) = - 2 \neq 0$ (note that sum of the coefficients is not zero)
$∴ (x - 1)$ is not a factor of p(x).
However, $p (- 1) = 2 (- 1)^{3} - 3 (- 1)^{2} - 3 (- 1) + 2 = 0$ .
So, x+1 is factor pf p(x).
We shall use synthetic division to find the other factors.
Thus, $p (x) = (x + 1) (2 x^{2} - 5 x + 2)$
Now, $2 x^{2} - 5 x + 2 = 2 x^{2} - 4 x - x + 2 = (x - 2) (2 x - 1)$
Hence, $2 x^{3} - 3 x^{2} - 3 x + 2 = (x + 1) (x - 2) (2 x - 1)$

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