Verify $f (x) = 2 x^{3} + 11 x^{2} - 7 x - 6$ is the factor of $(x - 1)$ using factor theorem.

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Solution

The factor theorem states that a polynomial $f (x)$ has a factor $(x - k)$ if and only if $f (k) = 0$ . If $x - 1$ is a factor, then $x = 1$ is a root.
To verify the root, substitute $x = 1$ in $f (x) = 0$ .
$f (x) = 2 x^{3} + 11 x^{2} - 7 x - 6$
$f (1) = 2 (1)^{3} + 11 (1)^{2} - 7 (1) - 6$
$0 = 2 + 11 - 7 - 6$
$0 = 0$
Therefore, the given factor $(x - 1)$ is a factor of $2 x^{3} + 11 x^{2} - 7 x - 6$

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