How do you determine whether x-1 is a factor of the polynomial $2 x^{3} - x^{2} - 3 x - 2$ $2x^3-x^2-3x-2$ ?

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Answer 1 · 2017-02-02T07:12:18

$(x - 1)$ $(x-1)$ is not a factor

We use the remainder theorem

When we divide a polynomial $f (x)$ $f(x)$ by $(x - c)$ $(x-c)$ ,

$f (x) = (x - c) q (x) + r (x)$ $f(x)=(x-c)q(x)+r(x)$

When $x = c$ $x=c$

$f (c) = r$ $f(c)=r$

If $(x - c)$ $(x-c)$ is a factor $f (c) = 0$ $f(c)=0$

So, here

$f (x) = 2 x^{3} - x^{2} - 3 x - 2$ $f(x)=2x^3-x^2-3x-2$

$f (1) = 2 - 1 - 3 - 2 = - 4$ $f(1)=2-1-3-2=-4$

Therefore,

the remainder is $= - 4$ $=-4$

So, $(x - 1)$ $(x-1)$ is not a factor of $f (x)$ $f(x)$

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