Question

If $p\left(x\right)=x^3-3x^2+6x-4$ and $p\left(\frac{\sqrt{3}}{2}\right)=0$ then by factor theorem the corresponding factor of $p\left(x\right)$ is

• A. $\sqrt{3}-1$
• B. $\sqrt{3}x+2$
• C. $x+\frac{\sqrt{3}}{2}$
• D. $x-\frac{\sqrt{3}}{2}$

Answer 1

Answer: (D)

$x-\frac{\sqrt{3}}{2}$

The factor theorem states that a polynomial $p\left(x\right)$ has a factor $(x-k)$ if and only if $p\left(k\right)=0$ (i.e. $k$ is a root).
So, the factor of $p\left(x\right)$ is $x-\frac{\sqrt{3}}{2}$