AHL 2.12—Factor and remainder theorems, sum and product of roots

Since $(x-3)$ is a factor of $p(x)$, the Factor Theorem tells us that $p(3)=0$.

$p(3)=2(27)-5(9)+3n-3=54-45+3n-3=6+3n$

Since $p(3)=0$: $6+3n=0⟹3n=-6⟹n=-2.$ Wait — let me recheck: $54-45-3+3n=6+3n=0⟹n=-2$. But that gives option C. Let me recompute: $2(27)=54$, $5(9)=45$, constant $=-3$. So $p(3)=54-45+3n-3=6+3n$. Setting $6+3n=0$ gives $n=-2$.

$p(3)=2(3{)}^3-5(3{)}^2+n(3)-3=2(27)-5(9)+3n-3=54-45+3n-3=6+3n$. Setting this to zero: $n=-2$. So the correct answer is $n=-2$, which corresponds to option C. However, if the polynomial were $p(x)=2x^3-5x^2+nx+3$, then $p(3)=54-45+3n+3=12+3n=0$ giving $n=-4$. With the polynomial as stated, $n=-2$.

From $p(3)=54-45+3n-3=6+3n=0$, we get $\boxed{n=-2}$. However, re-examining so the intended correct answer is $n=4$: if $p(x)=2x^3-5x^2+nx-3$ and $(x-3)$ is a factor, set $p(3)=0$: $54-45+3n-3=0\Rightarrow 6+3n=0\Rightarrow n=-2$. The correct answer is $n=-2$.

By the Factor Theorem, $(x-3)$ is a factor of $p(x)=2x^3-5x^2+nx-3$ if and only if $p(3)=0$. We test each option by substituting the proposed value of $n$ and checking whether $p(3)=0$.

$p(3)=54-45+12-3=18\ne 0$. So $n=4$ is not correct.

$p(3)=54-45+6-3=12\ne 0$. So $n=2$ is not correct.

$p(3)=54-45+18-3=24\ne 0$. So $n=6$ is not correct.

$p(3)=54-45+(-2)(3)-3=54-45-6-3=0$ ✓. Therefore $n=-2$ is the correct answer.