Question 207·Hard·Advanced Math

The factor $x-3$ becomes zero at $x=3$. Because the polynomial is divisible by that factor, the factor's zero must make the whole polynomial zero: $4(3)^3-k(3)^2-9(3)+6k=0$. Don't test $-3$; that would make $x+3$ zero, not $x-3$.

Type $4(3)^3-k(3)^2-9(3)+6k\sim0$. The $\sim$ makes this a regression: Desmos fits $k$ so the two sides agree. It shows $k=27$ under PARAMETERS.

The known factor starts with $x$, and the polynomial starts with $4x^3$. Since $x(4x^2)=4x^3$, the other factor must start with $4x^2$. Call its remaining coefficients $b$ and $c$: $4x^3-kx^2-9x+6k=(x-3)(4x^2+bx+c)$.

The constant term has no $x$. On the right, it comes from multiplying $-3$ by $c$; on the left, it's $6k$. Match them: $-3c=6k$. Divide by $-3$: $c=\frac{6k}{-3}$. Add $6k/(-3)$ in Desmos. It prints $-54$, so $c=-54$. Dividing by positive $3$ would lose the sign.

The $x^2$ terms on the right come from $x(bx)$ and $(-3)(4x^2)$. Their coefficient must match the original's $-k$, so $b-3(4)=-k$. Add $3(4)$: $b=-k+3(4)$. Type $-k+3(4)$ in Desmos; it prints $-15$. With $b=-15$ and $c=-54$, the equivalent polynomial is $(x-3)(4x^2-15x-54)$. Choice D.