Problem Statement
The cubic polynomial $P(x) = 2x^3 + ax^2 + bx - 6$ has a factor of $(x - 2)$, and when divided by $(x + 1)$ leaves a remainder of $-18$.
(a) Find the values of the constants $a$ and $b$. [4 Marks]
(b) Hence, factorize $P(x)$ completely into linear factors over $\mathbb{R}$. [2 Marks]
Verified Solution & Marking Scheme
Part (a): Set Up System using Remainder & Factor Theorems
Since $(x - 2)$ is a factor, $P(2) = 0$:
$$2(2)^3 + a(2)^2 + b(2) - 6 = 0 \implies 16 + 4a + 2b - 6 = 0 \implies 4a + 2b = -10 \implies 2a + b = -5 \quad \text{(Eq. 1)}$$
Remainder when divided by $(x + 1)$ is $-18$, so $P(-1) = -18$:
$$2(-1)^3 + a(-1)^2 + b(-1) - 6 = -18 \implies -2 + a - b - 6 = -18 \implies a - b = -10 \quad \text{(Eq. 2)}$$
Solve Simultaneous Equations
Adding (Eq. 1) and (Eq. 2):
$$(2a + b) + (a - b) = -5 + (-10) \implies 3a = -15 \implies a = -5$$
Substitute into (Eq. 2):
$$-5 - b = -10 \implies b = 5$$
So $P(x) = 2x^3 - 5x^2 + 5x - 6$.
Part (b): Factorize Completely
Since $(x - 2)$ is a factor, divide $P(x)$ by $(x - 2)$:
$$2x^3 - 5x^2 + 5x - 6 = (x - 2)(2x^2 - x + 3)$$
Check the quadratic $2x^2 - x + 3$ discriminant:
$$\Delta = (-1)^2 - 4(2)(3) = 1 - 24 = -23 < 0$$
Since $\Delta < 0$, $2x^2 - x + 3$ has no real linear factors. Thus the real factorization is:
$$P(x) = (x - 2)(2x^2 - x + 3)$$