Problem Statement
When a polynomial $f(x) = 2x^3 + ax^2 + bx - 6$ is divided by $(x - 1)$, the remainder is $0$, and when divided by $(x + 2)$, the remainder is $-12$.
(i) Find the values of the constants $a$ and $b$.
(ii) Hence, factorise $f(x)$ completely into linear factors.
Verified Solution & Marking Scheme
Form simultaneous equations using Remainder Theorem
By Remainder Theorem:
1) $f(1) = 0 \implies 2(1)^3 + a(1)^2 + b(1) - 6 = 0 \implies 2 + a + b - 6 = 0 \implies a + b = 4$ --- (1)
2) $f(-2) = -12 \implies 2(-2)^3 + a(-2)^2 + b(-2) - 6 = -12$
$-16 + 4a - 2b - 6 = -12 \implies 4a - 2b - 22 = -12 \implies 4a - 2b = 10 \implies 2a - b = 5$ --- (2)
Solve for a and b
Adding (1) and (2):
$(a + b) + (2a - b) = 4 + 5 \implies 3a = 9 \implies a = 3$
Substitute $a = 3$ into (1):
$3 + b = 4 \implies b = 1$
Factorise f(x) completely
$f(x) = 2x^3 + 3x^2 + x - 6$
Since $(x - 1)$ is a known factor, divide $f(x)$ by $(x - 1)$:
$2x^3 + 3x^2 + x - 6 = (x - 1)(2x^2 + 5x + 6)$
Let us check roots of $2x^2 + 5x + 6$: discriminant $D = 25 - 4(2)(6) = 25 - 48 = -23 < 0$.
$\text{Complete real factorization: } f(x) = (x - 1)(2x^2 + 5x + 6)$