If $r_1$ and $r_2$ are the remainder when $f(x)=4x^3+3x^2-12x+a$ is divided by $(x-1)$ and $(x+2)$, and $2r_1+r_2=6$, then $a=$
Step-by-Step Solution
Key Concept: $r_1=f(1)$, $r_2=f(-2)$
Step 1: Determine the remainders $r_1$ and $r_2$.
By the Remainder Theorem, when a polynomial $f(x)$ is divided by $(x-c)$, the remainder is $f(c)$.
For $r_1$, $f(x)$ is divided by $(x-1)$, so $c=1$.
$$r_1 = f(1) = 4(1)^3 + 3(1)^2 - 12(1) + a$$
$$r_1 = 4 + 3 - 12 + a$$
$$r_1 = 7 - 12 + a$$
$$r_1 = a - 5$$
For $r_2$, $f(x)$ is divided by $(x+2)$, which is $(x-(-2))$, so $c=-2$.
$$r_2 = f(-2) = 4(-2)^3 + 3(-2)^2 - 12(-2) + a$$
$$r_2 = 4(-8) + 3(4) - (-24) + a$$
$$r_2 = -32 + 12 + 24 + a$$
$$r_2 = -20 + 24 + a$$
$$r_2 = a + 4$$
Step 2: Use the given relationship between $r_1$ and $r_2$ to find $a$.
The problem states that $2r_1 + r_2 = 6$.
Substitute the expressions for $r_1$ and $r_2$ found in Step 1:
$$2(a-5) + (a+4) = 6$$
$$2a - 10 + a + 4 = 6$$
$$3a - 6 = 6$$
Add 6 to both sides:
$$3a = 6 + 6$$
$$3a = 12$$
Divide by 3:
$$a = \frac{12}{3}$$
$$a = 4$$
Correct Answer: 8