If the expression $f(x) = x^4 + 2x^3 + ax^2 + bx + 3$ has remainder $r(x) = 4x + 3$, when divided by $g(x) = x^2 + x - 2$, then:
Step-by-Step Solution
Step 1: Since $g(x) = x^2 + x - 2 = (x+2)(x-1)$, the roots are $x = 1$ and $x = -2$.
Step 2: When $f(x)$ is divided by $g(x)$, the remainder is $r(x) = 4x + 3$. So:
$$f(x) = g(x) \cdot q(x) + r(x)$$
This means $f(1) = r(1)$ and $f(-2) = r(-2)$.
Step 3: Apply $x = 1$:
$$f(1) = 1 + 2 + a + b + 3 = a + b + 6$$
$$r(1) = 4(1) + 3 = 7$$
$$a + b + 6 = 7 \Rightarrow a + b = 1 \quad \cdots (1)$$
Step 4: Apply $x = -2$:
$$f(-2) = 16 - 16 + 4a - 2b + 3 = 4a - 2b + 3$$
$$r(-2) = 4(-2) + 3 = -5$$
$$4a - 2b + 3 = -5 \Rightarrow 4a - 2b = -8 \Rightarrow 2a - b = -4 \quad \cdots (2)$$
Step 5: Solve equations (1) and (2):
Adding: $3a = -3 \Rightarrow a = -1$
From (1): $b = 1 - a = 1 - (-1) = 2$
Step 6: Verify each option:
- $a + b = -1 + 2 = 1$ ✓ (option 1)
- $a - b = -1 - 2 = -3$ ✓ (option 2)
- $|b| = 2$, $|2a| = 2$, so $|b| = |2a|$ ✓ (option 3)
- $3b - 2a = 6 - (-2) = 8$ ✓ (option 4)
Correct Answer: 1, 2, 3, 4