Question

The remainder when \(4x^3 + 4x^2 + x - 4\) is divided by \(2x - 1\) is

• A. 2
• B. -2
• C. 4
• D. -4

Hint:

For division by a linear polynomial like \(ax-b\), always find the remainder by putting \(x=\dfrac{b}{a}\) in the polynomial. Here, for \(2x-1\), use \(x=\dfrac{1}{2}\).

Answer 1

The Correct Option is C

Step 1: Use the Remainder Theorem for a linear divisor.}
When a polynomial \(f(x)\) is divided by a linear expression of the form \(ax-b\), the remainder can be found by substituting \(x = \dfrac{b}{a}\) into the polynomial.
Here, the divisor is \(2x - 1\). So, we put: \[ 2x - 1 = 0 \] \[ x = \frac{1}{2} \] Therefore, the remainder is: \[ f\left(\frac{1}{2}\right) \]
Step 2: Substitute \(x = \dfrac{1}{2}\) into the polynomial.}
Given polynomial: \[ f(x) = 4x^3 + 4x^2 + x - 4 \] Now substitute \(x = \dfrac{1}{2}\): \[ f\left(\frac{1}{2}\right) = 4\left(\frac{1}{2}\right)^3 + 4\left(\frac{1}{2}\right)^2 + \frac{1}{2} - 4 \]
Step 3: Simplify each term carefully.}
First term: \[ 4\left(\frac{1}{2}\right)^3 = 4 \cdot \frac{1}{8} = \frac{1}{2} \] Second term: \[ 4\left(\frac{1}{2}\right)^2 = 4 \cdot \frac{1}{4} = 1 \] Third term: \[ \frac{1}{2} \] So, \[ f\left(\frac{1}{2}\right) = \frac{1}{2} + 1 + \frac{1}{2} - 4 \] Now combine: \[ \frac{1}{2} + \frac{1}{2} = 1 \] \[ 1 + 1 = 2 \] \[ 2 - 4 = -2 \]
Step 4: Match with the options.}
Thus, the remainder is: \[ -2 \] So, the correct option is: \[ \text{(B) } -2 \] Final Answer:} \(-2\).