Question:medium

When \( 16x^{4} + 12x^{3} - 10x^{2} + 8x + 20 \) is divided by \( 4x - 3 \), the quotient and the remainder are, respectively

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Degree 4 divided by degree 1 must leave a degree 3 quotient and a constant remainder. Find the remainder fast by putting x = 3/4 into the dividend.
Updated On: Jul 17, 2026
  • \( 4x^{3} + 6x^{2} + 2x \) and \( \dfrac{61}{2} \)
  • \( 4x^{3} + 6x^{2} + \dfrac{7}{2} \) and \( \dfrac{51}{2} \)
  • \( 6x^{2} + 2x + \dfrac{2}{7} \) and \( \dfrac{61}{2} \)
  • \( 4x^{3} + 6x^{2} + 2x + \dfrac{7}{2} \) and \( \dfrac{61}{2} \)
Show Solution

The Correct Option is D

Solution and Explanation

Step 1: Use the remainder theorem first to cut the field.
The divisor $4x - 3$ vanishes at $x = \dfrac{3}{4}$. So the remainder is the dividend evaluated there.
\[ 16\left(\tfrac{3}{4}\right)^{4} + 12\left(\tfrac{3}{4}\right)^{3} - 10\left(\tfrac{3}{4}\right)^{2} + 8\left(\tfrac{3}{4}\right) + 20 \]
Term by term: $16 \times \dfrac{81}{256} = \dfrac{81}{16}$, $12 \times \dfrac{27}{64} = \dfrac{81}{16}$, $-10 \times \dfrac{9}{16} = -\dfrac{90}{16}$, $8 \times \dfrac{3}{4} = 6$.
Adding: $\dfrac{81 + 81 - 90}{16} + 6 + 20 = \dfrac{72}{16} + 26 = \dfrac{61}{2}$.
The remainder is $\dfrac{61}{2}$, which kills option (B) straight away.

Step 2: Build the quotient by comparing coefficients.
Write the quotient as $ax^{3} + bx^{2} + cx + d$ and expand:
\[ (4x - 3)(ax^{3} + bx^{2} + cx + d) + \tfrac{61}{2} = 16x^{4} + 12x^{3} - 10x^{2} + 8x + 20 \]

Step 3: Match one power at a time.
$x^{4}$: $4a = 16$, so $a = 4$.
$x^{3}$: $4b - 3a = 12$, so $4b = 12 + 12 = 24$ and $b = 6$.
$x^{2}$: $4c - 3b = -10$, so $4c = -10 + 18 = 8$ and $c = 2$.
$x^{1}$: $4d - 3c = 8$, so $4d = 8 + 6 = 14$ and $d = \dfrac{7}{2}$.

Step 4: Confirm with the constant term.
The constant on the left is $-3d + \dfrac{61}{2} = -\dfrac{21}{2} + \dfrac{61}{2} = \dfrac{40}{2} = 20$. This matches the 20 in the dividend, so every coefficient checks out.

Step 5: Compare with the options.
The quotient is $4x^{3} + 6x^{2} + 2x + \dfrac{7}{2}$ with remainder $\dfrac{61}{2}$, which is option (D).
Option (A) misses the $\dfrac{7}{2}$, and if the quotient really ended at $2x$ the remainder would not be $\dfrac{61}{2}$.
Option (C) has degree 2, impossible when a degree 4 polynomial is divided by a degree 1 one.

Final Answer:
Coefficient matching gives the same pair as long division. \[ \boxed{4x^{3} + 6x^{2} + 2x + \tfrac{7}{2}, \quad \tfrac{61}{2}} \]
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