Question:

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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For $ax-b$, synthetic division at $b/a$ is fast.
Updated On: Aug 11, 2025
  • $4x^{3}+6x^{2}+2x$ and $\dfrac{61}{2}$
  • $4x^{3}+6x^{2}+\dfrac{7}{2}$ and $\dfrac{51}{2}$
  • $6x^{3}+2x+\dfrac{2}{7}$ and $\dfrac{61}{2}$
  • $4x^{3}+6x^{2}+2x+\dfrac{7}{2}$ and $\dfrac{61}{2}$
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The Correct Option is D

Solution and Explanation

To solve the problem of dividing \(16x^4+12x^3-10x^2+8x+20\) by \(4x-3\), we use polynomial long division. The process involves dividing the terms one by one starting from the highest degree:

  1. Divide: Consider the leading term of the dividend \(16x^4\) and the divisor’s leading term \(4x\). Divide \(16x^4\) by \(4x\), which results in \(4x^3\).

  2. Multiply: Multiply the entire divisor \(4x-3\) by the result \(4x^3\), yielding:

    \(4x^3 \times (4x-3) = 16x^4 - 12x^3\).

  3. Subtract: Subtract the result from the original polynomial:

    \((16x^4 + 12x^3 - 10x^2 + 8x + 20) - (16x^4 - 12x^3)=24x^3-10x^2+8x+20\).

  4. Repeat: Take the new leading term \(24x^3\) and divide by \(4x\) to get \(6x^2\).

  5. Multiply: Multiply \(6x^2\) by \(4x-3\) to get:

    \(6x^2 \times (4x-3) = 24x^3 - 18x^2\).

  6. Subtract: Subtract again:

    \((24x^3 - 10x^2 + 8x + 20) - (24x^3 - 18x^2) = 8x^2+8x+20\).

  7. Continue: Divide \(8x^2\) by \(4x\) to get \(2x\), multiply \(2x\) by \(4x-3\):

    \(2x \times (4x-3) = 8x^2-6x\).

  8. Subtract: Perform subtraction:

    \((8x^2 + 8x + 20) - (8x^2 - 6x) = 14x+20\).

  9. Final step: Divide \(14x\) by \(4x\) to get \(\dfrac{7}{2}\). Multiply \(\dfrac{7}{2}\) by \(4x-3\):

    \(\dfrac{7}{2} \times (4x-3) = 14x - \dfrac{21}{2}\).

  10. Subtract: Subtract final time:

    \((14x + 20) - (14x - \dfrac{21}{2}) = \dfrac{61}{2}\).

The final quotient is \(4x^3 + 6x^2 + 2x + \dfrac{7}{2}\) and the remainder is \(\dfrac{61}{2}\). Therefore, the correct option is:

$4x^{3}+6x^{2}+2x+\dfrac{7}{2}$ and $\dfrac{61}{2}$
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