Question:
When \( 16x^{4} + 12x^{3} - 10x^{2} + 8x + 20 \) is divided by \( 4x - 3 \), the quotient and the remainder are, respectively
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 \( \dfrac{61}{2} \)
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The Correct Option is D
Solution and Explanation
Step 1: What long division of polynomials gives.
Dividing a polynomial by \( 4x - 3 \) leaves \[ \text{Dividend} = (\text{Divisor})(\text{Quotient}) + \text{Remainder} \]
The divisor has degree 1, so the remainder is a constant. The dividend has degree 4, so the quotient has degree 3. That already rules out option (C), whose quotient is only degree 2.
Step 2: Divide the leading terms.
\( 16x^{4} \div 4x = 4x^{3} \). Multiply back: \( 4x^{3}(4x - 3) = 16x^{4} - 12x^{3} \).
Subtract: \( 12x^{3} - (-12x^{3}) = 24x^{3} \), so we are left with \( 24x^{3} - 10x^{2} + 8x + 20 \).
Step 3: Next term of the quotient.
\( 24x^{3} \div 4x = 6x^{2} \). Multiply back: \( 6x^{2}(4x - 3) = 24x^{3} - 18x^{2} \).
Subtract: \( -10x^{2} + 18x^{2} = 8x^{2} \), leaving \( 8x^{2} + 8x + 20 \).
Step 4: Third term.
\( 8x^{2} \div 4x = 2x \). Multiply back: \( 2x(4x - 3) = 8x^{2} - 6x \).
Subtract: \( 8x + 6x = 14x \), leaving \( 14x + 20 \).
Step 5: Fourth term.
\( 14x \div 4x = \dfrac{7}{2} \). Multiply back: \( \dfrac{7}{2}(4x - 3) = 14x - \dfrac{21}{2} \).
Subtract: \( 20 + \dfrac{21}{2} = \dfrac{40 + 21}{2} = \dfrac{61}{2} \).
This is a constant, so the division stops.
Step 6: Collect the result and verify.
Quotient \( = 4x^{3} + 6x^{2} + 2x + \dfrac{7}{2} \), remainder \( = \dfrac{61}{2} \).
Check with the remainder theorem: the remainder equals the dividend evaluated at the root \( x = \dfrac{3}{4} \).
\[ 16\left(\frac{81}{256}\right) + 12\left(\frac{27}{64}\right) - 10\left(\frac{9}{16}\right) + 8\left(\frac{3}{4}\right) + 20 \]
\[ = \frac{81}{16} + \frac{81}{16} - \frac{90}{16} + 6 + 20 = \frac{72}{16} + 26 = 4.5 + 26 = 30.5 = \frac{61}{2} \]
The remainder matches.
Step 7: Why the other options are wrong.
Option (A) drops the constant \( \frac{7}{2} \) from the quotient.
Option (B) drops the \( 2x \) term and also misstates the remainder as \( \frac{51}{2} \).
Option (C) has a quotient of the wrong degree.
Final Answer:
Only option (D) carries both the full quotient and the correct remainder. \[ \boxed{4x^{3} + 6x^{2} + 2x + \tfrac{7}{2}, \quad \tfrac{61}{2}} \]
Dividing a polynomial by \( 4x - 3 \) leaves \[ \text{Dividend} = (\text{Divisor})(\text{Quotient}) + \text{Remainder} \]
The divisor has degree 1, so the remainder is a constant. The dividend has degree 4, so the quotient has degree 3. That already rules out option (C), whose quotient is only degree 2.
Step 2: Divide the leading terms.
\( 16x^{4} \div 4x = 4x^{3} \). Multiply back: \( 4x^{3}(4x - 3) = 16x^{4} - 12x^{3} \).
Subtract: \( 12x^{3} - (-12x^{3}) = 24x^{3} \), so we are left with \( 24x^{3} - 10x^{2} + 8x + 20 \).
Step 3: Next term of the quotient.
\( 24x^{3} \div 4x = 6x^{2} \). Multiply back: \( 6x^{2}(4x - 3) = 24x^{3} - 18x^{2} \).
Subtract: \( -10x^{2} + 18x^{2} = 8x^{2} \), leaving \( 8x^{2} + 8x + 20 \).
Step 4: Third term.
\( 8x^{2} \div 4x = 2x \). Multiply back: \( 2x(4x - 3) = 8x^{2} - 6x \).
Subtract: \( 8x + 6x = 14x \), leaving \( 14x + 20 \).
Step 5: Fourth term.
\( 14x \div 4x = \dfrac{7}{2} \). Multiply back: \( \dfrac{7}{2}(4x - 3) = 14x - \dfrac{21}{2} \).
Subtract: \( 20 + \dfrac{21}{2} = \dfrac{40 + 21}{2} = \dfrac{61}{2} \).
This is a constant, so the division stops.
Step 6: Collect the result and verify.
Quotient \( = 4x^{3} + 6x^{2} + 2x + \dfrac{7}{2} \), remainder \( = \dfrac{61}{2} \).
Check with the remainder theorem: the remainder equals the dividend evaluated at the root \( x = \dfrac{3}{4} \).
\[ 16\left(\frac{81}{256}\right) + 12\left(\frac{27}{64}\right) - 10\left(\frac{9}{16}\right) + 8\left(\frac{3}{4}\right) + 20 \]
\[ = \frac{81}{16} + \frac{81}{16} - \frac{90}{16} + 6 + 20 = \frac{72}{16} + 26 = 4.5 + 26 = 30.5 = \frac{61}{2} \]
The remainder matches.
Step 7: Why the other options are wrong.
Option (A) drops the constant \( \frac{7}{2} \) from the quotient.
Option (B) drops the \( 2x \) term and also misstates the remainder as \( \frac{51}{2} \).
Option (C) has a quotient of the wrong degree.
Final Answer:
Only option (D) carries both the full quotient and the correct remainder. \[ \boxed{4x^{3} + 6x^{2} + 2x + \tfrac{7}{2}, \quad \tfrac{61}{2}} \]
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