Quotient of (x³ + 6x² + 11x + 6) ÷ (x² + 4x + 3)x + 2x – 2x + 10x + 6
2026-02-28 11:48 Diff

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Last updated on October 4, 2025

The result we get when we divide one polynomial by another is called the quotient. The quotient can be a polynomial or a rational expression, depending on the polynomials involved. We will learn about the quotient of (x³ + 6x² + 11x + 6) ÷ (x² + 4x + 3)x + 2x – 2x + 10x + 6 below.

What is the Quotient of (x³ + 6x² + 11x + 6) ÷ (x² + 4x + 3)x + 2x – 2x + 10x + 6?

To find the quotient of (x³ + 6x² + 11x + 6) ÷ (x² + 4x + 3), we can follow the steps given below. These steps make the polynomial division process simple.

Step 1: Identify the divisor and dividend. In this case, the dividend is x³ + 6x² + 11x + 6, and the divisor is x² + 4x + 3.

Step 2: Divide the leading term of the dividend (x³) by the leading term of the divisor (x²) to get the first term of the quotient, which is x.

Step 3: Multiply the entire divisor by the first term of the quotient (x) and subtract from the dividend to find the new dividend: (x³ + 6x² + 11x + 6) - (x³ + 4x² + 3x) = 2x² + 8x + 6.

Step 4: Repeat the process with the new dividend. Divide 2x² by x² to get the next term of the quotient, which is 2.

Step 5: Multiply the entire divisor by 2 and subtract from the new dividend: (2x² + 8x + 6) - (2x² + 8x + 6) = 0.

Step 6: The quotient is x + 2, and since the remainder is 0, it confirms the division is exact.

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Important Glossaries of Polynomial Division

  • Quotient: The result obtained when dividing one polynomial by another.
  • Dividend: The polynomial being divided.
  • Divisor: The polynomial by which we divide.
  • Leading term: The term in a polynomial with the highest power of the variable.
  • Polynomial division: The process used to divide one polynomial by another, similar to long division with numbers.

Jaskaran Singh Saluja

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Jaskaran Singh Saluja is a math wizard with nearly three years of experience as a math teacher. His expertise is in algebra, so he can make algebra classes interesting by turning tricky equations into simple puzzles.

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