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CIE A-Level Maths Study Notes

2.1.2. Polynomial Division

Contents

In algebra, polynomial division is the process of dividing one polynomial, known as the dividend, by another, referred to as the divisor. This technique, essential for simplifying complex expressions and solving equations, is similar to long division in arithmetic but with algebraic elements. This section focuses on dividing polynomials (up to degree 4) by linear or quadratic polynomials and identifying the quotient and remainder, which can be zero.

Steps in Polynomial Division

1. Arrange in Descending Powers: Organise both the dividend and divisor in descending powers of x.

2. Divide Leading Terms: Divide the leading term of the dividend by the leading term of the divisor to find the first term of the quotient.

3. Multiply and Subtract: Multiply the entire divisor by this term and subtract it from the dividend.

4. Repeat the Process: Continue with the next term of the dividend, repeating the divide-multiply-subtract sequence.

5. Result: The final quotient and remainder are obtained from this process.

Examples

Example 1: Division by a Linear Polynomial

Divide x4+2x3−5x+6x^4 + 2x^3 - 5x + 6 by x−2x - 2.

Solution:

a. Set Up: x4+2x3−5x+6x^4 + 2x^3 - 5x + 6 as the dividend and x−2x - 2 as the divisor.

b. Divide First Terms: x4x^4 divided by xx gives x3x^3.

c. Multiply and Subtract: (x−2)×x3=x4−2x3(x - 2) \times x^3 = x^4 - 2x^3. Subtract this from the dividend to get 4x3−5x+64x^3 - 5x + 6.

d. Repeat: Divide 4x34x^3 by xx to get 4x24x^2. Multiply and subtract to get −5x+6-5x + 6.

e. Continue: Repeat the process until the degree of the polynomial after subtraction is less than the divisor.

Final Answer: Quotient is x3+4x2+8x+11x^3 + 4x^2 + 8x + 11 and remainder is 2828.

Example 2: Division by a Quadratic Polynomial

Divide 2x3−3x2+4x−52x^3 - 3x^2 + 4x - 5 by x2+x−1x^2 + x - 1.

Solution:

a. Set Up: 2x3−3x2+4x−52x^3 - 3x^2 + 4x - 5 as the dividend and x2+x−1x^2 + x - 1 as the divisor.

b. Divide First Terms: 2x32x^3 divided by x2x^2 gives 2x2x.

c. Multiply and Subtract: (x2+x−1)×2x=2x3+2x2−2x(x^2 + x - 1) \times 2x = 2x^3 + 2x^2 - 2x . Subtract this to get −5x2+6x−5-5x^2 + 6x - 5.

d. Repeat: Divide −5x2-5x^2 by x2x^2 to get −5-5. Multiply and subtract to find the remainder.

Final Answer: Quotient is 2x−52x - 5 and remainder is 11x−1011x - 10.

Example 3: Identifying Zero Remainder

Divide x4−6x3+11x2−6xx^4 - 6x^3 + 11x^2 - 6x by x2−3x+2x^2 - 3x + 2.

Solution:

a. Set Up: x4−6x3+11x2−6xx^4 - 6x^3 + 11x^2 - 6x as the dividend and x2−3x+2x^2 - 3x + 2as the divisor.

b. Divide First Terms: x4x^4 divided by x2x^2 gives x2x^2.

c. Multiply and Subtract: (x2−3x+2)×x2=x4−3x3+2x2(x^2 - 3x + 2) \times x^2 = x^4 - 3x^3 + 2x^2. Subtract to get −3x3+9x2−6x-3x^3 + 9x^2 - 6x.

d. Repeat: Continue the process until the remainder is zero.

Final Answer: Quotient is x2−3xx^2 - 3x and remainder is 00.

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