8 min readConcept sheetPart II (2 marks) by factorisation; Part III (5 marks) by the division method

GCD and LCM of polynomials

GCD takes the common factors with the lowest powers; LCM takes every factor with the highest power. Their product equals the product of the two polynomials.

GCDGreatest common divisor: the common factor of highest degree

LCMLeast common multiple: the lowest-degree polynomial that is divisible by both

Key relationf(x) × g(x) = GCD × LCM

Division methodDivide, then divide the divisor by the remainder, until the remainder is 0

GCD vs LCM by factorisation

GCDFactorise both. Take only the factors common to both, each with its lowest power. For 12x³y² and 18x²y⁴ the GCD is 6x²y².

LCMFactorise both. Take every factor that appears, each with its highest power. For 12x³y² and 18x²y⁴ the LCM is 36x³y⁴.

Check: 12x³y² × 18x²y⁴ = 216x⁵y⁶, and 6x²y² × 36x³y⁴ = 216x⁵y⁶. The relation GCD × LCM = product holds.

Worked example 1: by factorisation

Problem

Find the GCD and LCM of x² − 4 and x² + 5x + 6.

Working

Factorise the first, using a² − b² = (a − b)(a + b): x² − 4 = (x − 2)(x + 2).
Factorise the second: x² + 5x + 6 = (x + 2)(x + 3), since 2 × 3 = 6 and 2 + 3 = 5.
The only common factor is (x + 2). So GCD = x + 2.
All factors with their highest powers: (x − 2), (x + 2), (x + 3). So LCM = (x − 2)(x + 2)(x + 3).
Check: GCD × LCM = (x − 2)(x + 2)²(x + 3), which is the product of the two polynomials.

Answer

GCD = x + 2; LCM = (x − 2)(x + 2)(x + 3)

Division method for the GCD

This works like Euclid's division algorithm for the HCF of two numbers.

1
Write both polynomials in descending powers of x.
2
Divide the polynomial of higher degree by the other. If the degrees are equal, divide either one by the other.
3★
If the remainder is not zero, divide the divisor by the remainder.
4
Repeat until the remainder is zero.
5★
The last non-zero divisor is the GCD. You may drop a number factor of a remainder, like the 7 in 7(x² + x − 2).

★ key step

Worked example 2: by the division method

Problem

Find the GCD of f(x) = x³ + 4x² + x − 6 and g(x) = x³ − 3x² − 6x + 8.

Working

Both have degree 3. Divide f(x) by g(x): the quotient is 1.
Remainder = f(x) − g(x) = 7x² + 7x − 14 = 7(x² + x − 2).
Drop the number 7. Now divide g(x) by x² + x − 2.
x³ ÷ x² = x. Then x(x² + x − 2) = x³ + x² − 2x. Subtract from g(x): −4x² − 4x + 8.
−4x² ÷ x² = −4. Then −4(x² + x − 2) = −4x² − 4x + 8. Subtract: the remainder is 0.
The remainder is zero, so the last divisor is the GCD.

Answer

GCD = x² + x − 2

Worked example 3: find the other polynomial

Problem

The GCD of two polynomials is x − 3 and their LCM is x³ − x² − 14x + 24. One polynomial is x² − 5x + 6. Find the other.

Working

Formula: f(x) × g(x) = GCD × LCM, so g(x) = (GCD × LCM) ÷ f(x).
Factorise f(x): x² − 5x + 6 = (x − 2)(x − 3).
Factorise the LCM: x = 2 gives 8 − 4 − 28 + 24 = 0, so (x − 2) is a factor. The LCM = (x − 2)(x² + x − 12) = (x − 2)(x − 3)(x + 4).
g(x) = (x − 3) × (x − 2)(x − 3)(x + 4) ÷ [(x − 2)(x − 3)].
Cancel (x − 2)(x − 3): g(x) = (x − 3)(x + 4) = x² + x − 12.

Answer

The other polynomial is x² + x − 12

In real life

To pack 48 laddus and 60 murukkus into the largest number of identical boxes, you need their GCD, which is 12. For polynomials, factors play the part of prime numbers.

Exam angle

2 marks: “Find the GCD of 15x⁴y³z⁵ and 12x²y⁷z².” 15 = 3 × 5 and 12 = 2² × 3, so the number part is 3. Lowest powers: x², y³, z². GCD = 3x²y³z².

2 marks: “Find the LCM of x² − 1 and x² + 2x + 1.” x² − 1 = (x − 1)(x + 1) and x² + 2x + 1 = (x + 1)². LCM = (x − 1)(x + 1)².

5 marks: “Find the GCD of x³ + 2x² − x − 2 and x³ − 2x² − 5x + 6 by the division method.” Divide the first by the second: quotient 1, remainder 4x² + 4x − 8 = 4(x² + x − 2). Drop 4. Divide x³ − 2x² − 5x + 6 by x² + x − 2: quotient x − 3, remainder 0. GCD = x² + x − 2.

Remember it

Greatest takes the lowest, least takes the highest: GCD uses the lowest powers of common factors, LCM the highest powers of all factors.