Worksheet on H.C.F. and L.C.M. of Polynomials covers various types of questions. Learn the steps to solve the highest common factor and lowest common multiple of polynomials by checking the sample questions covering different models here. Start your practice using the L.C.M and G.C.F of Polynomials Worksheet as many times as possible and get good knowledge on the concept.

You can test your standard levels of preparation using the Worksheet for the least common multiple and greatest common factors of polynomials and plan preparation accordingly. Improve your score in the exam by practicing the different problems from our G.C.F and L.C.M of Polynomials Worksheet. This page also contains questions on the relation between H.C.F and L.C.M of two polynomials.

1. Find the greatest common factor (G.C.F) and Lowest Common Multiple (L.C.M) of two polynomials:

(a) x² – 2x – 35 and x² + 14x + 45

(b) x² + 6x+ 9, x² + 3x

(c) x³ – 9x² + 14x, x³ – 49x

## Solution:

(a) x² – 2x – 35 and x² + 14x + 45

First polynomial = x² – 2x – 35

= x² – 7x + 5x – 35 = x(x – 7) +5(x – 7)

= (x – 7) (x + 5)

Second Polynomial = x² + 14x + 45

= x² + 9x + 5x + 45 = x(x + 9) + 5(x + 9)

= (x + 9) (x + 5)

In both polynomials, the common factor is (x + 5) and extra factors are (x – 7) and (x + 9).

Therefore, required H.C.F = (x + 5)

Required L.C.M = (x + 5) (x – 7) (x + 9).

(b) x² + 6x+ 9, x² + 3x

First polynomial = x² + 6x+ 9

= x² + 3x + 3x + 9 = x(x + 3) + 3(x + 3)

= (x + 3) (x + 3)

Second Polynomial =x² + 3x

= x(x + 3).

In both polynomials, the common factor is (x + 3), extra common factorsare x, (x + 3).

Therefore, the required highest common factor is (x + 3).

Required least common factor is x (x + 3)².

(c) x³ – 9x² + 14x, x³ – 49x

First polynomial = x³ – 9x² + 14x

= x(x² – 9x + 14) = x(x² -7x -2x +14)

= x(x(x – 7) -2(x – 7)) = x (x – 2) (x – 7)

Second Polynomial = x³ – 49x

= x(x² – 49) = x(x² – 7²)

= x(x + 7) (x – 7)

In both polynomials, the common factors are x(x – 7), extra common factors are (x – 2) (x + 7).

Therefore, required G.C.F is (x(x – 7), L.C.M is x(x – 7) (x – 2) (x + 7).

2. Find the least common multiple (L.C.M) and highest common factor (H.C.F) of three polynomials:

(a) x⁶ + 9x⁴y + 27x²y² + 27y³, x²+3y, and x³ + x²y + 3xy + 3y²

(b) 2x² – 14x + 20, 6x² – 2x – 20, 3x² – 7x – 20

(c) a² + 3a – 4, a² + 5a + 4, and a² – 1

## Solution:

(a) x⁶ + 9x⁴y + 27x²y² + 27y³, x² + 3y, and x³ + x²y + 3xy + 3y²

First polynomial = x⁶ + 9x⁴y + 27x²y² + 27y³

According to the (a + b)³ = a³ + b³ + 3a²b + 3ab²

= (x²)³ + (3y)³ + 3 (x²)² (3y) + 3 x² (3y)²

= (x² + 3y)³

Second Polynomial = x²+3y

Third Polynomial = x³ + x²y + 3xy + 3y²

= x²(x + y) + 3y(x + y) = (x + y) (x² + 3y)

In three polynomials, the common factor is (x² + 3y), extra common factors are (x + y), (x² + 3y)².

Therefore, required, greatest common factor is (x² + 3y), lowest common multiple is (x² + 3y)³ (x + y).

(b) 2x² – 14x + 20, 6x² – 2x – 20, 3x² – 7x – 20

First polynomial = 2x² – 14x + 20

= 2x² – 4x – 10x + 20 = 2x(x – 2) – 10(x – 2)

= (2x – 10) (x – 2) = 2(x – 5) (x – 2)

Second Polynomial = 6x² – 2x – 20

= 6x² – 12x + 10x – 20 = 6x(x – 2) + 10(x – 2)

= (6x + 10) (x – 2) = 2(3x + 5) (x – 2)

Third Polynomial = 3x² – 7x – 20

= 3x² -12x + 5x – 20 = 3x(x – 4) +5 (x – 4)

= (3x + 5) (x – 4)

In three polynomials, the common factors between first and second polynomial is 2(x – 2), second and third polynomial is (3x + 5), extra common factors are (x – 4) (x – 5).

Therefore, required H.C.F is 1, required L.C.M is 2(x – 2) (3x + 5) (x – 4) (x – 5).

(c) a² + 3a – 4, a² + 5a + 4, and a² – 1

First polynomial = a² + 3a – 4

= a² + 4a – a – 4 = a(a + 4) -1(a + 4)

= (a + 4) (a – 1)

Second Polynomial = a² + 5a + 4

= a² + 4a + a + 4 = a(a + 4) +1(a + 4)

= (a + 1) (a + 4)

Third Polynomial = a² – 1

= a² – 1² = (a + 1) (a – 1)

In three polynomials, the common factors between first polynomial and second polynomial is (a + 4), the second and third polynomial is (a + 1), first and the third polynomial is (a + 1).

Therefore required H.C.F = 1, L.C.M = (a + 4) (a + 1) (a – 1)

3. Find H.C.F and L.C.M of the below-included polynomials:

(a) pq – np, pq – mq, q² – 3nq + 2n², pq – 2np – mq + 2mn, and pq – np – mq + mn

(b) x³ – 7x + 6, x³ + 2x² – 7x + 4

## Solution:

(a) pq – np, pq – mq, q² – 3nq + 2n², pq – 2np – mq + 2mn, and pq – np – mq + mn

First polynomial = pq – np

= p(q – n)

Second Polynomial = pq – mq

= q(p – m)

Third Polynomial = q² – 3nq + 2n²

= q² – 2nq – nq + 2n² = q(q – 2n) -n (q – 2n)

= (q – 2n) (q – n)

Fourth Polynomial = pq – 2np – mq + 2mn

= p(q – 2n) – m(q – 2n)

= (p – m) (q – 2n)

Fifth Polynomial = pq – np – mq + mn

= p(q – n) – m(q – n) = (p-m) (q-n)

In all polynomials, the common factors are zero.

Therefore, required lowest common multiple is pq(p – m) (q – n) (q – 2n), G.C.F is 1.

(b) x³ – 7x + 6, x³ + 2x² – 7x + 4

First Polynomial = x³ – 7x + 6

= (x – 2) (x + 3) (x – 1)

Second polynomial = x³ + 2x² – 7x + 4

= (x + 4) (x² – 2x + 1)

= (x + 4) (x – 1) (x – 1)

In both polynomials, the common factor is (x – 1), the extra common factors are (x – 1) (x + 4) ( x + 3) (x – 2).

Therefore, required L.C.M is (x – 1)² (x + 4) ( x + 3) (x – 2), H.C.F is (x – 1).

4. Find the LCM and GCD of (a² – 1), (a – 1) and verify that f(x) x g(x) = LCM x GCD.

## Solution:

Given that,

f(x) = (a² – 1) = (a² – 1²)

= (a + 1) (a – 1)

g(x) = (a – 1).

L.C.M of f(x), g(x) = (a – 1) (a + 1)

H.C.F of f(x), g(x) = (a – 1)

f(x) x g(x) = (a + 1) (a – 1) (a – 1)

LCM x GCD = (a – 1) (a + 1) (a – 1)

∴ Hence proved.

5. The product of two polynomials is a⁴ +53a³ + 957a² + 6435a +9450, one of them is a²+36a +315, and their greatest common factor is (a + 15). Find L.C.M, second polynomial.

## Solution:

Let f(a) = a²+ 36a +315

f(a) x g(a) = a⁴ +53a³ + 957a² + 6435a +9450

G.C.F of f(a), g(a) is (a + 15)

Factorize f(a)

= a²+ 36a +315 = a²+ 15a + 21a + 315

= a(a + 15) + 21(a + 15) = (a + 15) (a + 21)

f(a) x g(a) = (a + 15) (a + 21) (a + 15) * x = a⁴ +53a³ + 957a² + 6435a +9450

Factors of a⁴ +53a³ + 957a² + 6435a +9450 = (a + 15) (a + 21) (a + 15) (a + 2)

Therefore , g(a) = (a + 15) (a + 2)= a² + 17a + 30

L.C.M of polynomials = (a + 15) (a + 2) (a + 21).

6. If L.C.M and H.C.F of two unknown polynomials p(x), q(x) are x³ – 4x² – 19x – 14, (x – 7), find the polynomials.

## Solution:

Given that,

L.C.M of polynomials = x³ – 4x² – 19x – 14

H.C.F of polynomials = (x – 7)

Factorize the least common multiple,

= x³ – 4x² – 19x – 14 = (x – 7) (x + 1) (x + 2)

p(x) = (x – 7) (x + 1)

= x² – 7x + x – 7 = x² – 6x – 7

q(x) = (x – 7) (x + 2)

= x² – 7x + 2x – 14 = x² – 5x – 14

7. Find the LCM of the 2x² + 18x + 3xy + 27y, 6x² + 14x + 9xy + 21y polynomials whose GCD is (2x + 3y).

## Solution:

Given two polynomials are

f = 2x² + 18x + 3xy + 27y, g = 6x² + 14x + 9xy + 21y

The greatest common divisor of polynomials is (2x + 3y)

The relation between H.C.F and L.C.M of two polynomials says that the product of polynomials is equal to the product of their L.C.M and H.C.F

(2x² + 18x + 3xy + 27y) (6x² + 14x + 9xy + 21y) = (2x + 3y) * L.C.M

L.C.M = [(2x² + 18x + 3xy + 27y) (6x² + 14x + 9xy + 21y)] / (2x + 3y)

= [(2x + 3y) (x + 9y) (2x + 3y) (3x + 7y)] / (2x + 3y)

= (2x + 3y) (x + 9y) (3x + 7y) = 6x³ + 77x²y + 228xy² + 189y³.