Are you preparing for exams, here you can get an idea on the solving least common multiple of any number of polynomials? In this Worksheet on L.C.M. of Polynomials, we can see the detailed step by step solution for every question. That helps students to prepare the topic easily. The simple way to calculate the lowest common multiple of polynomials is by finding the factors of all polynomials. And multiply the common factors and extra factors to get the L.C.M of Polynomials.
1. Find the L.C.M of Two Polynomials:
(a) x² – 1, x² + 2x + 1
(b) x+8, x+2
(c) y² – 2², (y + 2)²
(d) p², p + 3
Solution:
(a) x² – 1, x² + 2x + 1
First polynomial = x² – 1
= x² – 1² = (x – 1) ( x + 1)
Second polynomial = x² + 2x + 1
= x² + x + x + 1, by splitting the middle term
= x(x + 1) + 1(x + 1) = (x + 1) (x + 1)
In both the polynomials, the common factors are (x+1), extra common factors are (x + 1) (x – 1).
Therefore, the required L.C.M. = (x + 1) (x + 1) (x – 1) =
(b) x+8, x+2
First polynomial = x + 8
Second polynomial = x + 2
There are no factors for the given polynomials.
Therefore, required lowest common multiple = (x + 8) (x + 2).
(c) y² – 2², (y + 2)²
First polynomial = y² – 2²
= (y – 2) (y + 2)
Second polynomial = (y + 2)²
= (y + 2) (y + 2)
In both the polynomials, the common factors are (y + 2), the extra common factor of the first polynomial (y – 2), the second polynomial is (y + 2).
Therefore, required the least common multiple = (y + 2) (y + 2) (y – 2)
(d) p², p + 3
First polynomial = p² = p * p
Second polynomial = p + 3
There are no factors for the given polynomials.
Therefore, required L.C.M = p * p (p + 3)
2. Find the Lowest Common Multiple (L.C.M) of Three Polynomials:
(a) x³ – y³, x² – xy + y², x² – 2xy + y²
(b) m(m – 1), m², (m – 1)²
(c) x² – 1, x² + 2x + 1, (x -1) (x + 2)
(d) x² + 8x + 12, x² + 2x – 24, and x² + 15x + 54
Solution:
(a) x³ – y³, x² – xy + y², x² – 2xy + y²
First Polynomial = x³ – y³
= (x – y)(x² + xy + y²)
Second Polynomial = x² – xy + y²
Third Polynomial = x² – 2xy + y²
= (x – y)²
There are no common factors in the polynomials.
Therefore, the Lowest common multiple of given polynomials is its product.
Hence, required L.C.M = (x – y)²(x² + xy + y²)
(b) m(m – 1), m², (m – 1)²
First Polynomial = m(m – 1)
Second Polynomial = m * m
Third Polynomial = (m – 1) (m – 1)
Least common multiple of m(m – 1), m², (m – 1)² = m² (m – 1)².
(c) x² – 1, x² + 2x + 1, (x -1) (x + 2)
First polynomial = x² – 1
= x² – 1² = (x + 1) (x – 1)
Second polynomial = x² + 2x + 1
= x² + x + x + 1 = x(x + 1) + 1(x + 1)
= (x+1) (x+1)
Third Polynomial = (x -1) (x + 2)
Therefore, required L.C.M = (x + 1) (x – 1) (x+1) (x + 2) = (x – 1) (x+1)² (x + 2)
(d) x² + 8x + 12, x² + 2x – 24, and x² + 15x + 54
First Polynomial = x² + 8x + 12
= x² + 6x + 2x + 12 = x(x + 6) + 2(x + 6)
= (x+2) (x+6)
Second Polynomial = x² + 2x – 24
= x² + 6x – 4x – 24 = x(x + 6) – 4(x + 6)
= (x-4)(x+6)
Third polynomial = x² + 15x + 54
= x² + 9x + 6x + 54 = x(x + 9) + 6(x + 9)
= (x+9) (x+6)
The common factor is (x+6), extra common factors are (x+2), (x-4), (x+9).
Therefore, the required least common factor = (x+6) (x+2) (x-4) (x+9).
3. Find the Least Common Multiple (L.C.M) of Polynomials:
(a) x² + 15x + 56, x² + 5x – 24, and x² + 8x
(b) x³ + 3x² + 4x + 12, x⁴ + x³ + 4x² + 4x
(c) 4(m² – 36), 6(m² + 4m – 12) and 12(m² – 12m + 36)
Solution:
(a) x² + 15x + 56, x² + 5x – 24, and x² + 8x
First Polynomial = x² + 15x + 56
= x² + 8x + 7x + 56 = x(x + 8) + 7(x + 8)
= (x + 7) (x + 8)
Second Polynomial = x² + 5x – 24
= x² + 8x – 3x – 24 = x(x + 8) – 3(x + 8)
= (x – 3) (x + 8)
Third Polynomial = x² + 8x
= x(x + 8)
The common factor is (x +8), extra common factors are x, (x – 3), (x + 7)
Therefore, the required L.C.M is x(x + 8) (x – 3) (x + 7).
(b) x³ + 3x² + 4x + 12, x⁴ + x³ + 4x² + 4x
First polynomial = x³ + 3x² + 4x + 12
= x²(x + 3) + 4(x + 3)
= (x + 3) (x² + 4)
Second Polynomial = x⁴ + x³ + 4x² + 4x
= x(x³ + x² + 4x + 4)
= x(x²(x + 1) + 4(x + 1)) = x(x² + 4) (x + 1)
The common factor is (x² + 4), extra common factors are x(x +1), (x + 3).
Therefore, required L.C.M is x(x² + 4) (x + 1) (x + 3).
(c) 4(m² – 36), 6(m² + 4m – 12) and 12(m² – 12m + 36)
First Polynomial = 4(m² – 36)
= 4(m² – 6²) = 4(m + 6) (m – 6)
Second polynomial = 6(m² + 4m – 12)
= 6(m² + 6m – 2m -12) = 6(m(m + 6) – 2(m + 6))
= 6(m – 2)(m + 6)
Third Polynomial = 12(m² – 12m + 36)
= 12(m² – 6m – 6m + 36) = 12(m(m – 6) – 6(m – 6))
= 12(m – 6) (m – 6)
Therefore, the required L.C.M is 24 (m – 6)² (m + 6) (m – 2).