Lowest Common Multiple of Polynomials
How
to find the lowest common multiple of polynomials?
To find the lowest common multiple (L.C.M.) of
polynomials, we first find the factors of polynomials by the method of
factorization and then adopt the same process of finding L.C.M.
Solved
examples to find lowest common factor of polynomials:
1. Find the L.C.M. of 4a
^{2} - 25b
^{2} and 6a
^{2} + 15ab.
Solution:
Factorizing 4a
^{2} - 25b
^{2} we get,
(2a)
^{2} - (5b)
^{2}, by using the identity a
^{2} - b
^{2}.
= (2a + 5b) (2a - 5b)
Also, factorizing 6a
^{2} + 15ab by taking the common factor '3a', we get
= 3a(2a + 5b)
Therefore, the L.C.M. of 4a
^{2} - 25b
^{2} and 6a
^{2} + 15ab is 3a(2a + 5b) (2a - 5b)
2. Find the L.C.M. of x
^{2}y
^{2} - x
^{2} and xy
^{2} - 2xy - 3x.
Solution:
Factorizing x
^{2}y
^{2} - x
^{2} by taking the common factor 'x
^{2}' we get,
x
^{2}(y
^{2} - 1)
Now by using the identity a
^{2} - b
^{2}.
x
^{2}(y
^{2} - 1
^{2})
= x
^{2}(y + 1) (y - 1)
Also, factorizing xy
^{2} - 2xy - 3x by taking the common factor 'x' we get,
x(y
^{2} - 2y - 3)
= x(y
^{2} - 3y + y - 3)
= x[y(y - 3) + 1(y - 3)]
= x(y - 3) (y + 1)
Therefore, the L.C.M. of x
^{2}y
^{2} - x
^{2} and xy
^{2} - 2xy - 3x is x
^{2}(y + 1) (y - 1) (y - 3).
3. Find the L.C.M. of x
^{2} + xy, xz + yz and x
^{2} + 2xy + y
^{2}.
Solution:
Factorizing x
^{2} + xy by taking the common factor 'x', we get
x(x + y)
Factorizing xz + yz by taking the common factor 'z', we get
z(x + y)
Factorizing x
^{2} + 2xy + y
^{2} by using the identity (a + b)
^{2}, we get
= (x)
^{2} + 2 (x) (y) + (y)
^{2}
= (x + y)
^{2}
= (x + y) (x + y)
Therefore, the L.C.M. of x
^{2} + xy, xz + yz and x
^{2} + 2xy + y
^{2} is xz(x + y) (x + y).
`
8th Grade Math Practice
From Lowest Common Multiple of Polynomials to HOME PAGE
Didn't find what you were looking for? Or want to know more information
about Math Only Math.
Use this Google Search to find what you need.
New! Comments
Have your say about what you just read! Leave me a comment in the box below. Ask a Question or Answer a Question.