H.C.F. of Polynomials by Factorization
Learn how to solve H.C.F. of polynomials by factorization splitting the middle term.
Solved examples on highest common factor of polynomials by factorization:
1. Find out the H.C.F. of x2 - 3x - 18 and x2 + 5x + 6 by factorization.Solution:
First expression = x2 - 3x - 18
= x2 - 6x + 3x - 18, by splitting the middle term - 3x = - 6x + 3x
= x(x - 6) + 3(x - 6)
= (x - 6) (x + 3)
Second expression = x2 + 5x + 6= x2 + 3x + 2x + 6, by splitting the middle term 5x = 3x + 2x
= x(x + 3) + 2(x + 3)
= (x + 3) (x + 2)
Therefore, in the two polynomials (x + 3) is the only common factors, so, the required H.C.F. = (x + 3).
Solution:
First expression = (2a2 - 8b2)
= 2(a2 - 4b2), by taking common 2
= 2[(a)2 - (2b)2], using the identity of a2 – b2
= 2(a + 2b) (a - 2b), we know a2 – b2 = (a + b) (a – b)
= 2 × (a + 2b) × (a - 2b)
Second expression = (4a2 + 4ab - 24b2)= 4(a2 + ab - 6b2), by taking common 4
= 4(a2 + 3ab - 2ab - 6b2), by splitting the middle term ab = 3ab - 2ab
= 4[a(a + 3b) - 2b(a + 3b)]
= 4(a + 3b) (a - 2b)
= 2 × 2 × (a + 3b) × (a - 2b)
Third expression = (2a2 - 12ab + 16b2)= 2(a2 - 6ab + 8b2), , by taking common 2
= 2(a2 - 4ab - 2ab + 8b2), by splitting the middle term - 6ab = - 4ab - 2ab
= 2[a(a - 4b) - 2b(a - 4b)]
= 2(a - 4b) (a - 2b)
= 2 × (a - 4b) × (a - 2b)
From the above three expressions ‘2’ and ‘(a - 2b)’ are the common factors of the expressions.
Therefore, the required H.C.F. is 2 × (a - 2b) = 2(a - 2b)
8th Grade Math Practice
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