The relation between H.C.F. and L.C.M. of two polynomials is the product of the two polynomials is equal to the product of their H.C.F. and L.C.M.
If p(x) and q(x) are two polynomials, then p(x) ∙ q(x) = {H.C.F. of p(x) and q(x)} x {L.C.M. of p(x) and q(x)}.
= a(a – 7) – 1(a – 7)
= (a – 7) (a – 1)
Therefore, the H.C.F. = (a – 7) and L.C.M. = (a – 7) (a – 5) (a – 1)
Note:
(i) The product of the two expressions is equal to the product of their factors.
(ii) The product of the two expressions is equal to the product of their H.C.F. and L.C.M.
Product of the two expressions = (a^{2} – 12a + 35) (a^{2} – 8a + 7)= (a – 7) (a – 5) (a – 7) (a – 1)
= (a – 7) (a – 7) (a – 5) (a – 1)
= H.C.F. × L.C.M. of the two expressions
= a(a + 9) + 1(a + 9)
= (a + 9) (a + 1)
Therefore, the H.C.F. = (a + 9)
Therefore, L.C.M. = Product of the two expressions/H.C.F.
= \(\frac{(a^{2} + 7a - 18) (a^{2} + 10a + 9)}{(a + 9)}\)
= \(\frac{(a + 9) (a - 2) (a + 9) (a + 1)}{(a + 9)}\)
= (a – 2) (a + 9) (a + 1)
Solution:
According to the problem,
Required Expression = \(\frac{L.C.M. × H.C.F.}{Given expression}\)
= \(\frac{(m^{3} - 10m^{2} + 11x + 70)(x - 7)}{x^{2} - 5x - 14}\)
= \(\frac{(m^{2} - 5m - 14)(x - 5)(x - 7)}{x^{2} - 5x - 14}\)
8th Grade Math Practice
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