find the hcf and lcm of the polinomial3(x2 + 2x - 8), 6(x2 + 9x + 20) and 15(x2 + x - 12)
Answers
Answer:
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)}.
1. Find the H.C.F. and L.C.M. of the expressions a2 – 12a + 35 and a2 – 8a + 7 by factorization.
Solution:
First expression = a2 – 12a + 35
= a2 – 7a – 5a + 35
= a(a – 7) – 5(a – 7)
= (a – 7) (a – 5)
Second expression = a2 – 8a + 7
= a2 – 7a – a + 7
= 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 = (a2 – 12a + 35) (a2 – 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
2. Find the L.C.M. of the two expressions a2 + 7a – 18, a2 + 10a + 9 with the help of their H.C.F.
Solution:
First expression = a2 + 7a – 18
= a2 + 9a – 2a – 18
= a(a + 9) – 2(a + 9)
= (a + 9) (a – 2)
Second expression = a2 + 10a + 9
= a2 + 9a + a + 9
= 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.
= (a2+7a−18)(a2+10a+9)(a+9)(a2+7a−18)(a2+10a+9)(a+9)
= (a+9)(a−2)(a+9)(a+1)(a+9)(a+9)(a−2)(a+9)(a+1)(a+9)
= (a – 2) (a + 9) (a + 1)
3. m2 – 5m -14 is an expression. Find out another similar expression such that their H.C.F. is (m – 7) and L.C.M. is m3 – 10m2 + 11m + 70.
Solution:
According to the problem,
Required Expression = L.C.M.×H.C.F.GivenexpressionL.C.M.×H.C.F.Givenexpression
= (m3−10m2+11x+70)(x−7)x2−5x−14(m3−10m2+11x+70)(x−7)x2−5x−14
= (m
Therefore, the required expression = m2 – 12m + 35
Answer:
Step-by-step explanation: