find the hcf and lcm of a2 b4 c and a4 b5 c7 Please i need step by step explanation
Answers
Full explanation
The following method is suggested for finding the HCF of several polynomials
(a) Write each polynomial as a product of prime factors.
(b) The HCF is the product obtained by taking each factor to the lowest power to which it occurs in both the polynomials. For example,
The HCF of 23 32 (x - y)3(x + 2y)2; 2233 (x - y)2 (x + 2y)3 and 32(x - y)2 (x + 2y) is 32(x - y)2 (x + 2y).
Note- Two or more polynomials are relatively prime if their HCF is 1.
LCM of Polynomials
The LCM of two or more given polynomials is the polynomial of lowest degree and smallest numerical coefficients for which each of the given polynomials will be its factor.
Method to find LCM of polynomial
The following procedure is suggested for determining the LCM of several polynomials.
(a) Write each polynomial as a product of prime factors.
(b) The LCM is the product obtained by taking each factor to the highest power to which it occurs. For example, The LCM of 2332 (x — y)3(x + 2y)2; 2233(x — y)2 (x + 2y)3; 32(x — y)2(x + 2y) is 2333 (X — y)3 (x + 2y)3.
Example1. Find the HCF and LCM of (I) 9X4y2 and 12X3y3
(ii) 6x — 6y and 4x2 — 4y2.
Sol. (i) 9x4y2 = 32 x x4 x y2 and
12X3y3 = 22x 3 x X3 x y3.
HCF = 3x3y2 and
LCM = 32 x 22 X x4 X y3 =36X4y3
So, HCF = 3x3y2, LCM = 36x4y3
(ii) 6x — 6y = 2 x 3 x (x — y) and
4x2 — 4y2 = 22(x2— y2) = 22 x (x + y) (x — y) HCF = 2(x — y) and
LCM = 22x 3 x (x — y)(x + y) = 12(x — y) (x + y). So, HCF = 2(x — y), LCM = 12(x + y) (x — y).
Special Products
The following are some of the products which occur frequently in mathematics and the student should become familiar with them as soon as possible.
Product of a monomial and a binomial a (c + d) = ac + ad.
Product of the sum and the difference of two terms (a + b)(a — b) = a2 — b2.
Square of a binomial
(a + b)2 = a2 + 2ab + b2.
(a — b)2 = a2 — 2ab + b2
Product of two binomials
(x + a)(x + b) = x2 + (a + b)x + ab.
(ax + b)(cx + d) = acx2 + (ad + bc)x + bd. ,(a+b)(c+d)=ac+bc+ad+bd.
Cube of a binomial
(a + b)3 = a3 + 3a2b + 3ab2 + b3.
(a — b)3 = a3 — 3a2b + 3ab2 — b3.
Square of a trinomial
(a + b + c)2 = a2 + b2 + c2 + 2(ab + ac + bc).
Products that give answers of the form a n ± bn.
It may be verified by multiplication that (a — b)(a2 + ab + b2) = a3 — b3
(a — b)(a3 + a2b + ab2 + b3) = a4 — b4
(a — b)(a4 + a3b + a2b2 + ab3 + b4) = a5 — b5
Answer:
From the given two expression
Highest power of a = 4
Highest power of b = 5
Highest power of c = 7
So LCM = a^4b^5c^7
From the given two expression
Lowest power of a = 2
Lowest power of b = 4
Lowest power of c = 1
So HCF = a^2b^4c