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Find the H.C.F of numerical coefficients and literal coefficients. And get the product of the highest common factor of numerical and literal coefficients to get the answer.

1. Find the Highest Common factor (H.C.F) of two monomials:

(a) 56a²bc, 24a²b²c²

(b) 45u²v, 60uv³

(c) 99x⁵y²z³, 66xy²z²

(d) 29d⁸, 18c

(e) 25a², 15ab

(f) 10s, 2s

(g) 36p³q⁵r, 6p²q³r⁵

(h) 84t²s, 28tu

## Solution:

(a) 56a²bc, 24ab²c²

The H.C.F. of numerical coefficients = The H.C.F. of 24 and 56.

Since, 24 = 2 * 2 * 2 * 3, 56 = 2 * 2 * 2 * 7

Therefore, the H.C.F. of 24 and 56 = 2³ = 8

The H.C.F. of literal coefficients = The H.C.F. of a²bc, ab²c² = abc

Since, in a²bc and ab²c², a, b and c are common.

The lowest power of a is a.

The lowest power of b is b.

The lowest power of c is c.

Therefore, the H.C.F. of a²bc, ab²c² is abc.

Thus, the H.C.F. of 56a²bc, 24ab²c²

= The H.C.F. of numerical coefficients × The H.C.F. of literal coefficients

= 8 × (abc)

= 8abc.

(b) 45u²v, 60uv³

The G.C.F of numerical coefficients = The G.C.F of 45, 60.

Since, 45 = 5 * 3 * 3, 60 = 3 * 2 * 5 * 2

Therefore, H.C.F of 45, 60 is 5 * 3 = 15

The H.C.F. of literal coefficients = The H.C.F. of u²v and uv³ = u²v

Since, in u²v, uv³ u, v are common

The lowest power of u is u²

The lowest power of v is v.

Thus, the G.C.F of 45u²v, 60uv³ = The H.C.F. of numerical coefficients × The H.C.F. of literal coefficients

= 15 x u²v

= 15u²v

(c) 99x⁵y²z³, 66xy²z²

The H.C.F of numerical coefficients = The H.C.F. of 99 and 66.

Since, 99 = 3 * 3 * 11, 66 = 3 * 2 * 11

The G.C.F. of 99 and 66 = 3 * 11 = 33

The H.C.F. of literal coefficients = The H.C.F. of x⁵y²z³, xy²z² = xy²z²

The common variables are x, y, z

The lowest power of x is x.

The lowest power of y is y².

The lowest power of z is z³.

Thus, the G.C.F of 99x⁵y²z³, 66xy²z² = 33 * xy²z² = 33xy²z²

(d) 29d⁸, 18c

The H.C.F of numerical coefficients = The H.C.F. of 29, 18

Since, 29 = 3 * 3 * 3, 18 = 2 * 3 * 3

The H.C.F. of 29, 18 = 9

The H.C.F. of d⁸, c = 1.

There are no common variables.

Thus, the G.C.F of 29d⁸, 18c = The H.C.F. of numerical coefficients × The H.C.F. of literal coefficients = 9 x 1

Therefore, G.C.F of 29d⁸, 18c = 9.

(e) 25a², 15ab

The highest common factor of numerical coefficients = The H.C.F. of 25, 15

Since, 25 = 5 * 5, 15 = 3 * 5

The H.C.F. of 25, 15 = 5

The H.C.F. of literal coefficients = The G.C.F of a², b = a.

The only common variable is a.

The lowest power of a is a.

Thus, the highest common factor of 25a², 15ab = The H.C.F. of numerical coefficients × The H.C.F. of literal coefficients

= 5 * a = 5a.

(f) 10s, 2s

The greatest common factor of numerical coefficients = The G.C.F of 10, 2

Since, 10 = 2 * 5, 2 = 2 * 1

The G.C.F of 10, 2 = 2

The H.C.F. of literal coefficients = The highest common factor of s, s is s.

The common variable is s.

The lowest power of s is s.

Thus, the G.C.F of 10s, 2s = The H.C.F. of numerical coefficients × The H.C.F. of literal coefficients

= 2 * s = 2s.

(g) 36p³q⁵r, 6p²q³r⁵

The highest common factor of numerical coefficients = The H.C.F of 36, 6

Since, 36 = 6 * 6, 6 = 1 * 6

The H.C.F of 36, 6 = 6

The H.C.F. of literal coefficients = The greatest common factor of p³q⁵r, p²q³r⁵ = p²q³r

The common variables are p, q, r

The lowest power of p is p².

The lowest power of q is q³.

The lowest power of r is r.

Thus, the H.C.F of 36p³q⁵r, 6p²q³r⁵ = The G.C.F. of numerical coefficients × The G.C.F. of literal coefficients

= 6 * p²q³r = 6p²q³r

(h) 84t²s, 28tu

The G.C.F. of numerical coefficients = The greatest common factor of 84, 28

Since, 84 = 2 * 2 * 3 * 7, 28 = 2 * 2 * 7

The H.C.F of 84, 28 = 14

The H.C.F. of literal coefficients = The highest common factor of t²s, tu = t

The common variable is t.

The lowest power of t is t.

Thus, the G.C.F of 84t²s, 28tu = The H.C.F of numerical coefficients x The G.C.F. of literal coefficients

= 14 * t = 14t

2. Find the greatest common factor (G.C.F.) of the three monomials:

(a) 35p²q³r, 42pq²r³ and 14p³qr²

(b) 49xy⁴z², 21xy², 7x²z⁴

(c) mno², mn, m²n

(d) 5abc, 25a²b²c², 50ac³

(e) m³n³a³, man, a²m²n²

(f) 6xy, 14x²y, 42xy²

(g) x²y³, x⁵, yz

(h) 8abc, 20abc, 28abc

## Solution:

(a) 35p²q³r, 42pq²r³ and 14p³qr²

The H.C.F of numerical coefficients = The H.C.F of 35, 42, 14

Since, 35 = 7 * 5, 42 = 2 * 3 * 7, 14 = 2 * 7

Therefore, the H.C.F. of 35, 42, 14 is 7

The highest common factor of literal coefficients = The G.C.F of p²q³r, pq²r³ and p³qr² = pqr

The common variables are p, q, r

The lowest power of p is p, the lowest power of q is q, the lowest power of r is r.

Thus, the H.C.F of 35p²q³r, 42pq²r³ and 14p³qr² = H.C.F of numerical coefficient x H.C.F of literal coefficients

= 7 x pqr = 7pqr.

(b) 49xy⁴z², 21xy², 7x²z⁴

The H.C.F of numerical coefficients = The H.C.F of 49, 21, 7

Since, 49 = 7 * 7, 21=7*3, 7=1*7

Therefore, the G.C.F of 49, 21, 7 is 7.

The greatest common factor of literal coefficients = The H.C.F of xy⁴z², xy², x²z⁴ = x

Thus, G.C.F of 49xy⁴z², 21xy², 7x²z⁴ = The H.C.F of numerical coefficients * literal coefficients

= 7 * x =7x.

(c) mno², mn, m²n

The H.C.F of literal coefficients = The greatest common factor of mno², mn, m²n = mn

The common variables are m, n.

The lowest power of m is m, n is n.

Thus, H.C.F of mno², mn, m²n = The H.C.F of numerical coefficients x The G.C.F. of literal coefficients

= 1xmn = mn.

(d) 5abc, 25a²b²c², 50ac³

The G.C.F of numerical coefficients = The H.C.F of 5, 25, 50

since , 5=1*5, 25=5*5, 50=5*5*2

Therefore, the H.C.F. of 5, 25, 50 = 5

The G.C.F of literal coefficients = The H.C.F of abc, a²b²c², ac³ = ac

The common variables are a, c.

Thus, the G.C.F of 5abc, 25a²b²c², 50ac³ = H.C.F of numerical coefficients x H.C.F of literal coefficients

= 5 x ac = 5ac.

(e) m³n³a³, man, a²m²n²

The greatest common factor of literal coefficients = The H.C.F of m³n³a³, man, a²m²n² = amn

The common variables are a, m, n.

The H.C.F of m³n³a³, man, a²m²n² = H.C.F of numerical coefficients x H.C.F of literal coefficients

= 1 x amn = amn.

(f) 6xy, 14x²y, 42xy²

The highest common factor of numerical coefficients = The H.C.F of 6, 14, 42

Since, 6=2*3, 14=7*2, 42=2*7*3

Thus, H.C.F of 6, 14, 42 = 2

The H.C.F of literal coefficients = G.C.F of xy, x²y, xy² = xy

Therefore, G.C.F of 6xy, 14x²y, 42xy² = H.C.F of numerical coefficients x H.C.F of literal coefficients

= 2 x xy = 2xy.

(g) x²y³, x⁵, yz

The G.C.F of literal coefficients = H.C.F of x²y³, x⁵, yz = 1.

There are no common variables.

Thus, H.C.f of x²y³, x⁵, yz = H.C.F of numerical coefficients x H.C.F of literal coefficients= 1* 1 = 1.

(h) 8abc, 20abc, 28abc

The G.C.F of numerical coefficients = The H.C.F of 8, 20, and 28 = 2

Since, 8=2*2*2, 20=10*2, 28=2*7*2

H.C.F of literal coefficients = G.C.F of abc, abc, abc = abc

Thus G.C.F of 8abc, 20abc, 28abc = H.C.F of numerical coefficients x H.C.F of literal coefficients = 2*abc = 2abc.