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1. Write all the possible factors of

(a) 4a

(b) 10ab

(c) 12mn^{2
}(d) 24m^{2}n

(e) 5ab^{2}c^{2}

## Solution:

(a) Given expression is 4a.

The possible factors of 4 are 1, 2, and 4.

The possible factors of a is a.

All possible factors of 4a are 1, a, 2, 2a, 4, 4a.

(b) Given expression is 10ab.

The possible factors of 10 are 1, 2, 5, and 10.

The possible factors of ab is a, b, and ab.

All possible factors of 10ab are 1, 2, 5, and 10, a, b, ab, 2a, 2b, 2ab, 5a, 5b, 5ab, 10a, 10b, 10a.

(c) Given expression is 12mn^{2}.

The possible factors of 12 are 1, 2, 3, 4, 6, and 12.

The possible factors of mn^{2 }are m, n, n^{2}, mn^{2}, mn.

All possible factors of 12mn^{2} are 1, 2, 3, 4, 6, 12, m, n, n^{2}, mn^{2}, mn, 2m, 2n, 2n^{2}, 2mn, 2mn^{2}, 3m, 3n, 3n^{2}, 3mn, 3mn^{2}, 4m, 4n, 4n^{2}, 4mn, 4mn^{2}, 6m, 6n, 6n^{2}, 6mn, 6mn^{2}, 12m, 12n, 12n^{2}, 12mn and 12mn^{2}

(d) Given expression is 24m^{2}n.

The possible factors of 24 are 1, 2, 3, 4, 6, 8, 12, and 24.

The possible factors of m^{2}n^{ }are m, m^{2}, n, mn, m^{2}n.

All possible factors of 24m^{2}n are 1, 2, 3, 4, 6, 8, 12, 24, m, m^{2}, n, mn, m^{2}n, 2m, 2m^{2}, 2n, 2mn, 2m^{2}n, 3m, 3m^{2}, 3n, 3mn, 3m^{2}n, 4m, 4m^{2}, 4n, 4mn, 4m^{2}n, 6m, 6m^{2}, 6n, 6mn, 6m^{2}n, 8m, 8m^{2}, 8n, 8mn, 8m^{2}n, 12m, 12m^{2}, 12n, 12mn, 12m^{2}n, 24m, 24m^{2}, 24n, 24mn, 24m^{2}n.

(e) Given expression is 5ab^{2}c^{2}.

The possible factors of 5 are 1, and 5.

The possible factors of ab^{2}c^{2}^{ }are a, b, b^{2}, c, c^{2}, ab, ab^{2}, bc, bc^{2}, ac, ac^{2}, b^{2}c^{2}.

All possible factors of 5ab^{2}c^{2} are 1, 5, a, b, b^{2}, c, c^{2}, ab, ab^{2}, bc, bc^{2}, ac, ac^{2}, b^{2}c^{2}, 5a, 5b, 5b^{2}, 5c, 5c^{2}, 5ab, 5ab^{2}, 5bc, 5bc^{2}, 5ac, 5ac^{2}, 5b^{2}c^{2}.

2. Find the H.C.F. of the monomials

(a) 4a^{2} and 10ab

(b) 21m^{2}n and 49mn^{2
}(c) a^{3}b^{2} and – 5b^{2
}(d) 4a^{3}, 6b^{2} and 8c

(e) 2a^{2}b^{3}, 10a^{3}b^{2} and 14ab

(f) 5m^{3}, -15m^{2} and 45m

## Solution:

(a) Given that 4a^{2} and 10ab

Firstly, find the HCF of given terms.

HCF of their numerical coefficients 4 and 10 is 2.

HCF of literal coefficients:

The lowest power of a is 1.

The lowest power of b is 0.

Therefore, the HCF of literal coefficients is a.

HCF of two or more monomials = (HCF of their numerical coefficients) × (HCF of their literal coefficients)

HCF of 4a^{2} and 10ab is 2a.

The final answer is 2a.

(b) Given that 21m^{2}n and 49mn^{2}

Firstly, find the HCF of given terms.

HCF of their numerical coefficients 21 and 49 is 7.

HCF of literal coefficients:

The lowest power of m is 1.

The lowest power of n is 1.

Therefore, the HCF of literal coefficients is a.

HCF of two or more monomials = (HCF of their numerical coefficients) × (HCF of their literal coefficients)

HCF of 21m^{2}n and 49mn^{2} is 7mn.

The final answer is 7mn.

(c) Given that a^{3}b^{2} and – 5b^{2}

Firstly, find the HCF of given terms.

HCF of their numerical coefficients 1 and -5 is 1.

HCF of literal coefficients:

The lowest power of a is 0.

The lowest power of b is 2.

Therefore, the HCF of literal coefficients is b^{2}.

HCF of two or more monomials = (HCF of their numerical coefficients) × (HCF of their literal coefficients)

HCF of a^{3}b^{2} and – 5b^{2} is b^{2}.

The final answer is b^{2}.

(d) Given that 4a^{3}, 6b^{2} and 8c

Firstly, find the HCF of given terms.

HCF of their numerical coefficients 4, 6, and 8 is 2.

HCF of literal coefficients:

The lowest power of a is 0.

The lowest power of b is 0.

The lowest power of c is 0.

Therefore, the HCF of literal coefficients is 0.

HCF of two or more monomials = (HCF of their numerical coefficients) × (HCF of their literal coefficients)

HCF of 4a^{3}, 6b^{2} and 8c is 2.

The final answer is 2.

(e) Given that 2a^{2}b^{3}, 10a^{3}b^{2} and 14ab

Firstly, find the HCF of given terms.

HCF of their numerical coefficients 2, 10, and 14 is 2.

HCF of literal coefficients:

The lowest power of a is 1.

The lowest power of b is 1.

Therefore, the HCF of literal coefficients is ab.

HCF of two or more monomials = (HCF of their numerical coefficients) × (HCF of their literal coefficients)

HCF of 2a^{2}b^{3}, 10a^{3}b^{2} and 14ab is 2ab.

The final answer is 2ab.

(f) Given that 5m^{3}, -15m^{2,} and 45m

Firstly, find the HCF of given terms.

HCF of their numerical coefficients 5, -15, and 45 is 5.

HCF of literal coefficients:

The lowest power of m is 1.

Therefore, the HCF of literal coefficients is m.

HCF of two or more monomials = (HCF of their numerical coefficients) × (HCF of their literal coefficients)

HCF of 5m^{3}, -15m^{2,} and 45m is 5m.

The final answer is 5m.

3. Find the common factors of the given terms.

(a) 3a^{2}, 15ab

(b) 33a^{2}b, – 77ab^{2
}(c) 9a^{2}b^{2}c, 54abc^{2
}(d) a^{3}b^{2}, – 5b^{2
}(e) 4a^{3}, 10b^{3}, 6c^{3
}(f) 2abc^{3}, 3a^{2}b^{2}c, 5abc

## Solution:

(a) Given that 3a^{2}, 15ab

Firstly, find the HCF of given terms.

HCF of their numerical coefficients 3 and 15 is 3.

HCF of literal coefficients:

The lowest power of a is 1.

The lowest power of b is 0.

Therefore, the HCF of literal coefficients is a.

HCF of two or more monomials = (HCF of their numerical coefficients) × (HCF of their literal coefficients)

HCF of 3a^{2 }and 15ab is 3a.

The final answer is 3a.

(b) Given that 33a^{2}b, – 77ab^{2}

Firstly, find the HCF of given terms.

HCF of their numerical coefficients 33 and -77 is 11.

HCF of literal coefficients:

The lowest power of a is 1.

The lowest power of b is 1.

Therefore, the HCF of literal coefficients is ab.

HCF of two or more monomials = (HCF of their numerical coefficients) × (HCF of their literal coefficients)

HCF of 33a^{2}b, and – 77ab^{2} is 11ab.

The final answer is 11ab.

(c) Given that 9a^{2}b^{2}c, 54abc^{2}

Firstly, find the HCF of given terms.

HCF of their numerical coefficients 9 and 54 is 9.

HCF of literal coefficients:

The lowest power of a is 1.

The lowest power of b is 1.

The lowest power of c is 1.

Therefore, the HCF of literal coefficients is abc.

HCF of two or more monomials = (HCF of their numerical coefficients) × (HCF of their literal coefficients)

HCF of 9a^{2}b^{2}c and 54abc^{2} is 9abc.

The final answer is 9abc.

(d) Given that a^{3}b^{2}, – 5b^{2}

Firstly, find the HCF of given terms.

HCF of their numerical coefficients 1 and -5 is 1.

HCF of literal coefficients:

The lowest power of a is 0.

The lowest power of b is 2.

Therefore, the HCF of literal coefficients is b^{2}.

HCF of two or more monomials = (HCF of their numerical coefficients) × (HCF of their literal coefficients)

HCF of a^{3}b^{2 }and – 5b^{2} is b^{2}.

The final answer is b^{2}.

(e) Given that 4a^{3}, 10b^{3}, and 6c^{3}

Firstly, find the HCF of given terms.

HCF of their numerical coefficients 4, 10, and 6 is 2.

HCF of literal coefficients:

The lowest power of a is 0.

The lowest power of b is 0.

The lowest power of c is 0.

Therefore, the HCF of literal coefficients is 1.

HCF of two or more monomials = (HCF of their numerical coefficients) × (HCF of their literal coefficients)

HCF of 4a^{3}, 10b^{3}, and 6c^{3} is 2.

The final answer is 2.

(f) Given that 2abc^{3}, 3a^{2}b^{2}c, 5abc

Firstly, find the HCF of given terms.

HCF of their numerical coefficients 2, 3, and 5 is 1.

HCF of literal coefficients:

The lowest power of a is 1.

The lowest power of b is 1.

The lowest power of c is 1.

Therefore, the HCF of literal coefficients is abc.

HCF of two or more monomials = (HCF of their numerical coefficients) × (HCF of their literal coefficients)

HCF of 2abc^{3}, 3a^{2}b^{2}c, and 5abc is abc.

The final answer is abc.

4. Find the H.C.F. of the terms of the expression

(a) – 2a^{2} – 8a^{3} – 20a^{4
}(b) 3mn^{2} + 9m^{2}n^{2} – 6m^{2}n

(c) 5a^{2} + 25a + 50

(d) a^{3}b^{2} – 8b^{2}

## Solution:

(a) Given that – 2a^{2} – 8a^{3} – 20a^{4}

Firstly, find the HCF of given terms.

HCF of their numerical coefficients -2, -8, and -20 is -2.

HCF of literal coefficients:

The lowest power of a is 2.

Therefore, the HCF of literal coefficients is a^{2}.

HCF of two or more monomials = (HCF of their numerical coefficients) × (HCF of their literal coefficients)

HCF of – 2a^{2} – 8a^{3} – 20a^{4} is -2a^{2}.

(b) Given that 3mn^{2} + 9m^{2}n^{2} – 6m^{2}n

Firstly, find the HCF of given terms.

HCF of their numerical coefficients 3, 9, and -6 is 3.

HCF of literal coefficients:

The lowest power of m is 1.

The lowest power of n is 1.

Therefore, the HCF of literal coefficients is mn.

HCF of two or more monomials = (HCF of their numerical coefficients) × (HCF of their literal coefficients)

HCF of 3mn^{2} + 9m^{2}n^{2} – 6m^{2}n is 3mn.

(c) Given that 5a^{2} + 25a + 50

Firstly, find the HCF of given terms.

HCF of their numerical coefficients 5, 25, and 50 is 5.

HCF of literal coefficients:

The lowest power of a is 0.

Therefore, the HCF of literal coefficients is 1.

HCF of two or more monomials = (HCF of their numerical coefficients) × (HCF of their literal coefficients)

HCF of 5a^{2} + 25a + 50 is 5.

(d) Given that a^{3}b^{2} – 8b^{2}

Firstly, find the HCF of given terms.

HCF of their numerical coefficients 1, and -8 is 1.

HCF of literal coefficients:

The lowest power of a is 0.

The lowest power of b is 2.

Therefore, the HCF of literal coefficients is b^{2}.

HCF of two or more monomials = (HCF of their numerical coefficients) × (HCF of their literal coefficients)

HCF of a^{3}b^{2} – 8b^{2} is b^{2}.

5. Find the factors of the expression

(a) 3a^{4} – 9a^{5} – 15a^{3
}(b) – 10mn^{3} + 30nm^{3} – 50m^{2}n^{3
}(c) 15x^{2} – 20y^{2} + 25z^{2
}(d) a^{3}bc + 5ab^{3}+ 9a^{3}b

(e) 3a^{2} – 9a^{2}b – 27a^{3}c

(f) 7a^{3} + 7ab^{2} + 7ac^{2}

## Solution:

(a) The given expression is 3a^{4} – 9a^{5} – 15a^{3}

Firstly, find the HCF of given terms.

HCF of their numerical coefficients 3, -9, and -15 is 3.

HCF of literal coefficients:

The lowest power of a is 3.

Therefore, the HCF of literal coefficients is a^{3}.

HCF of two or more monomials = (HCF of their numerical coefficients) × (HCF of their literal coefficients)

HCF of 3a^{4} – 9a^{5} – 15a^{3} is 3a^{3}.

Multiply and divide each term of the given expression 3a^{4} – 9a^{5} – 15a^{3} with 3a^{3}.

3a^{3 }(3a^{4}/3a^{3 }– 9a^{5}/3a^{3 }– 15a^{3}/3a^{3}) = 3a^{3} (a – 3a^{2} – 5)

The final answer is 3a^{3} (a – 3a^{2} – 5).

(b) The given expression is – 10mn^{3} + 30nm^{3} – 50m^{2}n^{3}

Firstly, find the HCF of given terms.

HCF of their numerical coefficients -10, 30, and -50 is -10.

HCF of literal coefficients:

The lowest power of m is 1.

The lowest power of n is 1.

Therefore, the HCF of literal coefficients is mn.

HCF of two or more monomials = (HCF of their numerical coefficients) × (HCF of their literal coefficients)

HCF of – 10mn^{3} + 30nm^{3} – 50m^{2}n^{3} is -10mn.

Multiply and divide each term of the given expression – 10mn^{3} + 30nm^{3} – 50m^{2}n^{3} with -10mn.

-10mn^{ }(– 10mn^{3}/-10mn+ 30nm^{3}/-10mn – 50m^{2}n^{3}/-10mn) = -10mn (n^{2} – 3m^{2} + 5mn^{2})

The final answer is -10mn (n^{2} – 3m^{2} + 5mn^{2}).

(c) The given expression is 15x^{2} – 20y^{2} + 25z^{2}

Firstly, find the HCF of given terms.

HCF of their numerical coefficients 15, -20, and 25 is 5.

HCF of literal coefficients:

The lowest power of x is 0.

The lowest power of y is 0.

The lowest power of z is 0.

Therefore, the HCF of literal coefficients is 1.

HCF of two or more monomials = (HCF of their numerical coefficients) × (HCF of their literal coefficients)

HCF of 15x^{2} – 20y^{2} + 25z^{2} is 5.

Multiply and divide each term of the given expression 15x^{2} – 20y^{2} + 25z^{2} with 5.

5^{ }(15x^{2}/5 – 20y^{2}/5 + 25z^{2}/5) = 5 (3x^{2} – 4y^{2} + 5z^{2})

The final answer is 5 (3x^{2} – 4y^{2} + 5z^{2}).

(d) The given expression is a^{3}bc + 5ab^{3}+ 9a^{3}b

Firstly, find the HCF of given terms.

HCF of their numerical coefficients 1, 5, and 9 is 1.

HCF of literal coefficients:

The lowest power of a is 1.

The lowest power of b is 1.

The lowest power of c is 0.

Therefore, the HCF of literal coefficients is ab.

HCF of two or more monomials = (HCF of their numerical coefficients) × (HCF of their literal coefficients)

HCF of a^{3}bc + 5ab^{3}+ 9a^{3}b is ab.

Multiply and divide each term of the given expression a^{3}bc + 5ab^{3}+ 9a^{3}b with ab.

ab^{ }(a^{3}bc/ab + 5ab^{3}/ab + 9a^{3}b/ab) = ab (a^{2}c + 5b^{2} + 9a^{2})

The final answer is ab (a^{2}c + 5b^{2} + 9a^{2}).

(e) The given expression is 3a^{2} – 9a^{2}b – 27a^{3}c

Firstly, find the HCF of given terms.

HCF of their numerical coefficients 3, -9, and -27 is 3.

HCF of literal coefficients:

The lowest power of a is 2.

The lowest power of b is 0.

The lowest power of c is 0.

Therefore, the HCF of literal coefficients is a^{2}.

HCF of two or more monomials = (HCF of their numerical coefficients) × (HCF of their literal coefficients)

HCF of 3a^{2} – 9a^{2}b – 27a^{3}c is 3a^{2}.

Multiply and divide each term of the given expression 3a^{2} – 9a^{2}b – 27a^{3}c with 3a^{2}.

3a^{2}^{ }(3a^{2}/3a^{2 }– 9a^{2}b/3a^{2} – 27a^{3}c/3a^{2}) = 3a^{2} (1 – 3b – 9ac)

The final answer is 3a^{2} (1 – 3b – 9ac).

(e) The given expression is 3a^{2} – 9a^{2}b – 27a^{3}c

Firstly, find the HCF of given terms.

HCF of their numerical coefficients 3, -9, and -27 is 3.

HCF of literal coefficients:

The lowest power of a is 2.

The lowest power of b is 0.

The lowest power of c is 0.

Therefore, the HCF of literal coefficients is a^{2}.

HCF of two or more monomials = (HCF of their numerical coefficients) × (HCF of their literal coefficients)

HCF of 3a^{2} – 9a^{2}b – 27a^{3}c is 3a^{2}.

Multiply and divide each term of the given expression 3a^{2} – 9a^{2}b – 27a^{3}c with 3a^{2}.

3a^{2}^{ }(3a^{2}/3a^{2 }– 9a^{2}b/3a^{2} – 27a^{3}c/3a^{2}) = 3a^{2} (1 – 3b – 9ac)

The final answer is 3a^{2} (1 – 3b – 9ac).

(f) The given expression is 7a^{3} + 7ab^{2} + 7ac^{2}

Firstly, find the HCF of given terms.

HCF of their numerical coefficients 7, 7, and 7 is 7.

HCF of literal coefficients:

The lowest power of a is 1.

The lowest power of b is 0.

The lowest power of c is 0.

Therefore, the HCF of literal coefficients is 7a.

HCF of two or more monomials = (HCF of their numerical coefficients) × (HCF of their literal coefficients)

HCF of 7a^{3} + 7ab^{2} + 7ac^{2} is 7a.

Multiply and divide each term of the given expression 7a^{3} + 7ab^{2} + 7ac^{2 }with 7a.

7a^{ }(7a^{3}/7a + 7ab^{2}/7a + 7ac^{2}/7a) = 7a (a^{2} + b^{2} + c^{2})

The final answer is 7a (a^{2} + b^{2} + c^{2}).