# factor the expressions a) 18a^3*b^2-27a^4 b) 25y^2-16 c) z^6-144 d)m^3-125

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a) 18a^3*b^2-27a^4

Let us factor 9a^3

==> 9a^3 ( 2b^2 - 3a)

b) 25y^2-16

The equation is perect square which we could factor as follow:

a^2 - b^2 = (a-b)(a+b)

a^2= 25y^2 ==> a= 5y

b^2 = 16 ==> b= 4

==> (5y-4)(5y+4)

c) z^6-144

We know that 144 = 12^2

==> z^6 - 12^2 = (z^3)^2 - 12^2)

since a^2 - b^2 = (a-b)(a+ b)

==> (z^3)^2 - 12^2 = (z^3 - 12)(z^3 + 12)

d)m^3-125

We know that 125 = 5^3

==> m^3 - 125= m^3 - 5^3

We know that :

a^3 - b^3 = (a-b)(a^2 + ab + b^2)

==> m^3 - 5^3 = ( m-5)(m^2 + 5m + 25)

To factor

a) 18a^3*b^2-27a^4. Both terms has the HCF 9a^3 .So we can factor out 9a^3 from both terms.

18a63*b^2-27a^4= 9a^3 (2b^2-3a)

b) 25y^2-16

= (5y)^2 -4^2 which is like a^2-b^2 = (a+b)(a-b)

= (5y+4)(5y-4)

c) z^6-144

=( Z^3)^2 - (12)^2

= (z^3+12)(z^3-12)

d)m^3-125

=m^3-5^3

= (m-5) (m^2+5m+5^2) , as a^3-b^3 = (a-b)(a^2+ab+b^2) is an identity.

=(m-5)(m^2+5m+25)

a) We notice that neither term is a perfect square nor a perfect cube, but they have the common factor 9a^3.

We'll factorize by 9a^3 and we'll get:

**18a^3*b^2-27a^4 = 9a^3*(2b^2 - 3a)**

b) Because the terms are perfect squares, we'll have in the given expression, a difference of squares which it could be written according to the rule:

a^2 - b^2 = (a-b)(a+b)

25y^2-16 = (5y)^2 - (4)^2

**(5y)^2 - (4)^2 = (5y - 4)(5y + 4) **

c) Because the terms are again perfect squares, we'll have in the given expression, a difference of squares which it could be written according to the rule:

a^2 - b^2 = (a-b)(a+b)

z^6-144 = (z^3)^2 - (12)^2

**(z^3)^2 - (12)^2 = (z^3 + 12)(z^3 - 12)**

Because of the fact that 12 is not a perfect cube, although z^3 is, the brackets cannot be factored further more.

d) Because the terms are perfect cubes, we'll have in the given expression, a difference of cubes which it could be written according to the rule:

a^3 - b^3 = (a-b)(a^2 + a*b + b^2)

a^3 = m^3 and b^3 = (5)^3

m^3-125 = m^3 - 5^3

**m^3 - 5^3 = (m-5)(m^2 + 5*m + 25)**