Page 23 - MODULE 1 Q1MATH8

To determine the 2nd term of the
d.
e.
f. Thus, the factor of 27y3 8z6 is
2nd factor, get the product of the
first factor then use the opposite
sign of the expression.
Enclose in a quantity unit the
terms of the second factor.
-1st term
-2nd term
-3rd term
 (3y)(2z2) = 6yz2
   
 (3y – 2z2) (9y2 + 6yz2 + 4z4)
9y2
6yz2
4z4
(9y2 6yz2 4z4)
Let’s factor more polynomials!
1. 64a3 b3
Writing the Cube Roots Writing the first factor Identifying the 2nd factor Factoring by sum of two cubes
2. 125x3 1
Given Expression Writing the Cube Roots
Writing the first factor
Identifying the 2nd factor
Factoring by difference of two cubes
(4a)(4a)(4a) + (b)(b)(b) (4a  b)
(16a2 4abb2 ) (4a  b) (16a2 4abb2 )
125x3 1 (5x)(5x)(5x) - (1)(1)(1)
(5x 1)
(25x2 5x1)
(5x 1) (25x2 5x1)
Given Expression 64a3 b3
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