2.3 Factoring Polynomials Math Worksheet
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72.3
Factoring Polynomials
GOAL
Review and extend factoring skills.
LEARN ABOUT the Math
Mai claims that, for any natural number n, the function
3
2
f (n) 5 n
1 3n
1 2n 1 6
always generates values that are not prime.
Is Mai’s claim true?
?
1
Selecting a factoring strategy: Testing values
EXAMPLE
of n to determine a pattern
3
2
Show that
f (n) 5 n
1 3n
1 2n 1 6
can be factored for any natural number, n.
Sally’s Solution
f (1) 5 12 5 4 3 3
After some calculations and guess
and check, I found a pattern.
f (2) 5 30 5 5 3 6
The first factor was of the form
f (3) 5 66 5 6 3 11
n 1 3
and the second factor was
2
of the form
n
1 2.
f (4) 5 126 5 7 3 18
f (5) 5 216 5 8 3 27
2
f (n) 5 (n 1 3) (n
1 2)
To confirm the pattern, I multiplied
2
(n 1 3) by (n
1 2).
3
2
5 n
1 3n
1 2n 1 6
Since both factors produce numbers
greater than 1,
f (n)
can never be
expressed as the product of 1 and itself.
So Mai’s claim is true.
Sometimes an expression that doesn’t appear to be factorable directly can be
factored by grouping terms of the expression and dividing out common factors.
98
2.3 Factoring Polynomials
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