College Algebra 9th txtbk

28 CHAPTER R BASIC ALGEBRAIC OPERATIONS
The signs must be opposite in the factors, because the third term is negative. We can
reverse our choice of signs later if necessary. We now write all factors of 6 and of 4:
6
2 3
3 2
1 6
6 1
4
2 2
1 4
4 1
and try each choice on the left with each on the right—a total of 12 combinations
that give us the first and last terms in the polynomial 6x 2 5xy 4y 2 . The question
is: Does any combination also give us the middle term, 5xy? After trial and error
and, perhaps, some educated guessing among the choices, we find that 3 2 matched
with 4 1 gives us the correct middle term.
6x 2 5xy 4y 2 (3x 4y)(2x y)
If none of the 24 combinations (including reversing our sign choice) had produced
the middle term, then we would conclude that the polynomial is not factorable using
integer coefficients.
MATCHED PROBLEM 7
Factor each polynomial, if possible, using integer coefficients:
(A)
(C)
x 2 8x 12 (B) x 2 2x 5
2x 2 7xy 4y 2 (D) 4x 2 15xy 4y 2
The special factoring formulas listed here will enable us to factor certain polynomial
forms that occur frequently.
Z SPECIAL FACTORING FORMULAS
1. u 2 2uv v 2 (u v) 2
Perfect Square
2. u 2 2uv v 2 (u v) 2
Perfect Square
3. u 2 v 2 (u v)(u v)
Difference of Squares
4. u 3 v 3 (u v)(u 2 uv v 2 ) Difference of Cubes
5. u 3 v 3 (u v)(u 2 uv v 2 ) Sum of Cubes
The formulas in the box can be established by multiplying the factors on the right.
ZZZ EXPLORE-DISCUSS 1
Explain why there is no formula for factoring a sum of squares u 2 v 2 into the
product of two first-degree polynomials with real coefficients.