Page 491 - Algebra 1

Example
2
Factoring when c is Negative Factor each trinomial.
a. x2+3x-10
SOLUTION
In this trinomial, b is 3 and c is −10.
Four pairs of positive and negative numbers have a product of −10. (-1)(10) (1)(-10) (-2)(5) (2)(-5)
The sum of only one of these pairs is 3.
(-1) + 10 = 9 1 + (-10) = -9 (-2) + 5 = 3 2 + (-5) = -3 The constant terms in the binomials are -2 and 5. So,
x2 +3x-10=(x-2)(x+5)
b. x2-7x-8
SOLUTION
In this trinomial, b = −7 and c = −8.
Four pairs of positive and negative numbers have a product of -8. (-1)(8) (1)(-8) (-2)(4) (2)(-4)
The sum of only one of these pairs is -7.
(-1) + 8 = 7 1 + (-8) = -7 (-2) + 4 = 2 2 + (-4) = -2 The constant terms in the binomial are 1 and -8. So,
x2 -7x-8=(x+1)(x-8)
Factoring with Two Variables
Factor each trinomial. a. x2 +5xy+6y2
SOLUTION In this trinomial, b and c have values of 5y and 6y2, respectively. Because both b and c are positive, b must be the sum of two positive terms that are factors of c.
Six pairs of positive terms have a product of 6y2.
(1y2)(6) (1y)(6y) (1)(6y2) (2y2)(3) (2y)(3y) (2)(3y2)
Eliminate pairs of terms that contain y2 because their sums cannot yield a term containing y. For example, (1y2)(6) has the sum (y2 + 6).
Only the pair 2y and 3y has a sum of 5y, which is the value of b. So, x2 +5xy+6y2 =(x+2y)(x+3y)
Math Reasoning
Analyze How can the absolute value of the factors of c be used to find the last terms of the binomial factors?
Example
3
Reading Math
The middle term of the trinomial is written 2xy, not 2yx, even though 2y is the value of b in the trinomial x2 + bx + c.
476 Saxon Algebra 1