Example M-5 Factoring a Second-Degree Polynomial
Factor the expression 6x 2 + 19xy + 10y 2.
Set Up
We examine the coefficients of the terms to see whether the expression can be factored without resorting to more advanced
methods. Remember that the multiplication (ax + by)(cx + dy) = acx 2 + (ad + bc)xy + bdy 2.
Solve
The coefficient of x2 is 6, which can be factored two ways.
ac = 6
3 # 2 = 6 or 6 # 1 = 6
The coefficient of y2 is 10, which can also be factored two
ways.
bd = 10
5 # 2 = 10 or 10 # 1 = 10
List the possibilities for a, b, c, and d in a table. Include a
column for ad + bc.
If a = 3, then c = 2, and vice versa. In addition, if
a = 6, then c = 1, and vice versa. For each value of a
there are four values for b.
a
b
c
d
ad + bc
3
3
3
3
2
2
2
2
6
6
6
6
1
1
1
1
5
2
10
1
5
2
10
1
5
2
10
1
5
2
10
1
2
2
2
2
3
3
3
3
1
1
1
1
6
6
6
6
2
5
1
10
2
5
1
10
2
5
1
10
2
5
1
10
16
19
23
32
19
16
32
23
17
32
16
61
32
17
61
16
Find a combination such that ad + bc = 19. As you can
see from the table there are two such combinations.
ad + bc = 19
3 # 5 + 2 # 2 = 19 and
2 # 2 + 5 # 3 = 19
It doesn’t matter which combination we choose. To finish
this problem we will use the combination in the second row
of the table to factor the expression in question:
6x 2 + 19xy + 10y 2 = (3x + 2y)(2x + 5y)
Reflect
As a check, expand (3x + 2y)(2x + 5y) to see if we return
to the original equation.
(3x + 2y)(2x + 5y) = 6x 2 + 15xy + 4xy + 10y 2
= 6x 2 + 19xy + 10y 2
You should be able to show that the combination in the
fifth row is also an acceptable factoring.
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