Transcript 12-4 - Fort Thomas Independent Schools

12-4 Factoring x² + bx + c
(x + 2)(x + 3) = x² + 3x + 2x + 2·3
x²
+ 5x + 6
(x – 4)(x + 3) = x² + 3x - 4x - 4·3
x² - 1x - 12
(x + a)(x + b) = x² + bx + ax + a·b
x² + (b + a)x + ab
In this section you will be working backwards to
create the 2 binomials, so you need to find sums &
products that will give the middle and last terms.
Factor the following trinomials.
1. x² + 50x + 96
(x + ?)(x + ?)
(x + 2)(x + 48)
x² + 48x + 2x + 96
x² + 50x + 96
All positive terms.
Find the combination
whose product is 96
and whose sum is 50.
1, 96 = 96, 97 no
2, 48 = 96, 50
Check mentally
2. y² - 3y - 54
(y + ?)(y + ?)
(y + 6)(y - 9)
Last term is negative so
there must be 1 postive
and 1 negative number.
1, -54 = -54, -53 no
-1, 54 = -54, 53 no
2, -27 = -54, -25 no
-2, 27 = -54, 25 no
3, -18 = -54, -15 no
-3, 18 = -54, 15 no
6, -9 = -54, -3 yes
3. z² - 9z + 20
(z + ?)(z + ?)
(z - 4)(z -5)
4. c² - 16
(c + ?)(c + ?)
(c + 4)(c -4)
Last term positive with
middle term negative,
so both must be negative.
-1, -20 = 20, -21 no
-2, -10 = 20, -12 no
-4, -5 = 20, -9 yes
This is the difference of
squares that were introduced previously.
4, -4 = -16, 0 yes
5. d³ - 16d
The terms have a gcf. Take
care of it first.
d(d² - 16)
Factor now into 2 binomials
d(d+4)(d-4)
6. Explain why c² + 16 is not factorable.
Because there are no 2 numbers
who’s product is positive and has
a sum of 0.