1. 10.5 Factoring Trinomials With a Lead Coefficient of 1 to Solve
Algebra
10.5 Factoring Trinomials With a Lead Coefficient of 1 to Solve

2. Factoring to solve…
a quadratic expression can be solved by factoring and then using the zero-product property.
Solve: x2 + 10x + 21 = 0
…and whose sum is 10.
Find two numbers whose product is 21...
(x + 7)(x + 3) = 0
x = -7 and -3 They should check!

3. Figuring out the signs! Frame it with the signs. x2 + bx + c = 0
(x )(x ) = 0
Frame it with the signs. x2 – bx + c = 0
(x )(x ) = 0
Frame it with the signs. x2 – bx – c = 0
(x )(x ) = 0
The larger # goes here.
Frame it with the signs. x2 + bx – c = 0
(x )(x ) = 0
The larger # goes here.

4. Methods Factor. x2 + 5x – 36 = 0 Factor. x2 – 14x = -48
Method 1: List out all the factors of the constant in the trinomial.
List factors of -36
-12 and 3
-18 and 2
6 and -6
Factor. x2 + 5x – 36 = 0
12 and -3
18 and -2
9 and -4
36 and -1
Frame it. (x )(x ) = 0
-9 and 4
-36 and 1
Which set of factors add to +5?
Solve. x = 4 and -9
Method 2: Do this process in your head!!!
Factor. x2 – 14x = -48
Put in standard form.
x2 – 14x + 48 = 0
Frame it with signs. (x )(x ) = 0
Solve. x = 6 and 8

5. Solve. x2 – 15x – 7 = -61 + 61 +61 Put in standard form!
Put in standard form!
x2 – 15x + 54 = 0
(x - )(x - ) = 0
x = 9 and 6

6. Solve. 1) x2 + 3x – 18 = 0 2) m2 + 11m = -10 3) x2 – 2x – 40 = 8
x = 3 and -6
2) m2 + 11m = -10
m2 + 11m + 10 = 0
m = -10 and -1
(m + 10)(m + 1) = 0
3) x2 – 2x – 40 = 8
x2 – 2x – 48 = 0
x = -6 and 8
(x + 6)(x – 8) = 0
4) a2 – 33a = 280
a2 – 33a – 280 = 0
a = 40 and -7
(a – 40)(a + 7) = 0

7. Solve. 5) x2 + 3x = 6 Then how do you solve the equation?
If you think the quadratic equation cannot be factored, check the discriminant.
If the discriminant is a perfect square: The equation can be factored.
If the discriminant is not a perfect square: The equation cannot be factored.
b2 – 4ac
32 – 4(1)(-6)
9 + 24
Not a perfect square,
the trinomial cannot be factored.

8. HW
P #15-47, 52-56