Algebra

10.5 Factoring Trinomials With a Lead Coefficient of 1 to

Solve

Factoring to solve…

• a quadratic expression can be solved by factoring and then using the zero-product property.

Solve: x2 + 10x + 21 = 0

Find two numbers whose product is 21...…and whose sum is 10.

(x + 7)(x + 3) = 0 x = -7 and -3 They should check!

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- +

Frame it with the signs. x2 + bx – c = 0

(x )(x ) = 0- +

The larger # goes here.

The larger # goes here.

Methods

Method 1: List out all the factors of the constant in the trinomial.

Factor. x2 + 5x – 36 = 0

List factors of -36

-12 and 3

Frame it. (x )(x ) = 0

12 and -3

9 and -4

-9 and 4

-18 and 2

18 and -2

36 and -1

-36 and 1

6 and -6

Which set of factors add to +5?

+ 9 - 4

Method 2: Do this process in your head!!!

Factor. x2 – 14x = -48

Frame it with signs. (x )(x ) = 0- - 6 8

Put in standard form. x2 – 14x + 48 = 0

Solve. x = 4 and -9

Solve. x = 6 and 8

Solve.

x2 – 15x – 7 = -61

Put in standard form! + 61 +61

x2 – 15x + 54 = 0

(x - )(x - ) = 0 9 6

x = 9 and 6

Solve.

1) x2 + 3x – 18 = 0

2) m2 + 11m = -10

3) x2 – 2x – 40 = 8

4) a2 – 33a = 280

(x – 3)(x + 6) = 0 x = 3 and -6

m2 + 11m + 10 = 0 (m + 10)(m + 1) = 0

m = -10 and -1

x2 – 2x – 48 = 0 (x + 6)(x – 8) = 0

x = -6 and 8

a2 – 33a – 280 = 0 (a – 40)(a + 7) = 0

a = 40 and -7

Solve.

5) x2 + 3x = 6

x2 + 3x – 6 = 0

(x – )(x + ) = 0

If you think the quadratic equation cannot be factored, check the discriminant.

b2 – 4ac

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.

32 – 4(1)(-6)

9 + 24

33 Not a perfect square, the trinomial cannot be factored.

Then how do you solve the equation?

x = -3 +√33 x = -3 - √332 2

HW

• P. 607-609 #15-47, 52-56