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 zeroproduct 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!
Figuring out the signs!
Frame it with the signs.
x2 + bx + c = 0
(x +
Frame it with the signs.
)(x +
)=0
x2 – bx + c = 0
(x -
)(x -
Frame it with the signs.
)=0
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.
Methods
Method 1: List out all the factors of the constant in the trinomial.
List factors of -36
x2 + 5x – 36 = 0
Factor.
Frame it.
Solve.
-12 and 3
-18 and 2
12 and -3
18 and -2
9 and -4
36 and -1
-9 and 4
-36 and 1
(x + 9)(x - 4 ) = 0
Which set of factors add to +5?
x = 4 and -9
Method 2: Do this process in your head!!!
x2 – 14x = -48
Factor.
Put in standard form. x2 – 14x + 48 = 0
Frame it with signs.
6 and -6
(x - 6 )(x - 8 ) = 0
Solve.
x = 6 and 8
Solve.
x2 – 15x – 7 = -61
+ 61 +61
Put in standard form!
x2 – 15x + 54 = 0
(x - 9 )(x - 6 ) = 0
x = 9 and 6
Solve.
1)
x2 + 3x – 18 = 0
(x – 3)(x + 6) = 0
2)
m2 + 11m = -10
m2 + 11m + 10 = 0
(m + 10)(m + 1) = 0
m = -10 and -1
3)
x2 – 2x – 40 = 8
x2 – 2x – 48 = 0
(x + 6)(x – 8) = 0
x = -6 and 8
4)
a2 – 33a = 280
a2 – 33a – 280 = 0
(a – 40)(a + 7) = 0
a = 40 and -7
x = 3 and -6
Solve.
5)
x2
+ 3x = 6
x2 + 3x – 6 = 0
(x – )(x + ) = 0
Then how do you solve the equation?
x = -3 +√33
2
x = -3 - √33
2
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
33 Not a perfect square,
the trinomial cannot be factored.
HW
• P. 607-609 #15-47, 52-56
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10.5 Solving Quadratics with a Lead Coefficient of 1 by Factoring

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