Transcript Factoring & Solving Quadratics Equations

Factoring & Solving
Quadratics Equations
Intermediate Algebra
Final Exam Review
1. Factor
16a b - 24ab
3 5
9
= 8ab (2a - 3b )
5
2
4
GCF
2. Factor
6ax + 6bx - 5a - 5b
= 6x(a + b) - 5(a + b)
= (6x - 5)(a + b)
grouping
3. Factor
56
x +15x + 56
2
= x + 8x + 7x + 56
= x(x + 8) + 7(x + 8)
2
= (x + 8)(x + 7)
AC method
grouping
factor
8
7
15
4. Factor
32
a -12a + 32
2
= a - 8a - 4a + 32
= a(a - 8) - 4(a - 8)
2
= (a - 8)(a - 4)
AC method
grouping
factor
-8 -4
-12
5. Factor
90
m - m - 90
2
= m -10m + 9m - 90 AC method
= m(m -10) + 9(m -10) grouping
2
= (m -10)(m + 9)
factor
-10 9
-1
6. Factor
-30
15x + x - 2
2
= 15x - 5x + 6x - 2 AC method
= 5x(3x -1) + 2(3x -1) grouping
2
= (3x -1)(5x + 2)
factor
-5 6
1
7. Factor
100x +140x + 49
2
= (10x)2 + 2(10x)(7) + (7)2
= (10x + 7)
2
factor
Special Products
8. Factor
81y - 36y + 4
2
= (9y)2 - 2(9y)(2) + (2)2
= (9y - 2)2
factor
Special Products
9. Factor
81x -100y
2
= (9x) - (10y)
2
= (9x -10y)2
2
2
Special Products
factor
10. Factor
3x + 24x + 45x
3
2
= 3x(x 2 + 8x +15)
GCF
= 3x(x 2 + 3x + 5x +15)
AC method
= 3x(x(x + 3) + 5(x + 3))
grouping
= 3x(x + 3)(x + 5)
factor
11. Factor
x + 3x - 4x -12
3
2
= x (x + 3) - 4(x + 3)
2
= (x + 3)(x - 4)
2
factor
= (x + 3)((x) - (2) )
2
grouping
2
= (x + 3)(x + 2)(x - 2)
special products
factor
1. Solve: Zero Product Property
x +11x + 27 = 3
2
x +11x + 24 = 0
2
standard form
x 2 + 3x + 8x + 24 = 0
AC method
x(x + 3) + 8(x + 3) = 0 grouping
(x + 3)(x + 8) = 0 factor
x + 3 = 0 x + 8 = 0 zero product property
x = -3
x = -8
solve
1. Solve: Square Root Method
x - 75 = 0
2
x 2 = 75
standard form
x = ± 75
square root
x = ±5 3
simplify
2. Solve: Square Root Method
(x + 3) - 60 = 4
2
(x + 3) = 64 isolate the square term
2
x +3 = ± 8
square root
x = -3 ± 2 2
simplify
1. Solve: Quadratic Formula
x + 6x + 2 = 0
2
-6 ± 36 - 4(1)(2)
x=
2(1)
a =1
b=6
-6 ± 28
x=
discriminant
2
-6 ± 2 7
x=
= -3 ± 7 simplify
2
{-3 +
7,-3 - 7
}
solution
c=2
2. Solve: Quadratic Formula
2x + 3x - 20 = 0
2
-3 ± 9 - 4(2)(-20)
x=
2(2)
a=2
b=3
c = -20
-3 ± 169
discriminant
x=
4
-3 +13 10
-3 ±13
x=
=
=5
simplify
x=
2
2
2
-3 -13 -16
x=
=
= -8
2
2
{5,-8} solution
Discriminant
Discriminant
32
Number of Solutions Type of Solutions
Two
Irrational
0
One
Rational
Negative Number
Two
Imaginary
25
Two
Rational
1. Radical Equations
2x + 7 + 8 = x +12
2x + 7 = x + 4
2x + 7 = (x + 4)2
isolate the radical
square
2x + 7 = x + 8x +16
2
0 = x 2 + 6x + 9
FOIL
standard form
0 = (x + 3)2
factor special products
0 = x + 3 zero product property
x = -3
solve
2. Radical Equations
-x - 3 = x + 5
-x - 3 = (x + 5)2
square
-x - 3 = x +10x + 25
2
FOIL
0 = x +11x + 28 standard form
0 = (x + 4)(x + 7) factor AC method
0 = x + 4 0 = x + 7 zero product property
x = -4 x = -7 solve
2
{-4},-7 is extraneous