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Algebra 2

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2x2 4x 2 = x2 7x 12 foil method -x2 -7x -12 -x2 -7x -12. x2 3x 10 = 0 ... n2 4n 4 = 2n 1 foil method -2n 1 -2n 1. n2 6n 5 = 0 ... – PowerPoint PPT presentation

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Title: Algebra 2


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2
Algebra 2 Ch s 1-8 Cumulative Review (WS 1-8
Test B)
Solve the equation.
1.
cross-multiply
(x 1)(2x 2) (x 3)(x 4) 2x2 4x 2
x2 7x 12 foil method -x2 -7x
-12 -x2 -7x -12 x2 3x 10
0 (x 5)(x 2) 0 factor into 2
binomials x 5 0 or x 2 0 set each
factor equal to 0
5 5 x 5 or
-2 -2 x -2
3
Solve for x.
4.
252x-1 125
get both to have a base of 5
52(2x-1) 53
bases are equal...set exponents equal to each
other
2(2x 1) 3
4x 2 3 distributive property 2
2 4x 5
Solve for n.
6.
(n 2)2 2n 1 n2 4n 4 2n 1
foil method -2n 1 -2n 1 n2 6n
5 0 (n 5)(n 1) 0 factor into 2
binomials n 5 0 or n 1 0 set each
factor equal to 0
square both sides
5 5 n 5 or
1 1 n 1
4
Solve for x.
8. a(2x b) c(3x d) 2ax ab 3cx
cd distributive property -3cx ab
-3cx ab 2ax 3cx ab cd x(2a
3c) ab cd factor x out
divide both sides by 2a 3c
Find the value.
9.
5
Multiply.
11.
factor 1 out before multiplying
(5 2i)(2 3i) 10 11i 6i2 foil
method 10 11i 6(-1) 10 11i 6 16
11i
Simplify.
3
12a6 b-3 c-2 8a-2 b-6 c4
12.
2
subtract exponents when dividing
3a8 b3 c-6 2
3a8 b3 2c6
move negative exponents to make positive
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