Regents Review #1
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Transcript of Regents Review #1
Regents Review #1
Expressions &
Equations(x – 4)(2x + 5)
3x3 – 4x2 + 2x – 1
(4a – 9) – (7a2 + 5a + 9)
4x2 + 8x + 1 = 0 (x – 5) 2 = 25
(10x3)2
5x5 x – 3 = 2x 5x 7
Evaluating and WritingAlgebraic Expressions
3) Express the cost of y shirts bought at x dollars each.
2) Express “three times the quantity of 4 less than a number” as an expression.
xy
3(n – 4)
1) Evaluate x2 – y when x = -2 and y = -5
x2 – y (-2)2 – (-5) 4 + 5 9
Simplifying Exponential Expressions
1) xy0 2) (2x2y)(4xy3) 3) (2x3y5)4
x(1)
x
8x3y4 24(x3)4(y5)4
16x12y20
Any nonzero number raised to the zero power equals 1.
Multiply coefficients and add exponents.
Raise each factor to the power.
Simplifying Exponential Expressions
4)3
3
2
y
x2
52
2
6
ab
ba 6)5)
33ab
Divide coefficientssubtract exponents.
9
6
y
x
Raise the numerator and denominator to the power of the fraction.
3
4
24
10
y
y
12
5y
Simplify numerator and denominator coefficients by dividing by a common factor.
Simplifying Exponential Expressions7) 534 yx
3
54
x
y
Move powers with negative exponents to the other part of the fraction. Rewrite using positive exponents.
8)6
59
9
15
ab
ba
3
5 18 ba
b
a
3
5 8
Simplify coefficients. Subtract exponents (all results appear in the numerator). Move powers with negative exponents to the denominator and rewrite with positive exponents.
Simplifying Exponential Expressions
When simplifying exponential expressions, remember…
1) Use exponent rules to simplify.
2) When dividing, all results appear in the numerator. Change negative exponents to positive by moving them to the other part of the fraction.
3) No decimals or fractions are allowed in any part of the fraction.
PolynomialsWhen adding polynomials, combine like terms.
1) Represent the perimeter of a rectangle as a simplified polynomial expression if the width is 3x – 2 and the length is 2x2 – x + 11.
3x – 2 3x – 2
2x2 – x + 11
2x2 – x + 11
(3x – 2) + (3x – 2) + (2x2 – x + 11) + (2x2 – x + 11)
2x2 + 2x2 + 3x + 3x – x – x – 2 – 2 + 11 + 11
4x2 + 4x + 18
Can also simplify 2(3x – 2) + 2(2x2 – x + 11)
PolynomialsWhen subtracting polynomials, distribute the minus sign before combining like terms.
2) Subtract 5x2 – 2y from 12x2 – 5y
(12x2 – 5y) – (5x2 – 2y)
12x2 – 5y – 5x2 + 2y
12x2 – 5x2 – 5y + 2y
7x2 – 3y
PolynomialsWhen multiplying polynomials, distribute each term from one set of parentheses to every term in the other set of parentheses.
3) (3x – 4)2
4) Express the area of the rectangle as a simplified polynomial expression.
9x2 – 12x – 12x + 16 9x2 – 24x + 16
2x2 – 4x + 1
2x2
-4x
1x 2x3 -4x2 x
5 10x2 -20x 5
2x3 + 6x2 – 19x + 5
x + 5
(3x – 4)(3x – 4)
PolynomialsWhen dividing polynomials, each term in the numerator is divided by the monomial that appears in the denominator.
24
2
23
2
42
2
2342
4
3
12
3
3
3
123
xyy
x
yx
x
yx
x
yxyx
5)
Factoring
What does it mean to factor?Create an equivalent expression that is
a “multiplication problem”.
Remember to always factor completely.Factor until you cannot factor anymore!
Factoring
1) Factor out the GCF
2) AM factoring
3) DOTS
)12(224 2 xxxx
)2)(3(652 xxxx
)43)(43(169 2242 yxyxyx
“Go to Methods”
FactoringWhat about ax2 + bx + c when a 1?
Factor 5n2 + 9n – 2
5n2 + 10n – 1n – 2
5n(n + 2) – 1(n + 2) (5n – 1)(n + 2)
GCF is 1, Factor by Grouping!
Find two numbers whose product = ac and whose sum = b
ac = (5)(-2) = -10 b = 9 The numbers are -1 and 10
Rewrite the polynomial with 4 terms
Factor out the GCF of each group
Write the factors as 2 binomials
Create two groups5n2 + 10n - 1n – 2
FactoringWhen factoring completely, factor until you cannot factor anymore! xxx 12102 23 1)
44 8116 yx 2)
)94)(32)(32( 22 yxyxyx )94)(94( 2222 yxyx
)65(2 2 xxx)2)(3(2 xxx
A polynomial expression is factored completely when all the factors are prime.
Solving EquationsWhat types of equations do we need to know how to solve?
1) Proportions2) Quadratic Equations3) Square Root Equations4) Literal Equations (solving for another variable)
Solving Proportions
5(3x – 2) = 10(x + 3)
15x – 10 = 10x + 30
5x – 10 = 30
5x = 40
x = 8
Always check solution(s) to any equation10
23
5
3
xx
5
11
5
1110
22
5
1110
2)(3
5
3
88
Solving Quadratic Equations1) x2 = a Example: x2 = 16 Take the square root of both sides
x = x = 4 or x = {4,-4}2) x2 + bx + c = 0 Example: x2 – 5x = -6
x2 – 5x + 6 = 0 Set all terms equal to zero (x – 2)(x – 3)= 0 Factor
x – 2 = 0 x – 3 = 0 Set each factor equal to zero x = 2 x = 3 Solve
x = {2,3}
16
Solving Quadratic EquationsWhat happens when a quadratic equation cannot be factored?
Example: Find the roots of x2 – 2x – 5 = 0.
)1(2
)5)(1(4)2()2( 2 x
2
242 x
2
622 x
1
611x
}61,6{1 x
61x
Use the quadratic formula:a
acbbx
2
42
a = 1, b = -2, c = -5
626424
Solving Quadratic EquationsThis equation can also be solved by completing the square.Find the roots of x2 – 2x – 5 = 0.
x2 – 2x – 5 = 0x2 – 2x = 5x2 – 2x _____ = 5 _______
x2 – 2x + 1 = 5 + 1(x – 1)(x – 1)
1)1(2
2
22
22
b
(x – 1)2 = 6
61)(x 2
61x61x
Solving Square Root EquationsExample: Solve 31935 x
4035 x
83 x
2283 x
643 x
3.21x
Literal EquationsWhen solving literal equations, isolate the indicated variable using inverse operations
candxa,oftermsinyforSolve
cyxay
xa
c
xa
xay
)(
xa
cy
hforSolve
bhA2
1
bhA2
b
bh
b
A
2
hb
2A
cxay )(
Now it’s your turn to review on your own!
Using the information presented today and the study guide posted on halgebra.org,
complete the practice problem set.
Regents Review #2 Friday, May 9th
BE THERE!