Regents Review #1 Expressions & Equations (x – 4)(2x + 5) 3x 3 – 4x 2 + 2x – 1 (4a – 9) – (7a 2 + 5a + 9) 4x 2 + 8x + 1 = 0 (x – 5) 2 = 25 10x 3 5x 5 x – 3 = 2x 5x 7

Exponential Expressions

1) xy 0 2) (2x 2 y)(4xy 3 ) 3) x(1) x 8x 3 y 4 3xy 3 Any nonzero number raised to the zero power equals 1. Multiply coefficients and add exponents. Divide coefficients and subtract exponents.

Polynomials A polynomial is a sum of terms. Each term is separated by either a + or – sign.

Polynomials The degree of a polynomial with one variable is the highest power to which the variable is raised.

Polynomials A polynomial is written in standard form when the degrees (exponents) are listed from highest to lowest.

Polynomials When 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 2x 2 – x x – 2 2x 2 – x + 11 (3x – 2) + (3x – 2) + (2x 2 – x + 11) + (2x 2 – x + 11) 2x 2 + 2x 2 + 3x + 3x – x – x – 2 – x 2 + 4x + 18 The expression can also be simplified this way. 2(3x – 2) + 2(2x 2 – x + 11)

Polynomials When subtracting polynomials, distribute the minus sign before combining like terms. 2)Subtract 5x 2 – 2y from 12x 2 – 5y (12x 2 – 5y) – (5x 2 – 2y) FROM COMES FIRST 12x 2 – 5y – 5x 2 + 2y 12x 2 – 5x 2 – 5y + 2y 7x 2 – 3y

Polynomials When 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. 9x 2 – 12x – 12x x 2 – 24x x 2 – 4x + 1 2x 2 -4x 1 x 2x 3 -4x 2 x 5 10x 2 -20x 5 2x 3 + 6x 2 – 19x + 5 x + 5 (3x – 4)(3x – 4) Expand 3x - 4

Polynomials When dividing polynomials, each term in the numerator is divided by the monomial that appears in the denominator. 5)

Factoring Polynomials 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 “Go to Methods” Binomial difference of two perfect squares Trinomial ax 2 + bx + c, a = 1

Factoring What about ax 2 + bx + c when a 1? Factor 5n 2 + 9n – 2 5n n – 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 = 9The 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 groups 5n n – 1n – 2

Factoring When factoring completely, factor until you cannot factor anymore! 1)2) The factored form of the polynomial expression is equivalent to the standard form of the polynomial expression.

Equations What types of equations (in one variable) do we need to know how to solve? 1)Equations with rational expressions (fractions) 2)Quadratic Equations 3) Square Root Equations 4) Literal Equations (solving for another variable) Remember: When solving any type of equation, always use properties of equality and check solution(s).

Rational Equations (proportions) 5(3x – 2) = 10(x + 3) 15x – 10 = 10x x – 10 = 30 5x = 40 x = 8 Always check solution(s) to any equation

Rational Equations How do we solve a rational equation with more than one fraction? Option 1: Combine fractions and create a proportion Option 2: Multiply by the LCD ( least common denominator ) Example: Solve for x.

Rational Equations FOO 5x + 10 = 70 Create a ProportionMultiply by the LCD

Quadratic Equations 1) ax 2 + c = 0 Ex: 2x 2 – 32 = 0 2x 2 = 32 x 2 = 16 x = x = 4 or x = {4,-4} 2) ax 2 + bx + c = 0 Ex: x 2 – 5x = -6 Isolate x 2 and take the square root. Set the equation equal to zero. Factor. Set each factor equal to zero and solve. Zero Product Property x 2 – 5x + 6 = 0 (x – 2)(x – 3)= 0 x – 2 = 0 x – 3 = 0 x = 2 x = 3 x = {2,3}

Quadratic Equations How do we solve a quadratic equation that cannot be factored? Example: Find the roots of x 2 – 2x – 5 = 0. Use the quadratic formula: a = 1, b = -2, c = -5

Quadratic Equations The equation can also be solved by completing the square. Find the roots of x 2 – 2x – 5 = 0. x 2 – 2x – 5 = 0 x 2 – 2x = 5 x 2 – 2x _____ = 5 _______ x 2 – 2x + 1 = (x – 1)(x – 1) (x – 1) 2 = 6

Square Root Equations Solve Isolate the Square both sides of the equation to eliminate the

Literal Equations When solving literal equations, isolate the indicated variable using properties of equality. y(a + x) = c Factor out the variable that you are solving for.

Literal Equations Multiply both sides of the equation by 2 to eliminate the fractional coefficient.

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 15 th Be there!