1.4 Polynomials 1.5 Equations ALGEBRA 1.4 Polynomials 1.5 Equations

1.4 A polynomial is an algebraic expression that has many terms connected by only the operations of +, -, and • of variables. 2x + 5 2x3y4z 5x2 - 7x + 19 The terms are the parts of the expression that are added or subtracted. Number of Terms Name Example 1 monomial 3x2y5 2 binomial 3x2 + y5 3 trinomial 3 - x2 + y5 4 or more no special name 3x5 - 2x4 + 5x3 - 6x2 + 2x

1.4 The degree of a polynomial is the maximum number of variables that appear as factors in any one term. Degree Name Example Memory Aid constant 4 Constants do not vary A line has one dimension linear 1st 1st linear 4x; 4x+4 A square is a quadrilateral 2nd quadratic 2nd quadratic 4x2; 4x2+5x+2 A cube has 3 dimensions cubic 3rd 3rd cubic 4x3; 4xyz+2 A quarter is a fourth of a dollar quartic 4th 4th quartic 4x4; 8x2y2+4x3 Quintuplets are 5 children 5th quintic 5th quintic 4x5; 3x+4tx2y2 6th or more no special name 4x44

1.4 x2 is the power 3x2 3x2 x2 is the power 3x2 2 is the exponent 3x2 x is the base 3x2 3x2 3 is the numerical coefficient Zero could have many different degrees, therefore it is considered to be a polynomial with no degree. 0 = 0x = 0x2 = 0xy = 0x92 = 0x3y9z999

1.4 Multiplying binomials can be done by two different procedures. FOIL Double use of the distributive property (2x - 3)(3x + 1) FOIL mean First Outside Inside Last (2x - 3)(3x + 1) First 6x2 Outside (2x - 3)(3x + 1) + 2x (2x - 3)(3x + 1) Inside - 9x - 3 (2x - 3)(3x + 1) Last

1.4 Multiplying binomials can be done by two different procedures. FOIL Double use of the distributive property (2x - 3)(3x + 1) To proceed through double use of the distributive property, think of (2x - 3) as one number,and distribute this number over the other binomial. (2x - 3) • 3x + (2x - 3) • 1 Then use the commutative property of multiplication to get 3x(2x - 3) + 1(2x - 3) Now distribute each of these new values 6x2 - 9x + 2x - 3

1.4 In both cases, the polynomial can be simplified by combining like terms. 6x2 - 7x - 3

1.5 Equations are two expressions connected by an equal sign. Solving an equation means writing its solution set. Solve: x2 = 25 There are two answers, x = 5 and x = -5, so the solution is S = {-5, 5} Solve: 5x - 8 = 2x + 4 3x = 12 x = 4 x = 4 This equation is equivalent to the original one. 5x - 8 = 2x + 4 S = {4} They have the same solution set.

1.5 Before writing a solution set, you should check your answer by substituting the value(s) back into the original equation. An equation is not solved until you write the solution set. Adding (or subtracting) both members of an equation by the same number is justified by the Addition Property of Equality. Multiplying (or dividing) both members of an equation by the same number is justified by the Multiplication Property of Equality. While it is not necessary to justify each step when solving an equation, it will be necessary to do so when completing algebraic proofs - therefore learn the properties. No Work = No Credit, and the more work you show, the more partial credit you can earn.

1.5 Given the equation (x - 3)(x - 7) = 0: S = {3, 7} Write the solution set. This equation contains a product which equals zero. The only way a product of two real numbers can equal zero is for one of the factors to equal zero. This fact is expressed as the converse of the Multiplication Property of Zero. x - 3 = 0 or x - 7 = 0 x = 3 or x = 7 S = {3, 7}

No, the solution sets are different. 1.5 Given the equation (x - 3)(x - 7) = 0: S = {3, 7} x - 7 = 0 S = {7} Divide both members of the equation by (x - 3), and write the solution set for the transformed equation. (x - 3)(x - 7) = 0 . (x - 3) (x - 3) x - 7 = 0 S = {7} Is the transformed equation equivalent to the original equation? Explain. No, the solution sets are different.

1.5 Given the equation (x - 3)(x - 7) = 0: S = {3, 7} x - 7 = 0 S = {7} (x - 3)(x - 7)(x - 2) = 0 S = {2, 3, 7} Multiply both members of the original equation by (x - 2), then solve it. (x - 3)(x - 7)(x - 2) = 0(x - 2) (x - 3)(x - 7)(x - 2) = 0 x - 3 = 0 or x - 7 = 0 or x - 2 = 0 x = 3 or x = 7 or x = 2 S = {2, 3, 7} What number is a solution for the transformed equation that was not a solution of the original one? The new solution is 2.

1.5 Given the equation (x - 3)(x - 7) = 0: S = {3, 7} x - 7 = 0 S = {7} (x - 3)(x - 7)(x - 2) = 0 S = {2, 3, 7} How can all three of these equations be equal to zero? Since x is a variable, there is the possibility that any of the factors could equal zero. When you multiply or divide by zero, you are performing an operation which seems perfectly correct, but which changes the solution set. Since it is the original equation you are trying to solve, you must be aware of these issues.

1.5 You should never divide an equation by something that can equal zero. DIVISION BY ZERO IS UNDEFINED and therefore NOT ALLOWED!!!!! When you divide by zero you can lose a valid solutions. When you multiply by zero, you are performing an irreversible step. This can lead to an extraneous solution. An extraneous solution is a solution which satisfies the transformed equation, but not the original equation. It is an extra solution. extraneous You will find which solutions are extraneous by checking your answer. You can have equations with no solutions. In this case the solution set is empty. S = Ø or S = { }

1.5 Solve: | x - 2 | = 3 Remember: absolute value is distance from zero. | x - 2 | = 3 means (x - 2) is three units form zero. This is three units to the left and three units to the right. x - 2 = 3 or x - 2 = -3 x = 5 or x = -1 S = {-1, 5}