P4 Polynomials and P5 Factoring Polynomials Professor Tim Busken Graduate T.A. Dynamical Systems Program Department of Mathematics San Diego State University June 22, 2011 Professor Tim Busken (Graduate T.A.Dynamical P4 Polynomials Systems ProgramDepartment and P5 Factoringof Polynomials Mathematics San Diego State JuneUniversity) 22, 2011 1 / 36
P4 POLYNOMIALS Definition (Term) A term is either a single number or variable, or the product or quotient of several numbers or variables separated from another term by a plus or minus sign in an overall expression. For example, the following algebraic expression has terms 100,3x, 5yz 2 w 3, and 2 3 x. Definition (Monomial) 100+3x +5yz 2 w 3 2 3 x A term that is a product of constants and/or variable is called a monomial. Terms having variables in the denominator and terms having a variable raised to a power which is not a non-negative integer are not monomials. Professor Tim Busken (Graduate T.A.Dynamical P4 Polynomials Systems ProgramDepartment and P5 Factoringof Polynomials Mathematics San Diego State JuneUniversity) 22, 2011 2 / 36
P4 POLYNOMIALS 11 monomial 3x 4 34x 4 y 4 2x 2 +1 5x 3 + x 1/2 +5 2x 3 1 x 1 23x 3 +x 2 monomial monomial binomial trinomial is not a polynomial is not a polynomial is not a polynomial Note: Polynomials are sums of monomials. Professor Tim Busken (Graduate T.A.Dynamical P4 Polynomials Systems ProgramDepartment and P5 Factoringof Polynomials Mathematics San Diego State JuneUniversity) 22, 2011 3 / 36
P4 POLYNOMIALS Defn: Polynomial Expression Definition (Polynomial Expression) Suppose n is a non-negative integer. A polynomial expression (in x) has the form: a n x n +a n 1 x n 1 + +a 3 x 3 +a 2 x 2 +a 1 x +a 0 where a n,a n 1,...,a 1,a 0 are constants (just numbers) and a n 0. 1 The numbers a n,a n 1,...,a 3,a 2,a 1,a 0 are called the COEFFICIENTS of the polynomial. In plain terms, coefficients are the numbers in front of the variables. 2 a n x n is called the LEADING TERM of the polynomial. 3 a n is called the LEADING COEFFICIENT of the polynomial. 4 n is called the DEGREE of the polynomial. Professor Tim Busken (Graduate T.A.Dynamical P4 Polynomials Systems ProgramDepartment and P5 Factoringof Polynomials Mathematics San Diego State JuneUniversity) 22, 2011 4 / 36
P4 POLYNOMIALS more ex s Examples: For the polynomials given below, find the leading term, leading coefficient, and the degree of the polynomial: Polynomial Degree Leading Leading Constant Expression Term Coefficient term 2x 4 3x 5 4 2x 4 2 5 x 5 3x 6 10x 4 6 3x 6 3 4 5x 10 8x 3 10x +5 10 5x 10 5 5 17x +4 1 17x 17 4 24 0 24 24 24 Professor Tim Busken (Graduate T.A.Dynamical P4 Polynomials Systems ProgramDepartment and P5 Factoringof Polynomials Mathematics San Diego State JuneUniversity) 22, 2011 5 / 36
P4 POLYNOMIALS Poly Add/Subtract Polynomial addition and subtraction is as one would expect. Example Compute the difference (x 2 5x) (3x 2 4) (x 2 5x) (3x 2 4) = = (x 2 5x) 1(3x 2 4) since a = ( 1) a = (x 2 5x)+( 1)(3x 2 4) since a b = a+( b) = (x 2 5x) 3x 2 +4x +1 distr. prop = x 2 5x 3x 2 +4x +1 assoc. prop = (x 2 3x 2 )+( 5x +4x)+1 comm. and assoc. props = 2x 2 x +1 addn closure prop Professor Tim Busken (Graduate T.A.Dynamical P4 Polynomials Systems ProgramDepartment and P5 Factoringof Polynomials Mathematics San Diego State JuneUniversity) 22, 2011 6 / 36
P4 POLYNOMIALS Combining Like Terms Definition (Like Terms) Like terms are terms that contain the same variable(s) raised to the same power(s). Like terms can be combined or collected together. Example Identify the like terms in 4x 3 +5x 7x 2 +2x 3 +x 2 Solution: like terms: 4x 3 and 2x 3 like terms: 7x 2 and x 2 same variable and exponent same variable and exponent Example Identify the like terms in 8x 2 y 2 +4x 6x 5 +2x 2 y 2 Solution: like terms: 8x 2 y 2 and 2x 2 y 2 same variables and exponents Professor Tim Busken (Graduate T.A.Dynamical P4 Polynomials Systems ProgramDepartment and P5 Factoringof Polynomials Mathematics San Diego State JuneUniversity) 22, 2011 7 / 36
P4 POLYNOMIALS Multiplying Poly s Multiplying Polynomials Professor Tim Busken (Graduate T.A.Dynamical P4 Polynomials Systems ProgramDepartment and P5 Factoringof Polynomials Mathematics San Diego State JuneUniversity) 22, 2011 8 / 36
P4 POLYNOMIALS Multiplying Poly s The product of two binomials results in four terms before the like terms are combined. The acronym Foil stands for FIRST, OUTER, INNER, LAST, and should remind you how to compute the product of two binomials. Consider the following product: F {}}{{}}{{}}{{}}{ (a+b)(c +d) = a(c +d)+b(c +d) = ac + ad + bc + bd The product of the two binomials consists of four terms: the product of the FIRST term of each (ac), the product of the OUTER term of each (ad), the product of the INNER term of each (bc), and the product of the LAST term of each (bd). 2 Examples: a.) (x +4) (2x 3) b.) (3 6 2 5) 2 O I L Professor Tim Busken (Graduate T.A.Dynamical P4 Polynomials Systems ProgramDepartment and P5 Factoringof Polynomials Mathematics San Diego State JuneUniversity) 22, 2011 9 / 36
P4 POLYNOMIALS Trinomial times Binomial Example: Multiply (x 2 3x +4) (2x 3) Solution (x 2 3x +4) (2x 3) = = (2x 3) (x 2 3x +4) comm prop = 2x (x 2 3x +4)+( 3) (x 2 3x +4) distr. prop = 2x 3 6x 2 +8x 3x 2 +92 distr. prop = 2x 3 +( 6x 2 3x 2 )+(8x +9x) 12 comm., assoc. + = 2x 3 9x 2 +172 addn closure prop Professor Tim Busken (Graduate T.A.Dynamical P4 Polynomials Systems ProgramDepartment and P5 Factoringof Polynomials Mathematics San Diego State June 22, University) 2011 10 / 36
Example: Simplify Solution 3 5 2 = 3 5 2 (1) P4 POLYNOMIALS Conjugate the Denominator 3 5 so that no radicals are in the denominator. 2 3 = 5 2 5+ 2 5+ 2 since 5+ 2 5+ 2 = 1 = 3(5+ 2) 5 2 ( 2) 2 since (a b) (a+b) = a 2 b 2 = 15+3 2 25 2 = 15+3 2 23 Distr. prop & n an = a closure = 15 23 + 3 2 23 since a+b c = a c + b c Professor Tim Busken (Graduate T.A.Dynamical P4 Polynomials Systems ProgramDepartment and P5 Factoringof Polynomials Mathematics San Diego State June 22, University) 2011 11 / 36
p5 Factoring Polynomials p5 Factoring Polynomials Professor Tim Busken (Graduate T.A.Dynamical P4 Polynomials Systems ProgramDepartment and P5 Factoringof Polynomials Mathematics San Diego State June 22, University) 2011 12 / 36
p5 Factoring Polynomials #1 RULE: FACTOR OUT THE GCF Factoring reverses multiplication. You should already be familiar with factoring second degree polynomials (quadratics). Consider the polynomial expression 6x 2 3x. The two terms of the given expression have a greatest common factor, 3x. 6x 2 3x = (2x) (3x) (3x) (1) = 3x (2) Distr. prop We call this process factoring out the gcf. Definition (The #1 Rule of Factoring Anything) The first step to any factoring any algebraic expression is to identify if there is a greatest common factor (gcf) amongst all the terms given. If the gcf 1, then you must always factor out the gcf before using any other tricks in your factoring toolbox. Professor Tim Busken (Graduate T.A.Dynamical P4 Polynomials Systems ProgramDepartment and P5 Factoringof Polynomials Mathematics San Diego State June 22, University) 2011 13 / 36
p5 Factoring Polynomials Factor By Grouping Method Polynomials with four terms can sometimes be factored by grouping. Example Factor x 4 2x 3 8x +16 Solution: x 4 2x 3 8x +16 = (x 4 2x 3 )+( 8x +16) (assoc. prop +) [ ] [ ] = x (x 3 )+( 2) (x 3 ) + ( 8) x +( 8) ( 2) = x 3 (x 2) 8(x 2) (distr. prop.) = x 3 (x 2) 8(x 2) (identify common factor) = (x 2)(x 3 8) (distr. prop) **Technically we are not done. This is not the prime factorization of the given polynomial since x 3 8 can be factored with the difference of cubes formula. Professor Tim Busken (Graduate T.A.Dynamical P4 Polynomials Systems ProgramDepartment and P5 Factoringof Polynomials Mathematics San Diego State June 22, University) 2011 14 / 36
p5 Factoring Polynomials ac-grouping method for quadratics Problem: Factor ax 2 +bx +c Problem: Factor 10x 2 11x 6 (1) Multiply a times c. (1) a = 10, b = 11, c = 6, so clearly a c = 60 (2) List all possible pairs of 60 60 numbers whose product is ac ւց ւց 6 10 1 60 6 ( 10) 1 ( 60) 3 ( 20) 5 12 3 20 5 ( 12) 15 4 15 ( 4) (3) Box the pair whose sum is bր (3) b = 11, and 15+4 = 11 Professor Tim Busken (Graduate T.A.Dynamical P4 Polynomials Systems ProgramDepartment and P5 Factoringof Polynomials Mathematics San Diego State June 22, University) 2011 15 / 36
p5 Factoring Polynomials ac-grouping method for quadratics Problem: Factor ax 2 +bx +c Problem: Factor 10x 2 11x 6 (4) Replace b with the sum (4) 10x 2 11x 6 of the circled pair. Distribute = 10x 2 +( 15x +4x) 6 x into this quantity = 10x 2 15x +4x 6 (5) Now factor by grouping: (5) (10x 2 15x)+(4x 6) Use parenthesis to group the first two terms, and another () = 5x(2x 3)+2(2x 3) to group the second two terms. = 5x(2x 3)+2(2x 3) = (2x 3) (5x +2) Professor Tim Busken (Graduate T.A.Dynamical P4 Polynomials Systems ProgramDepartment and P5 Factoringof Polynomials Mathematics San Diego State June 22, University) 2011 16 / 36
Special Case: Division by a Monomial Example: Divide 6x3 9x 2 +12x 3x Solution 6x 3 9x 2 +12x 3x = 6x3 3x + 9x2 + 12x 3x 3x = 2x2 3x +4 Professor Tim Busken (Graduate T.A.Dynamical P4 Polynomials Systems ProgramDepartment and P5 Factoringof Polynomials Mathematics San Diego State June 22, University) 2011 17 / 36
Division Algorithm Theorem (Division Algorithm:) Suppose D(x) and P(x) are polynomial functions (chapter 2) of x with D(x) 0, and suppose that D(x) is less than the degree of P(x). Then there exist unique polynomials Q(x) and R(x), where R(x) is either 0 or has degree less than the degree of D(x), such that P(x) = Q(x)D(x)+R(x) or, equivalently P(x) R(x) = Q(x)+ D(x) D(x) In words, we have or dividend = (quotient)(divisor) + remainder dividend divisor = quotient + remainder divisor. Professor Tim Busken (Graduate T.A.Dynamical P4 Polynomials Systems ProgramDepartment and P5 Factoringof Polynomials Mathematics San Diego State June 22, University) 2011 18 / 36
Division Algorithm In any case, x3 +2x 2 x 2 problem and solution: is equivalent to the following long division x 2 + 3x + 2 Divisor, D(x) x 3 + 2x 2 x 2 x 3 x 2 և Quotient, Q(x) և Dividend, P(x) 3x 2 x 2 3x 2 3x 2x 2 2x 2 0 և Remainder, R(x) Professor Tim Busken (Graduate T.A.Dynamical P4 Polynomials Systems ProgramDepartment and P5 Factoringof Polynomials Mathematics San Diego State June 22, University) 2011 19 / 36
Long Division Example Example: Divide x3 +2x 2 x 2 Solution: Step 1: set the problem up for long division: x 3 + 2x 2 x 2 Professor Tim Busken (Graduate T.A.Dynamical P4 Polynomials Systems ProgramDepartment and P5 Factoringof Polynomials Mathematics San Diego State June 22, University) 2011 20 / 36
Long Division Example Example: Divide x3 +2x 2 x 2 Solution: Step 2: Compute x3 x = x2 x 2 x 3 + 2x 2 x 2 Professor Tim Busken (Graduate T.A.Dynamical P4 Polynomials Systems ProgramDepartment and P5 Factoringof Polynomials Mathematics San Diego State June 22, University) 2011 21 / 36
Long Division Example Example: Divide x3 +2x 2 x 2 Solution: Step 3: Compute the product x 2 () and list the result below x 3 +2x 2 x 2 x 2 x 3 + 2x 2 x 2 Professor Tim Busken (Graduate T.A.Dynamical P4 Polynomials Systems ProgramDepartment and P5 Factoringof Polynomials Mathematics San Diego State June 22, University) 2011 22 / 36
Long Division Example Example: Divide x3 +2x 2 x 2 Solution: Step 3: Compute the product x 2 ()= x 3 x 2 and list the result below x 3 +2x 2 x 2 x 2 x 3 + 2x 2 x 2 x 3 x 2 Professor Tim Busken (Graduate T.A.Dynamical P4 Polynomials Systems ProgramDepartment and P5 Factoringof Polynomials Mathematics San Diego State June 22, University) 2011 23 / 36
Long Division Example Example: Divide x3 +2x 2 x 2 Solution: Step 4: CHANGE THE SIGNS AND ADD THE RESULT x 2 x 3 + 2x 2 x 2 x 3 + x 2 0 + 3x 2 Professor Tim Busken (Graduate T.A.Dynamical P4 Polynomials Systems ProgramDepartment and P5 Factoringof Polynomials Mathematics San Diego State June 22, University) 2011 24 / 36
Long Division Example Example: Divide x3 +2x 2 x 2 Solution: Step 5: Bring -x 2 down. x 2 x 3 + 2x 2 x 2 x 3 + x 2 3x 2 x 2 Professor Tim Busken (Graduate T.A.Dynamical P4 Polynomials Systems ProgramDepartment and P5 Factoringof Polynomials Mathematics San Diego State June 22, University) 2011 25 / 36
Long Division Example Example: Divide x3 +2x 2 x 2 Solution: Repeat Steps 2-6 now on 3x 2 x 2. Step 2: Compute 3x2 x = 3x and add the result to the quotient, Q(x). x 2 + 3x x 3 + 2x 2 x 2 x 3 + x 2 3x 2 x 2 Professor Tim Busken (Graduate T.A.Dynamical P4 Polynomials Systems ProgramDepartment and P5 Factoringof Polynomials Mathematics San Diego State June 22, University) 2011 26 / 36
Long Division Example Example: Divide x3 +2x 2 x 2 Solution: Repeat Step 3: Compute the product 3x () and list the result below 3x 2 x 2. x 2 + 3x () x 3 + 2x 2 x 2 x 3 + x 2 3x 2 x 2 Professor Tim Busken (Graduate T.A.Dynamical P4 Polynomials Systems ProgramDepartment and P5 Factoringof Polynomials Mathematics San Diego State June 22, University) 2011 27 / 36
Long Division Example Example: Divide x3 +2x 2 x 2 Solution: Repeat Step 3: Compute the product 3x ()= 3x 2 3x and list the result below 3x 2 x 2. x 2 + 3x () x 3 + 2x 2 x 2 x 3 + x 2 3x 2 x 2 3x 2 3x Professor Tim Busken (Graduate T.A.Dynamical P4 Polynomials Systems ProgramDepartment and P5 Factoringof Polynomials Mathematics San Diego State June 22, University) 2011 28 / 36
Long Division Example Example: Divide x3 +2x 2 x 2 Solution: Repeat Step 4: CHANGE THE SIGNS AND ADD THE RESULT x 2 + 3x () x 3 + 2x 2 x 2 x 3 + x 2 3x 2 x 2 3x 2 + 3x 0 + 2x Professor Tim Busken (Graduate T.A.Dynamical P4 Polynomials Systems ProgramDepartment and P5 Factoringof Polynomials Mathematics San Diego State June 22, University) 2011 29 / 36
Long Division Example Example: Divide x3 +2x 2 x 2 Solution: Repeat Step 5: Bring 2 down. x 2 + 3x () x 3 + 2x 2 x 2 x 3 + x 2 3x 2 x 2 3x 2 + 3x 0 + 2x 2 Professor Tim Busken (Graduate T.A.Dynamical P4 Polynomials Systems ProgramDepartment and P5 Factoringof Polynomials Mathematics San Diego State June 22, University) 2011 30 / 36
Long Division Example Example: Divide x3 +2x 2 x 2 Solution: Repeat Steps 2-6 now on 2x 2. Step 2: Compute 2x x = 2 and add the result to the quotient, Q(x). x 2 + 3x + 2 (x 1) x 3 + 2x 2 x 2 x 3 + x 2 3x 2 x 2 3x 2 + 3x 2x 2 Professor Tim Busken (Graduate T.A.Dynamical P4 Polynomials Systems ProgramDepartment and P5 Factoringof Polynomials Mathematics San Diego State June 22, University) 2011 31 / 36
Long Division Example Example: Divide x3 +2x 2 x 2 Solution: Repeat Step 3: Compute the product 2 () and list the result below 2x 2. x 2 + 3x + 2 () x 3 + 2x 2 x 2 x 3 + x 2 3x 2 x 2 3x 2 + 3x 2x 2 Professor Tim Busken (Graduate T.A.Dynamical P4 Polynomials Systems ProgramDepartment and P5 Factoringof Polynomials Mathematics San Diego State June 22, University) 2011 32 / 36
Long Division Example Example: Divide x3 +2x 2 x 2 Solution: Repeat Step 3: Compute the product 2 ()= 2x 2 and list the result below 2x 2. x 2 + 3x + 2 () x 3 + 2x 2 x 2 x 3 + x 2 3x 2 x 2 3x 2 + 3x 2x 2 2x 2 Professor Tim Busken (Graduate T.A.Dynamical P4 Polynomials Systems ProgramDepartment and P5 Factoringof Polynomials Mathematics San Diego State June 22, University) 2011 33 / 36
Example: Divide x3 +2x 2 x 2 Long Division Example Solution: Repeat Step 4: CHANGE THE SIGNS AND ADD THE RESULT x 2 + 3x + 2 () x 3 + 2x 2 x 2 x 3 + x 2 3x 2 x 2 3x 2 + 3x 2x 2 2x +2 0 Hence the remainder, R(x) = 0, and x 3 +2x 2 x 2 = x 2 +3x +2 Professor Tim Busken (Graduate T.A.Dynamical P4 Polynomials Systems ProgramDepartment and P5 Factoringof Polynomials Mathematics San Diego State June 22, University) 2011 34 / 36
Long Division Example Moreover x 3 +2x 2 x 2 has the following factorization: x 3 +2x 2 x 2 = ()(x 2 +3x +2) = ()(x +1)(x +2) Professor Tim Busken (Graduate T.A.Dynamical P4 Polynomials Systems ProgramDepartment and P5 Factoringof Polynomials Mathematics San Diego State June 22, University) 2011 35 / 36
Long Division Example x 2 + 3x + 2 x 3 + 2x 2 x 2 x 3 x 2 3x 2 x 2 3x 2 3x 2x 2 2x 2 0 Professor Tim Busken (Graduate T.A.Dynamical P4 Polynomials Systems ProgramDepartment and P5 Factoringof Polynomials Mathematics San Diego State June 22, University) 2011 36 / 36