Chapter 4 Polynomials TSWBAT determine if an expression is a monomial, binomial, or polynomial; determine the degree of a monomial or polynomial; add and subtract two polynomials.

Section 1 – Polynomials cont. Monomial – a constant, a variable, or the product of a constant and one or more variables. Ex. -7, u, 1/3m 2, -s 2t 3, 6xy 3 Binomial – a polynomial that has two terms. Ex. 2x +3 Trinomial – a polynomial that has three terms. Ex. X2 + 4x – 5

Section 1 – Polynomials cont. Degree of a variable – the number of times the variable occurs as a factor in the monomial or the exponent of the variable. Ex. 6xy 3 has variables x and y. X has degree 1 and y has degree 3.

Section 1 – Polynomials cont. Degree of a monomial – the sum of the degrees of the variables in the monomial. A nonzero constant has degree 0. The constant 0 has no degree. Ex. 6xy 3 has degree 0+1+3=4 -s 2t 3 has degree 0+2+3=5 u has degree 1 -7 has degree 0

Section 1 – Polynomials cont. Similar (or like) Monomials – monomials that are identical or that differ only in their coefficients. Ex. -s 2t 3 and 2s 2t 3 are similar 6xy 3 and 6x 3y are not similar

Section 1 – Polynomials cont. Polynomial – a monomial or a sum of monomials. The monomials in a polynomial are called the terms of the polynomial. Ex. x2 + (-4)x + 5 or x2 – 4x + 5 The terms are x2, -4x, and 5

Section 1 – Polynomials cont. Simplified polynomial – a polynomial in which no two terms are similar. The terms are usually arranged in order of decreasing degree of one of the variables, usually x. Ex. 2x3 – 5 + 4x + x3 is not simplified but 3x3 + 4x – 5 is simplified.

Section 1 – Polynomials cont. Degree of a polynomial – the greatest of the degrees of its terms after it has been simplified. Ex. x4 – 2x2y3 + 6y – 11 has terms with degree 4, 5, 1, and 0. Thus the polynomial has degree 5 the largest degree of its terms.

Section 1 – Polynomials cont. Addition of Polynomials – to add two or more polynomials, write their sum and then simplify by combining similar terms. Ex. Add 2x2 – 3x + 5 and x3 – 5x2 + 2x – 5 2x2 – 3x + 5 + x3 – 5x2 + 2x – 5 x3 – 3x2 - x

Section 1 – Polynomials cont. Subtracting Polynomials – to subtract one polynomial from another, add the opposite of each term of the polynomial you are subtracting. Ex. Subtract 2x2 – 3x + 5 from x3 – 5x2 + 2x – 5 x3 – 5x2 + 2x – 5 x3 – 5x2 + 2x – 5 - (2x2 – 3x + 5) - 2x2 + 3x – 5 x3 – 7x2 + 5x – 10

Multiplying Polynomials TSWBAT multiply Polynomials using the FOIL method, the Rules of Special Products, and the distributive Property.

Section 3 – Multiplying Polynomials If you are multiplying two binomials together it is best to use the FOIL method of multiplication. FOIL Method – F - Multiply the two first terms O – Multiply the two outer terms I – Multiply the two inner terms L – Multiply the two last terms

Section 3 – Multiplying Polynomials cont. Combining the four terms into one polynomial and combining the like terms of O and I we get the following as a result: 6a2 + 7ab – 5b2 Example of FOIL (2a – b)(3a + 5b) F – (2a)(3a) = 6a2 O – (2a)(5b) = 10ab I – (-b)(3a) = -3ab L – (-b)(5b) = -5b2

Section 3 – Multiplying Polynomials cont. Besides the FOIL Method there are also three special products to know, that will help you multiply polynomials and their terms faster. Special Products (a+b)2 = a2 + 2ab + b2 (a-b)2 = a2 -2ab + b2 (a+b)(a-b) = a2 + b2

Section 3 – Multiplying Polynomials cont. Besides these methods of FOIL and Special Products there is also the old fashioned Distribution Method. Ex. (2x+3)(x2+4x-5) 2x(x2+4x-5)+3(x2+4x-5) 2x(x2)+2x(4x)-2x(5)+3(x2)+3(4x)-3(5) 2x3+8x2-10x+3x2+12x-15 Combine Similar Terms 2x3+11x2+2x-15

Exponents TSWBAT Use the laws of exponents in polynomial multiplication and simplification.

Section 2 – Laws of Exponents There are five laws of exponents – Let a and b be real numbers and m and n be positive integers. Then: 1. am * an = am+n 2. (ab)m = ambm 3. (am)n = am*n 4. (a/b)n = an/bn 5. am/an = am-n

Section 2 – Laws of Exponents Examples

Factoring TSWBAT find the prime factorization of a number and the greatest common factor or least common multiple of two or more numbers or monomials.

Section 4 – Prime Factorization Factor Set – The set of all numbers that are a factor of a given number. Factor – One of two numbers that multiplied together form a given number. Prime Number or Prime – a number whose only factors are 1 and itself. Composite Number – any number that has more than one set of factors. Activity – Primes Chart

Section 4 – Prime Factorization Greatest Common Factor (GCF) – of two or more integers is the greatest integer that is a factor of each. Ex. GCF of 15 and 30 1. Find all the factors of each number. 2. Find the greatest number in both sets. This is your GCF. Factors of 15 Factors of 30 1 15 30 2 3 5 10 6

Section 4 – Prime Factorization 12 24 36 48 60 30 90 120 150 Least Common Multiple (LCM) – of two or more integers is the least positive integer having each as a factor. Ex. Find the LCM of 12 and 30. 1. Find the multiples of the two numbers. 2. Find the least common number among both numbers. That is your LCM.

Section 4 – Prime Factorization The GCF and LCM also hold true for Monomials with a slight variation as you now need to be aware of the variables. GCF of Monomials – the common factor with the greatest degree and greatest numerical coefficient. LCM of Monomials – the common multiple that has the least degree and least positive numerical coefficient.

Section 4 – Prime Factorization Ex. of GCF of Monomials 48u2v2 and 60uv3w. Find GCF of Coefficients 48 – 1,48, 2,24, 3,16, 4,12, 6,8 60 – 1,60, 2,30, 3,20, 4,15, 5,12, 6,10 2. Compare the variables. Only variables in both can be used in GCF. Then use the smaller exponent variable. u2 or u – use u v2 or v3 – use v2 3. Therefore our GCF is 12uv2

Section 4 – Prime Factorization Ex. of LCM of Monomials 48u2v2 and 60uv3w Find LCM of coefficients. 48,96,144,192,240,288 60,120,180,240,300,360 2. Compare the variables. All variables in both must be used in LCM. Then use the greater exponent variable. u2 or u – use u2 v2 or v3 – use v3 use w 3. Therefore the LCM is 240u2v3w

Section 4 – Prime Factorization Prime Factorization – is the factorization of a number down to the product of a set of prime numbers. The most common way to find the prime factorization is to use a factor tree working down to all prime factors. Ex. of Prime Factorization 936=2*468 2*2*234 2*2*2*117 2*2*2*3*39 2*2*2*3*3*13 Therefore the Prime Factorization is 23*32*13

Factoring Quadratic Polynomials Quadratic Polynomials – or second-degree polynomials are polynomials of the form ax2+bx+c where (a≠0). Ex. X2+0x+25 Quadratic Polynomials have three terms each with its own name. Quadratic Term – is the ax2 term Linear Term – is the bx term Constant Term – is the c term

Factoring Quadratic Polynomials Quadratic Trinomial – a quadratic polynomial in which a, b, and c are all nonzero integers. Ex. x2+2x-15 In this section we will factor a special type of quadratic polynomial the perfect square trinomial. In the next section we will look at how to factor any Quadratic Polynomial.

Section 5 – Factoring Polynomials 5 Special Factorings – In section 3 we had 3 special products. Those products play a role here in factoring as well. Besides those three there are also two other special cases in factoring. On the following slide you will find these 5 cases. (Hint – Take an index card and write these 5 cases on it to help you when doing Problems!!!)

Section 5 – Factoring Polynomials Special Factorings Perfect Square Trinomials Positive a2 + 2ab + b2 = (a + b)2 Negative a2 - 2ab + b2 = (a - b)2 Difference of Squares a2 – b2 = (a + b)(a – b) Sum and Difference of Cubes Sum a3 + b3 = (a + b)(a2 – ab + b2) Difference a3 - b3 = (a - b)(a2 + ab + b2)

Section 5 – Factoring Polynomials Example of a Perfect Square Trinomial Positive –> z2+6z+9= (z+3)2 Negative -> 4s2-4st+t2 = (2s-t)2 Example of a Difference of Squares 25x2-16a2 = (5x+4a)(5x-4a) Example of Cubes Sum ->8u3+v3 = (2u+v)(4u2-2uv+v2) Difference -> y3-1 = (y-1)(y2+y+1)

Section 5 – Factoring Polynomials Common Numerical Powers Activity You can also factor a polynomial by factoring out the Greatest Common Monomial Factor. Ex. 2x4-4x3+8x2 Step 1 – Find the GCF, which here it is 2x2 Step 2 – Divide out 2x2 from each term. Step 3 – Write out your answer. 2x2(x2-2x+4)

Section 5 – Factoring Polynomials One last way to factor a polynomial is to rearrange and group terms that you might be able to find a GCF from. Ex. 3xy-4-6x+2y If we group the 1st and 3rd terms together they have a common factor of 3x and then grouping the 2nd and 4th terms they have a common factor of 2. (3xy-6x)+(2y-4) -> 3x(y-2)+2(y-2) Now we have a common factor of (y-2) that we can pull out and we are left with (y-2)(3x+2).

Section 6 – Factoring Quadratic Polynomials Factoring a Quadratic Polynomial – ax2+bx+c can be factored into the product (px+q)(rx+s) where p, q, r, and s are integers. Then ax2+bx+c = prx2+(ps+qr)x+qs. So a = pr, b=(ps+qr), and c = qs

Section 6 – Factoring Quadratic Polynomials Lets look at an example of how to factor a Quadratic Polynomial. 15t2-16t+4 -> 1. Lets begin by looking at the pr term which is 15 here. There are two ways in which to factor 15t2. They are (t___)(15t____) or (3t____)(5t____).

Section 6 – Factoring Quadratic Polynomials 2. We need to look at the qs term of 4. Since 4 is positive, both factors are either positive or negative. To determine this we look at the middle term or -16t. Since this is negative both of our factors of 4 will be negative. Thus there are 3 negative combinations that factor 4. They are (___-1)(_-4), (_-2)(_-2), or (_-4)(__-1).

Section 6 – Factoring Quadratic Polynomials If we take our two possible answers for the first terms and our three for our second term, we find we have 6 possible factoring possibilities. These are shown in the table. (t-1)(15t-4) (3t-1)(5t-4) (t-2)(15t-2) (3t-2)(5t-2) (t-4)(15t-1) (3t-4)(5t-1)

Section 6 – Factoring Quadratic Polynomials To find out which of these possibilities work, we check each of them by doing the O and I part of FOIL on each pair to find the grouping that gives us -16t. (t-1)(15t-4) -15t-4t=-19t (t-2)(15t-2) -30t-2t=-32t (t-4)(15t-1) -60t-t=-61t (3t-1)(5t-4) -5t-12t=-17t (3t-2)(5t-2) -10t-6t=-16t (3t-4)(5t-1) -20t-3t=-23t

Section 6 – Factoring Quadratic Polynomials Therefore our quadratic polynomial 15t2-16t+4 factors down into (3t-2)(5t-2). This polynomial is now factored completely as it is written as the product of factors that are prime polynomials. It could also be factored completely if we had monomials 2x(3y), or a power of a prime polynomial like (x-2)2 .

Section 6 – Factoring Quadratic Polynomials Prime Polynomial – a polynomial that is not reducible and the greatest common factor of its coefficients is 1. Ex. X2+4x-3 is a Prime Polynomial, since it can not be factored into two lower degree polynomials with integers and the GCF of its coefficients is 1. 2x2+8x-6 is not Prime because the GCF of its coefficients is 2. Thus a 2 can be factored out.

Section 6 – Factoring Quadratic Polynomials Irreducible polynomial – a polynomial that contains more than one term and cannot be expressed as a product of lower degree polynomials taken from its factor set. Ex. X2+4x-3 Why? Factor sets are (x-1)(x+3) and (x+1)(x-3) and neither of these sets gives a middle term of +4. (They give +2x and -2x.)

Section 7 – Solving Polynomial Equations Polynomial Equation – an equation that is equivalent to an equation with a polynomial as one side and 0 as the other. Ex. x2-5x-24=0 and x2=5x+24 Root – is a solution to the polynomial equation. Solution – a value of the variable that satisfies the equation. Ex. Roots and solutions for the above problem are -3 and 8.

Section 7 – Solving Polynomial Equations To solve a polynomial equation we use the Zero-Product Property. Step 1 – Write the equation in Zero Equals Form. Ex. If we take the problem x2=5x+24, we need to rewrite it as x2-5x-24=0

Section 7 – Solving Polynomial Equations Step 2 – Factor the polynomial side of the equation. Ex. From our equation x2-5x-24=0 we factor the left side into (x-8)(x+3)=0 Step 3 – Solve the equation by setting each factor equal to zero. Ex. Taking our factor of (x-8)(x+3) we set each equal to zero and solve for x. x-8=0 and x+3=0 -> x=8 and x=-3

Section 7 – Solving Polynomial Equations Double Root or Double Zero – of an equation is a solution of the equation that appears twice. Ex. If we have (x-2)(x-2) as factors we have the solution of 2 twice. This means that 2 is a double root of that equation and can be written x=2d.r.

Section 7 – Solving Polynomial Equations Multiple Root or Multiple Zero - of an equation is a solution of the equation that appears more than twice. Ex. If we have (t-1)(t-1)(t-1)(t-1) as factors we have the solution of 1 four times. This means that 1 is a multiple root of that equation and can be written t=1m.r.4.

Section 7 – Solving Polynomial Equations Using the Zero-Product Property we can also find the zeros of a polynomial function. That means find the values for when f(x)=0 Ex. f(x)=x2-9 following the steps we factor the polynomial side to (x-3)(x+3) (special products) and then solve to x=3 and x=-3.

Section 8 – Problem Solving Using Polynomial Equations The seven steps of word problems are: 1. Read and Write the problem 2. Draw a picture 3. Define the variables 4. Re-Read the problem and set up the basic problem 5. Fill-in values from the problem 6. Solve the equation 7. Check your solutions (this is critical now because not all values will be accurate solutions)

Section 8 – Problem Solving Using Polynomial Equations Example – Step 1 – Read and Write the problem. - A graphic artist is designing a poster that consists of a rectangular prism with a uniform border. The print is to be twice as tall as it is wide, and the border is to be 3 inches wide. If the area of the poster is to be 680 inches squared, find the dimensions of the print.

Section 8 – Problem Solving Using Polynomial Equations Step 2 – Draw a picture – See overhead. Step 3 – Define the variable – w = width of print Step 4 – Label drawing – See Overhead Step 5 – Create equation – using area formula – A = L*W -> 680 = (w+6)(2w+6)

Section 8 – Problem Solving Using Polynomial Equations Step 6 – Solve for variable. 680 = (w+6)(2w+6) -> 680 = 2w2+18w +36 -> 0 = 2w2+18w-644 -> Divide by 2 -> 0 = w2 +9w -322 -> (w-14)(w+23) = 0 -> w-14 = 0 and w+23 = 0 -> w = 14 or w = -23 Step 7 – Check solutions – We cannot have a negative length thus x = -23 does not work and our solution is x = 14.

Section 9 – Solving Polynomial Inequalities Polynomial Inequality – an inequality that is equivalent to an inequality with a polynomial as one side and zero as the other side. Ex. x2-x-6>0 and x2>x+6

Section 9 – Solving Polynomial Inequalities To solve a polynomial inequality we start by following the first two steps of that of an equation. Step 1 - We make one side zero and the other side the polynomial. Ex. x2>x+6 becomes x2-x-6>0

Section 9 – Solving Polynomial Inequalities Step 2 - We factor the polynomial side of the inequality. Ex. x2-x-6>0 becomes (x-3)(x+2)>0. Step 3 is actually a two part step. Part A – if the factors of the polynomial are >0 we will use the same inequality sign in Step 4. Part B – if the factors of the polynomial are <0 we will use opposite inequality signs in Step 4.

Section 9 – Solving Polynomial Inequalities Step 3 cont. – Ex. (x-3)(x+2)>0 our factors are >0 and thus for Step 4 we will use the same inequality sign. Step 4 – is another two part process. If you remember when we first talked about inequalities we had the terms Conjunction and Disjunction, well they return here in this step when we set the inequality factors to zero.

Section 9 – Solving Polynomial Inequalities Step 4 cont. – In this step we have two conjunctions joined together by a disjunction. So in our example, (x-3)(x+2)>0 is possible only if both factors are positive (conjunction) or (disjunction) both factors are negative (conjunction).

Section 9 – Solving Polynomial Inequalities Step 4 cont. – Ex. we have the following possibilities… x-3>0 and x+2>0 OR x-3<0 and x+2<0 Step 5 – solve for x… x>3 and x>-2 OR x<3 and x<-2 so… x>3 OR x<-2. Step 6 – Graph the solution.

Section 9 – Solving Polynomial Inequalities Example 2 – 3t<4 - t2 Step 1 – t2 + 3t – 4 <0 Step 2 – (t+4)(t-1)<0 Step 3 – Use Part B as factors are <0. Step 4 – (t+4)>0 and (t-1)<0 OR (t+4)<0 and (t-1)>0 Step 5 – t>-4 and t <1 OR t<-4 and t >1 -4<t<1 OR no solution Step 6 graph the one remaining solution.