Chapter 3: Factors and Products 3.1 Factors and Multiples of Whole Numbers In this chapter we will look at the topic of factors and products. In previous years, we examined these with only numbers, whereas in this chapter we will also take a look at factors and multiples of polynomials. We begin, however, with the concept of prime factorization. Definition: Prime factorization - We can determine the prime factorization of a number by slowly breaking it down, taking out one prime factor at a time. Recall that prime numbers include 2, 3, 5, 7, 11, 13, 17, Example: Determine the prime factorization of 12. Example: Determine the prime factorization of 18000. Example: Determine the prime factorization of 55125.
We can also employ factor trees, where we can break down a number into the product of two factors, and keep breaking that down until we reach prime values. Example: Use a factor tree to determine the prime factorization of 12. Example: Use a factor tree to determine the prime factorization of 18000. We can use prime factorization as a means to find the greatest common factor and least common multiple of a group of numbers. Definitions: Greatest common factor - Least common multiple -
To find the greatest common factor (GCF) of a group of numbers, we start by looking only at the prime factors that appear in all numbers of the group. The greatest number of times each prime factor can be removed is limited by the lowest exponent we see on that factor. Example: Determine the GCF of 28 and 20. Example: Determine the GCF of 18000 and 420. To find the least common multiple (LCM) of a group of numbers, we start by looking at all of the prime factors that appear in any numbers of the group. The least number of times that factor must be included is given by the highest exponent we see on that factor. Example: Determine the least common multiple of 28 and 20. Example: Determine the least common multiple of 72 and 108. Homework: textbook pages 139-141, #3-7, 8-10 (a, c, e only), 11 (a, c only), 12-18, 21-22
Chapter 3: Factors and Products 3.2 Perfect Squares, Perfect Cubes, and Their Roots A perfect square is a whole number that can be expressed as the product of a whole number by itself. For example, 5 5 = 25, therefore 25 is a perfect square. The square root of a number is that value that gets multiplied by itself. Just as 25 is the square of 5, we can say that 5 is the square root of 25. We write 25 = 5. A perfect cube is a whole number that can be expressed as a whole number to the power of 3. For example, 5! = 5 5 5 = 125, therefore 125 is a perfect cube. The cube root of a number is that value that we take the third power of. Just as 125 is the! cube of 5, we can say that 5 is the cube root of 125. We write 125 = 5. We can use prime factorization to determine whether a number is a perfect square or not. If a number is a perfect square, then If a number is a perfect cube, then Example: Determine whether 576 is a perfect square, perfect cube, or neither. If it is a perfect square or cube, state its square or cube root. Example: Determine whether 3375 is a perfect square, perfect cube, or neither. If it is a perfect square or cube, state its square or cube root.
Example: Determine whether 64 is a perfect square, perfect cube, or neither. If it is a perfect square or cube, state its square or cube root. Example: Determine whether 785 is a perfect square, perfect cube, or neither. If it is a perfect square or cube, state its square or cube root. Homework: textbook pages 146-147, #1-8, 11, 13 Quiz review: textbook page 149, #1-9
Chapter 3: Factors and Products 3.3 Common Factors of a Polynomial We will now begin introducing ourselves to factoring polynomials. To fully factor a polynomial there can be a few things to check for, but the first should be to see if there exists a common factor. Example: Determine the GCF of 6x! and 15x. Example: Factor the binomial 6x! + 15x. Example: Determine the GCF of 4, 16y, and 8y!. Example: Factor the trinomial 4 16y + 8y!. Let s try a few without the extra step! Example: Factor 4x! + 4x! + 8x. Example: Factor 2x! y + 8x! y! 2x! y.
Example: Factor 7xy + 7xz + 7yz. Example: Factor 27r! s! 18r! s! 36rs!. Example: Factor 81a! b! c! + 27a! bc! 24a! b! c! 9a! b! c!. Note: if the terms of a binomial have a GCF of only 1, then the binomial cannot be factored. Homework: textbook pages 155-156, #5-6, 8, 10, 12-18, 20, 22
Chapter 3: Factors and Products 3.5 Polynomials of the Form x 2 + bx + c When expanding a product such as 2x(x + 4) we would use the distributive property as follows: What we have above is the product of a monomial and a binomial. Suppose, however, that we are tasked with expanding the product of two binomials. We then end up using the distributive property twice. For example, expand (x + 2)(x + 3): What we see here is that each term in the first binomial must get multiplied by each term in the second binomial. We can use a process called FOIL to ensure this all gets taken care of. FOIL ensures we multiply the first terms, the outside terms, the inside terms, and the last terms. Example: Expand and simplify: (x + 4)(x + 1) (x 2)(x + 7) (a 6)(a 3) t + 5 t 5 In one of the examples we see that the product x + 4 x + 1 simplifies to x! + 5x + 4. If x! + 5x + 4 is the result of multiplying those two expressions, we can then declare that x + 4 and x + 1 must be factors of it. It logically follows that some trinomials can be factored.
We will look at polynomials of the form x! + bx + c - ones where the leading coefficient is 1. If they can be factored, the factors will be of the form (x ± an integer)(x ± an integer). What we seek in factoring these will be values that multiply to c and sum to b. With the above example of x! + 5x + 4, we can look at various factor pairs of 4 and determine which pair sums to 5: Hence we declare x! + 5x + 4 factors to (x + 4)(x + 1). Example: Factor the following: x! + 5x + 6 x! + 13x + 12 x! 2x 8 x! + 2x 8 x! x 20 Recall from the previous section that when factoring trinomials, we should first check for a common factor. This could make the process of fully factoring much easier.
Example: Factor the following: 3x! + 15x! 18x 128 16t 4t! It is worth noting that just as 3 4 is the same as 4 3, the product (x + a)(x + b) is the same as (x + b)(x + a). This is by the commutative property of multiplication. Similarly, we were also able to rearrange the order of the terms in the second example above. Just as 2 + 5 is the same as 5 + 2, a + b is the same as b + a since addition is also commutative. There exists another way to factor trinomials into a product of binomials. It involves a process of breaking up the middle term, partially factoring, then fully factoring. It can be a fairly involved process but has the advantage of potentially being more direct, though less efficient. Examples: Factor x! + 10x + 9 Factor x! + 14x + 24 Factor x! 5x 36 Factor x! + 3x 180 Homework: textbook pages 165-167, #3, 7, 9-15, 17-21, 23
Chapter 3: Factors and Products 3.6 Polynomials of the Form ax 2 + bx + c We continue factoring trinomials by examining those similar to the ones in the previous section, but such that there is a leading coefficient that cannot be removed by simple factoring (finding a common factor). In these cases we will want to closely examine factors of the product a c that sum to b. Example: Factor each of the following. 2x! + 5x + 2 3x! + 8x + 4 6x! 5xy + y! 10x! x 3 2x! + 11x + 15 12x! 4x 5 Alternately, we can use logical reasoning. We know that the first terms in the binomials must multiply to the first term of the trinomial, and we know that the last terms in the binomials must multiply to the last term of the trinomial, so we can try a few options to see what works. If no options work, then the trinomial is unfactorable. Example: Factor each of the following. 4x! + 20x + 9 6x! 11x 35
Once again, remember that we should check for a GCF before attempting other methods of factoring trinomials. Example: Factor each of the following. 3x! + 3x! 18x 8x! yz! 12x! yz! 80x! yz! Note that if you want to verify your answer, you can always expand your resultant answer to ensure it gets the trinomial you want. Homework: textbook pages 176-178, #2-4, 8-10, 12, 13, 15, 16, 18, 19, 21-23 Quiz review: textbook pages 180-181, #2a, 5-9
Chapter 3: Factors and Products 3.7 Multiplying Polynomials When we expanded the product of two binomials, we ensured that each term in the first binomial was multiplied by each term of the second binomial. We can expand this so as to apply it to higher-order polynomials, always ensuring that each term in the first polynomial gets multiplied by each term in the second. With binomial multiplication, we had two terms (in the first binomial) that each had to multiply with two terms (in the second binomial). Therefore, prior to simplification, the initial expansion had 2 2 = 4 terms. Similarly, the initial expansion of the product of a trinomial and a binomial would have 3 2 = 6 terms. We can use this as a quick check to make sure we ve done all the multiplication necessary. Example: Expand and simplify each of the following: (2x + 1)(x! 6x + 3) (x! 2x + 5)(2x! x 1)
When there are multiple operations being performed on polynomials, we must remember to follow the order of operations. Example: Simplify x 1 x + 2 + (x + 4)(x + 1) Example: Simplify 2x 3 x + 1 x(x + 2) Example: Simplify 2x + 5! (x + 15)(x + 5) Example: Simplify x + 2y! Homework: textbook pages 185-187, #1, 4-7, 8-10 (a, c only), 11, 13, 14, 15 (a, c, e), 17-19, 21-22
Chapter 3: Factors and Products 3.8 Factoring Special Trinomials Simplify each expression below: (x + 2)(x 2) (x 5)(x + 5) (2x + 3)(2x 3) (x! 4y)(x! + 4y) What is special about these particular cases of binomial multiplication? The result is known as a difference of squares. It is called a difference of squares because the two monomials present are square terms, and the operation between them is subtraction. We note that the product (a + b)(a b) is equal to a! b!, and we can thus conclude that it is possible to factor: a! b! = (a + b)(a b) Example: Determine if each expression below is a difference of squares. If it is, factor. x! 9 25x! 1 8x! 49 36x! y! 121y!
We now look at another case. Simplify each expression below. x + 3! x + 1! 2x 3! x 1! The results here are referred to as perfect square trinomials. They are called such because they are produced by squaring binomials. We note that since a + b! = a! + 2ab + b!, it logically follows that a! + 2ab + b! factors into a + b!. Example: Determine if each expression below is a perfect square trinomial. If it is, factor. x! + 4x + 4 x! 4x + 4 9x! + 30x + 25 4x! + 9 9x! + 30x 25 Note that general factoring practices still apply check for a GCF first, then use other factoring methods. Homework: textbook pages 194-195, #2, 4-9, 10-13 (odd letters only), 15, 17-21 Test review: textbook pages 198-200, #1-9, 11-14, 18-21, 24-36