Chris Kirkpatrick, Barbara Alldred, Andrew Dmytriw, Shawn Godin, Angelo Lillo, David Pilmer, Susanne Trew, Noel Walker · Grade 11 · Study material generated by AI from this edition. Not affiliated with the publisher.
Chapter 2 · pp. 90–145
This chapter shows how to decide whether two algebraic expressions mean the same thing. It covers adding, subtracting, multiplying, and factoring polynomials, and then extends these skills to rational expressions, their restrictions, graphs, and operations. Factoring is the reverse of expanding, and many later ideas depend on both.
Two polynomial expressions or functions are equivalent if they simplify to the same expression or give identical graphs. Equivalent expressions give the same output for every allowed input. For example, and both simplify to . If they simplify to different expressions, they are not equivalent. (pp. 94, 97)
There are three ways to test this:
If two expressions give different results at one input, they cannot be the same function. Fred's rocket problem shows this. His original difference function gives , but his simplified function gives 2200 at . One mismatch proves they are not equivalent. Using is handy because the arithmetic is easy. (pp. 95, 97)
The reverse does not work. If two functions give the same value at one input, they might still differ elsewhere. This leads to a common wrong belief: "same value at one input means equivalent." It is false because two different graphs can cross. For example, and both equal 1 at , but at they give 2 and 4. Equivalence needs matching values for every input. So one evaluation can prove non-equivalence, but it can never prove equivalence. (pp. 95, 97)
For expressions with several variables, try simple values such as 0, 1, and −1. For and , use , , . The first gives 1 and the second gives 0, so they are not equivalent. (p. 97)
Polynomials behave like numbers, because substituting numbers for the variables gives a number. So the rules for numbers carry over.
To add polynomials, write them as a sum and collect like terms. In the banquet problem, Nigel's cost is . Petra's cost is . Both simplify to , so the cost functions are equivalent. Regrouping terms does not change the value. (pp. 96–97)
To subtract one polynomial from another, add the opposite of the second polynomial, then collect like terms. This works like with integers. For change every sign in the second bracket. Then combine like terms to get . The most common error is changing only the first sign in the second bracket. (pp. 94, 97)
To multiply polynomials, use the distributive property. Multiply each term of one polynomial by each term of the other, then collect like terms. For :
Before collecting, the number of terms in the product equals the number of terms in the first polynomial times the number in the second. A binomial times a trinomial gives products. For example, has six products. (p. 107)
A linear polynomial times a quadratic gives a cubic. The degree of a product is the sum of the degrees of the factors.
For any polynomials , , and :
So you can multiply the first factor by the second or the second by the first and get the same result. (pp. 105, 115)
Subtraction is different. In general , and , except in special cases. (p. 115)
Multiply two of the binomials first to get a trinomial. Then multiply that trinomial by the third binomial. Because multiplication is associative, any order gives the same answer. For the box volume :
Multiplying the last two first gives the same result. (pp. 102–103)
A wrong belief is that the square of a sum is the sum of the squares, as in . Squaring means multiplying the whole expression by itself, so cross terms appear. The correct expansion is A quick check uses . The left side is and the right side of the wrong claim is . The values differ, so the claim is false. In the same way, , not . (pp. 103, 105)
To represent a change in area, write expressions for the original and new areas. Then subtract: change = new area − original area. (p. 104)
Suppose a rectangle has width and length , so its area is . Now make the length and the width . The new area is The change is . Since is positive, this is always negative, so the area decreases. (p. 104)
This corrects another wrong belief: that adding an amount to one side and subtracting the same amount from the other leaves the area unchanged. The two changes act on different dimensions, so they do not cancel. Here the area drops by . (p. 104)
Factoring a polynomial means writing it as a product of simpler polynomials. It undoes expanding. For example, expands to , and factoring brings you back to . You can always check a factoring by expanding your answer and comparing it with the original. (p. 111)
A polynomial is factored fully when the only common factors left in the product are 1 and −1. Take . It is not finished, because is still a difference of squares. The full factoring is . (p. 111)
Always begin by dividing out the greatest common factor (GCF) of all the terms. For the GCF is 2, which gives . This step makes the later steps easier and often reveals a hidden pattern. (pp. 110, 116)
To factor , look for two numbers with sum and product . For , the numbers 5 and −6 have sum −1 and product −30. So . Checking: . (pp. 109–110)
For with , use decomposition:
Example: . Here , and 5 and −6 have sum −1. Then
. (p. 110)
Sometimes no pair of integers works. For you need sum 1 and product 6. The pairs (1, 6) and (2, 3) add to 7 and 5, and negative pairs give negative sums. So no integer pair exists, and this trinomial cannot be factored over the integers. Test this first, so you do not waste time on other methods. (p. 110)
Difference of squares: . Take the square root of each term to find and . For , and , so the result is . A sum of squares such as does not factor this way. (pp. 110–111)
Perfect square trinomial: and . Check that the first and last terms are squares and that the middle term is twice the product of their square roots. In the roots are and 5, and . So it equals . (pp. 110–111)
With four or more terms, try grouping. Split the terms into groups, take the common factor out of each group, then take out the common binomial. For :
. (p. 111)
This only works when each group leaves the same binomial factor. If it does not, try different groupings.
Another kind of grouping builds a perfect square first. In , the first three terms are , and . The expression becomes a difference of squares:
. (p. 111)
Consider for natural numbers . Testing values gives , , and . The first factor looks like and the second like . Multiplying confirms the pattern. Grouping gives the same result:
.
For any natural number , and . So is always a product of two numbers greater than 1, and it is never prime. (pp. 108–109)
Testing values can help you spot a pattern, but the factoring is what proves it. Grouping is faster and works for any . Testing needs guessing and checking. (pp. 108–109)
The polynomial does not factor by the usual trinomial methods. To factor it, add and subtract to create a perfect square:
.
This is a difference of squares:
.
Each quadratic factors again, so the full result is . (p. 114)
Example: . (pp. 111, 114, 116)
The degree of a product is the sum of the degrees of the factors. A linear polynomial times a quadratic polynomial therefore gives a cubic. The highest-degree terms multiply as . For example, . (p. 117)
Polynomials follow the same rules as numbers, but subtraction is an exception. Subtraction means adding the opposite, so order and grouping matter. In general, and , except in special cases. With numbers, but . Also, but . Addition and multiplication do not have this problem. (p. 115)
A rational function is a function that is the ratio of two polynomials. It can be written , where and are polynomials and is not the zero polynomial. An example is . A rational expression is a quotient of polynomials, such as . The word "function" is used when the quotient defines an output for each input. (p. 118)
Rational functions come up in real problems. In the "2 and 1" charity game, a player's chance of winning is
where numbers are drawn from to . For , . Computing this from the original form is slow, so the goal is to find a simpler, equivalent expression. (p. 118)
Dividing by zero is not allowed. So any value of a variable that makes the denominator zero must be excluded. These excluded values are called restrictions. They are the zeros of the denominator. (p. 119)
Example: in , solving gives the restriction . (p. 119)
The restrictions tell you the domain of a rational function. The domain is all real numbers except the zeros of the denominator. For , the denominator is zero only when . So the domain is . (pp. 121–122)
Simplifying works like reducing a fraction. With , you factor out the common factor 3 to get . For rational expressions, the steps are:
(p. 119)
Monomials and polynomials over a monomial. Divide out the GCF.
Quadratics in two variables. Take . Factor the numerator as . Factor the denominator as . Divide out :
Both restrictions come from setting the factored denominator equal to zero. (p. 121)
Restrictions must be found before any factors are divided out. A factor that disappears during simplifying still made the original expression undefined.
For the game, factor . Factor . Dividing out and gives
The simplified form only forbids . But and made the original denominator zero, so they stay excluded. (pp. 119, 122)
The restrictions are solved from . Set each factor equal to zero: , , . (p. 119)
The factors and are opposites. You can write , and then divide out the common factor.
For :
Don't forget the negative sign that comes from rewriting . (p. 121)
Mistake 1: cancelling any matching terms. Only common factors can be divided out. Terms joined by or cannot. In , you cannot remove the 2 and the 3, and you cannot remove either. Factor first. Then cancel only whole factors that multiply the rest of the numerator or denominator. (pp. 119, 121)
Mistake 2: thinking the same restrictions mean the functions are equivalent. Restrictions only show where a function is undefined. They say nothing about its values. For example, and both have the restriction . At , though, the first gives and the second gives . Two rational functions are equivalent only if they simplify to the same expression, with the same restrictions. (p. 133)
A rational function has restricted values, which are x-values that make the denominator zero. At a restricted value the graph does one of two things, depending on what happens after you simplify.
Hole. A hole occurs at a restricted value whose factor also appears in the numerator, so the factor divides out. Take . It simplifies to . The graph is the line with one point missing at , because is undefined. (pp. 125–126)
Vertical asymptote. A vertical asymptote occurs at a restricted value that is still a zero of the denominator after simplifying. For nothing cancels, so is a vertical asymptote. Near the outputs grow without bound, positive on one side and negative on the other. (pp. 125–126)
The basic example. has domain and a vertical asymptote at . For very large positive or negative x, the outputs get close to 0 but never reach it, so is a horizontal asymptote. (p. 125)
Vertical translations. Shifting a graph up moves every point up, including its horizontal asymptote. So has horizontal asymptote . Its vertical asymptote stays at . (p. 125)
End behaviour. End behaviour describes what the output approaches as x gets very large () or very negative (). You can make a conjecture by calculating the function at larger and larger x-values. For the outputs approach 0, which we write as . The same is true as . (p. 124)
Units as rational expressions. Units such as , and are rational expressions. They are related by multiplication:
The mol and the L divide out, and writing the negative exponent as a reciprocal shows the link. (pp. 127–128)
Multiplying. Use the same steps as for fractions:
For , multiplying gives . Dividing by the GCF leaves , with , . (pp. 128–130)
Sometimes a factor and its opposite appear, such as and . Rewrite so the factors match and can divide out. For example, simplifies to , with . (p. 129)
Dividing. Multiply by the reciprocal of the divisor:
For example, . (pp. 130–131)
Restrictions in division. After taking the reciprocal, the divisor's numerator becomes a denominator. So you need , and also . In the example , the answer is . The restrictions are . (pp. 130–131)
Why factor first. If you multiply first, you get large polynomials that are hard to factor. Factoring first shows common factors immediately and keeps the numbers small. Dividing by a rational function gives the same result as multiplying by its reciprocal. (p. 133)
A false "proof". An argument can seem to show that , or that implies . The hidden step divides both sides by . If , then , and dividing by zero is not allowed. This is why restrictions matter at every step. (p. 133)
Travel-time model. Time is distance divided by speed, so travel problems give rational expressions. A wind adds to the speed or subtracts from it. With still-air speed v and a 540 km trip, the time is with no wind. With an 80 km/h tailwind it is , and against the wind it is . (p. 134)
Toronto–Montreal round trip. With no wind, the round trip takes . With wind, add the two legs:
Rewrite as . Both times now have the same numerator, and is smaller than . Dividing by a smaller number gives a larger result, so . The no-wind trip is faster. (pp. 134–135)
Method. Rational expressions need a common denominator before you add or subtract them.
For the LCD is . The terms become , with . (pp. 136–137)
The LCD is not always the product of the denominators. It contains every common factor once and every unique factor. For the denominators factor as and . The LCD is , using only once. The result is . (p. 137)
Restrictions. They come from the zeros of every denominator, which are the zeros of the LCD. For , we need and . Keep these restrictions even if the final form looks as if it allows those values. (p. 137)
A telescoping identity. The reciprocal of a product of consecutive integers is a difference of reciprocals:
To see why, use the LCD . The difference is . Examples: and . (p. 140)
