Read the following definitions and match them with the appropriate example(s) using the lines provided.

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Algebraic Expressions Prepared by: Sa diyya Hendrickson Name: Date: Read the following definitions and match them with the appropriate example(s) using the lines provided. 1.

Algebraic expression: An expression that involves real numbers and at least one variable using addition, subtraction, multiplication, division and exponentiation or radicals. 3.

1 Algebraic Expressions Prepared by: Sa diyya Hendrickson Name: Date: Read the following definitions and match them with the appropriate example(s) using the lines provided. 1. Variable: A letter that is used to: (1) describe properties of a mathematical object, (2) represent a quantity that we dont know or (3) be a placeholder for the input of one or more numbers. 2. Algebraic expression: An expression that involves real numbers and at least one variable using addition, subtraction, multiplication, division and exponentiation or radicals. 3. Monomial: A real number or an expression that involves the product of real numbers and variables with positive integer exponents. 4. Polynomial: A monomial or an expression that involves the addition and/or subtraction of monomials. 6. Degree of a monomial: The sum of the exponents of the variables in a monomial. By convention, if the monomial is a real number, we say that it has degree zero. 7. Degree of a polynomial: The highest degree amongst the monomials in a polynomial. 8. Coefficient: The real number in a monomial. 9. Leading coefficient of a polynomial: The coefficient of the monomial with the highest degree. (a) 5xy xz 3 (b) t (c) 5x y 3xz (d) From (a): 4 (e) 7 (f) 4y 3 + x (g) 8k 2 (h) From (a): 1 (i) From (e): 0 (j) From (a): 5 (k) 9x 4 2x (l) From (k): 9 (m) 3 4 (n) From (g): k (o) From (b): 1 (p) 2 x Making Math Possible 1 of 9 c Sa diyya Hendrickson

Notations Consider the following general polynomial form: p(x) = a n x n + a n 1 x n 1 + + a 2 x 2 + a 1 x + a 0 (i) The expression p(x) reads p of x or p at x and is one way to refer to a polynomial

2 Notations Consider the following general polynomial form: p(x) = a n x n + a n 1 x n a 2 x 2 + a 1 x + a 0 (i) The expression p(x) reads p of x or p at x and is one way to refer to a polynomial that is defined in terms of the variable x. Note that we could have replaced x with any variable of our choosing. (ii) Because p is a polynomial, n represents some nonnegative integer (by definition). (iii) The variables a n, a n 1,..., a 2, a 1, a 0 represent coefficients. Check-In a) What does the variable n represent in the polynomial p (above)? b) What does a n represent in the polynomial p (above)? c) Consider the polynomial p(t) = 3t 5 4t 2 +2t 1. Determine a 0, a 1, a 2, a 3, a 4 and a 5. Solution: Based on the general form above, p(t) = a 5 t 5 + a 4 t 4 + a 3 t 3 + a 2 t 2 + a 1 t + a 0. So, a 0 = 1, a 1 = 2 a 2 = 4, a 3 = 0, a 4 = 0 and a 5 = 3. d) Suppose p(z) = 3z 2 4z + 1. What does p(1) mean? Determine the resulting value. Solution: p(1) reads p at 1 and represents the number produced by the polynomial when we replace z with the number 1. The resulting value is given by: p(1) = 3(1) 4(1) + 1 = = = 0 e) Consider the following one-degree polynomials, p(x) = a 1 x + a 0 and q(x) = b 1 x + b 0. Under which circumstances will p and q represent the same polynomial? Under which circumstances will they represent different polynomials? Solution: p and q will represent the same polynomial if a 1 = b 1 and a 0 = b 0. If a 1 = b 1 or a 0 = b 0, then p and q will represent different polynomials. f) Based on the information given in (i), what do you think p(x, y) might represent? If p(x, y) = y 2 3xy, determine the value of p( 1, 2). Solution: p(x, y) simply represents a polynomial defined in terms of two variables, x and y. Since p(x, y) = y 2 3xy, p( 1, 2) requires that we replace x with 1 and y with 2. This gives: p( 1, 2) = ( 2) 2 3( 1)( 2) = 4 6 = 2. Making Math Possible 2 of 9 c Sa diyya Hendrickson

3 Expanding Expressions The most popular polynomials, defined in terms of one variable (at most), are as follows: a) Degree = 0: constant polynomials (e.g. 5, 1, 2, etc.) b) Degree = 1: mx + b, where m, b R c) Degree = 2: quadratic polynomials (e.g. ax 2 + bx + c, where a, b, c R) d) Degree = 3: cubic polynomials (e.g. ax 3 + bx 2 + cx + d, where a, b, c, d R) ** It s all about the Distributive Property (DP)! ** Exercise E1: Expand the following expression: (x 1)(x + 4) a! "# $ b c!"#$!"#$ Solution: For the given expression: (x 1)( x + 4 ), we can let a = x + 1, b = x and c = 4. This results in the following expansion: Notice that each term in the first pair of brackets was simply multiplied with each term in the 2nd pair of brackets! In general, we have: Many people are familiar with the acronym FOIL (First, Outer, Inner, Last). But this is just a special case of the Distributive Property, when we have exactly two elements in each pair of brackets! Making Math Possible 3 of 9 c Sa diyya Hendrickson

4 Expanding Expressions The Distributive Property tells us to simply take each term in the 1st pair of brackets (one at a time) and multiply them with each term in the 2nd pair of brackets (i.e. distribute them). Unlike FOIL, this works for any number of elements in each pair of brackets! For the above exercise, our strategy gives: (x 1)(x + 4) DP = (x)(x) + (4)(x) (1)(x) (1)(4) as before, but in one step! = x 2 + 4x x 4! "# $ like terms = x 2 + 3x 4 Exercise E2: Expand (a b) 3 Solution: First recall that (a b) 3 = (a b)(a b)(a b), by def n of exponentiation for a positive integer exponent. So, our approach is to use the associative property of multiplication to group and expand two at a time! (a b) 3 = (a b)(a b) (a b)! "# $ DP = (a 2 ab ba + b 2 )(a b) Notice that ab = ba When expanding, it s a good idea to write the letters in alphabetical order because it makes it easier to identify like terms!! = (a 2 2ab + b 2 )(a b) DP = (a 2 )(a) + (a 2 )( b) (2ab)(a) (2ab)( b) +(b 2 )(a) + (b 2 )( b) = a 3 a 2 b 2a 2 b + 2ab 2 + ab 2 b 3 = a 3 3a 2 b + 3ab 2 b 3 Try the following exercises on your own, using the same strategy: 1) Prove that (a + b) 3 = a 3 + 3a 2 b + 3ab 2 + b 3. 2) Prove that (a b)(a 2 + ab + b 2 ) = a 3 b 3. Making Math Possible 4 of 9 c Sa diyya Hendrickson

5 Expanding Expressions Exercise E3: Expand the expression 2ab 2 c (5a 3 b + 3b 2 c) As you distribute each term, it may be helpful to: first record the product of the constants, while carefully determining the sign (i.e. is the product positive or negative?). Then, determine and record the product of each variable, one by one. (The chart below describes how to organize your mental math, as you expand.) 2ab 2 c (5a 3 b + 3b 2 c) Distribution Constants a b c Summary First ( 2)(5) = 10 a a 3 = a 4 b 2 c 10a 4 b 2 c Second ( 2)( 1) = 2 a Third b 2 b 2 = b 4 c c = c 2 Solution: 2ab 2 c (5a 3 b + 3b 2 c) DP = 10a4 b 2 c + 2ab 3 c 6ab 4 c 2 Exercise E4: Expand the expression 4(x 2 3x + 5) 3(x 2 2x + 1) Solution: 4(x 2 3x + 5) 3(x 2 2x + 1) DP = 4x 2 12x x 2 + 6x 3 = 4x 2 3x 2 12x + 6x+20 3 = x 2 6x + 17 Exercise E5: Expand and evaluate (2 3)(2 + 3) Solution: (2 3)(2 + 3) DP = 2(2) ( 3)( 3) = 4 ( 3) 2 = 4 3 since ( n a) n = a, for a 0 = 1 Making Math Possible 5 of 9 c Sa diyya Hendrickson

6 Factoring Expressions ** It s still all about the Distributive Property (DP)! ** On the previous pages, we were going from factored form to expanded form. Now, we will begin with expanded form and end in factored form. We ll know that we ve achieved factored form once the original expression is a product of terms and/or smaller expressions. Exercise F1: Factor the expression 12x 4 y xy 3 18xy 4 z. S1 Decide if you d like to factor out 1 from all terms. This is recommended if your leading term is negative. Because our leading term is 12x 4 y 2, which is negative, we will pull 1 as a common factor. Note that we can always do this! e.g. (12x 4 y 2 30xy 3 +18xy 4 z) S2 Determine the GCF (Greatest Common Factor) among the constants and each variable, one at a time. Note that for each variable, we must factor the lowest power among all terms. Otherwise, we cannot guarantee that we have a common factor. Expression: 12x 4 y xy 3 18xy 4 z Elements GCF Factor 1? N/A Yes/No: 1 Constants 12, 30, = 6 x x 4, x 1, x 1 x 1 y y 2, y 3, y 4 z z 0, z 0, z 1 z 0 = 1 Summary N/A 6xy 2 This chart gives us a visual representation of how we can organize our approach if we were to complete this strategy using mental math. Making Math Possible 6 of 9 c Sa diyya Hendrickson

7 Factoring Expressions Exercise F1 (Continued) S3 Factor out the GCF of each element, starting with the constants and then each variable. Then complete the expression in the brackets by writing in the missing part of the original term. 12x 4 y xy 3 18xy 4 z = (12x 4 y 2 30xy 3 +18xy 4 z) = 6 ( 2!"#$ 12 6 x 4 y 2 5!"#$ 30 6 xy 3 + xy 4 z) = 6x 1 (2 x 4 1! "# $ x 4 x 1 y 2 5 y y 4 z) = 6xy 2 (2x 3 y z) = 6xy 2 (2x 3 5y + 3y 2 z) With practice, you will be able to do this in one line! Let s begin by recalling the following definition: Perfect square: a perfect square is an expression of form a 2, where a is any integer. Note that there is nothing specific about our choice of the letter a. Any letter will do! e.g. b 2, c 2, s 2, t 2, x 2, y 2, etc. Below are some square numbers that appear quite often. Therefore, it may be worthwhile to commit them to memory. 0 2 = = = = = = = = = = = 100 I. Difference of Squares: a 2 b 2 = (a + b)(a b) proof : (a b)(a + b) DP = a 2 + ab ab b! "# $ 2 = 0 = a 2 b 2 Exercise F2: Factor the expression x 2 1. S1 Determine if there are exactly two perfect squares, seperated by a sign. Then, express the terms as powers of 2. x 2 " 1 = 1 2 " Making Math Possible 7 of 9 c Sa diyya Hendrickson

8 Factoring Expressions Exercise F2 (Continued) S2 Identify a and b based on the given identity. x 2 1 = x a = x and b = 1 S3 Substitute into the identity to obtain the factored form. x 2 1 = x a = x and b = 1 = (x a 1)(x a 1) = (x + 1)(x 1) Exercise F3: Factor the following expressions. a) 9 z 2 b) 4k 2 49 c) 16u 2 25v 2 a) Solution : 9 z 2 = 3 2 z 2 a = 3 and b = z = (3 a z)(3 a z) = (3 + z)(3 z) b) Solution : 4k 2 49 = 2 2 k = (2k) a = 2k and b = 7 = (2k a 7)(2k a 7) = (2k + 7)(2k 7) c) Solution : 16u 2 25v 2 = 4 2 u v 2 = (4u) 2 (5v) 2 a = 4u and b = 5v = (4u a 5v)(4u a 5v) = (4u + 5v)(4u 5v) Making Math Possible 8 of 9 c Sa diyya Hendrickson

9 Factoring Expressions II. Perfect Square: (a ± b) 2 = a 2 ± 2ab + b 2 proof : (a ± b) 2 = (a ± b)(a ± b) DP = a 2 ± ab ± ab + (±b) 2 = a 2 ± 2ab + b 2 Exercise F4: Factor the expression 4x 2 12x + 9. S1 Determine if the expression is a trinomial with perfect squares as its first and last terms. Then, express the first and last terms as powers of 2. 4x 2 = 2 2 x 2 = (2x) 2 " 9 = 3 2 " S2 Identify what a and b may be based on the identity and info from S1. Note: we still have to verify the middle term! 4x 2 12x + 9 = (2x) 2 12x + (3) 2 a = 2x and b = 3 S3 Verify that the middle term equals ±2ab. For a = 2x and b = 3 2ab = 2(2x)(3) = 12x " S4 Substitute into the identity to obtain the factored form. 4x 2 12x + 9 = (2x) 2 2(3)(2x) + (3) 2 a = 2x and b = 3 = (2x a 3) 2 by identity = (2x 3) 2 since the middle term is negative Exercise F5: Factor 9s 2 t 2 6st + 1. Solution : 9s 2 t 2 6st + 1 = (3st) 2 2(3)st + (1) 2 a = 3st and b = 1 = (3st a 1) 2 by identity = (3st 1) 2 Making Math Possible 9 of 9 c Sa diyya Hendrickson

(2) Dividing both sides of the equation in (1) by the divisor, 3, gives: =

(2) Dividing both sides of the equation in (1) by the divisor, 3, gives: = Dividing Polynomials Prepared by: Sa diyya Hendrickson Name: Date: Let s begin by recalling the process of long division for numbers. Consider the following fraction: Recall that fractions are just division