Definition 1.1 If x is any real number and n is a positive integer, then. x 1 = x. x 2 = x x x 3 = x x x

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1 1.2 Exponents and Radicals Definition 1.1 If x is any real number and n is a positive integer, then x 1 = x x 2 = x x x = x x x... x n = x x... x }{{} n times Here, the number x is called the base and the number n is called the exponent. Additionally, if x 0, we have x 0 = 1 x 1 = 1 x x 2 = 1 x 2... x n = 1 x n = 7 7 = 49 ( 1 2 )5 = = 1 2 Properties of Exponents Let a, b 0 be real numbers and let m, n Z. Then Property Example 1. a n a m = a n+m 2 5 = 2+5 = 7 am m n 5 = a = 5 2 = = 27 a n 2. (a m ) n = a mn ( 2 ) 5 = 2 5 = (ab) n = a n b n ( 4) 2 = = 9 16 = ( ) a n ( b = a n ) 2 b n 4 = 2 = ( ) a n ( b = b n ( 2 ( a) 4) = 4 ) 2 7. a m = bn 2 = 45 b n a n 4 5 2

2 Simplify the following expressions: 1. 6st 4 2s 2 t 2 ( y z ) 2 6st 4 2s 2 t = 6ss2 t 4 2 2t 2 = 6ss2 2t 2 t = 6s 4 2t = s 6 t 6 ( y z ) 2 = ( ) z 2 = 2 (z ) 2 y y 2 = 9z6 y 2 Radicals Definition 1.2 Let x R and n N. Then, the nth root of x is defined as follows: n x = y means y n = x where x, y 0. If n is not written, n = 2 ie 2 x = x. Example: 4 = ±2 since ( 2) 2 = 4 and 2 2 = 4. In general, if x > 0 and n is an even number, there are two nth roots of x, say y and y, since y n = ( y) n = x. If x > 0 and n is even, there is no nth root of x (this doesn t depend on the sign of x or evenness/oddness of n) does not exist since x = 8 < 0 and n = 2 is even. 8 = 2 since ( 2) = 8. Further, since n = is odd there is only one root = 16 = 4, but notice that ( 4) 2 = 16, thus 4 2 = 16 = ±4. Note: For us, when there are two possibilities we will always choose the positive one. So x2 = x. Properties of Radicals Let a, b be real numbers and let m, n Z. Then

3 Property Example 1. n ab = n a n b 8 27 = 8 27 = ( 2)(9) = 18 n a = n a 4 16 b n = 4 16 b = 4 m. n a = mn a 729 = = 4. n an = a if n is odd ( 5) = 5 5. n 4 an = a if n is even ( ) 4 = = WARNING: Note that, in general, a + b a + b For instance, = 25 = 5, but = + 4 = 7 5. Now with our knowledge of integer exponents and roots we can define exponents for any rational number. Definition 1. Let m Z and n N. We define a rational exponent as a m/n = ( n a ) m or a m/n = n a m If n is even, this is only defined if a 0. Note that the properties of exponents also hold for rational exponents. 1. a 1/ a 7/ = a 1/+7/ = a 8/ ( ) 2x /4 ( ) y 4 = 2 (x /4 ) y 1/ x 1/2 (y 1/ ) (y4 x 1/2 ) = 2 x 9/4 (y 4 x 1/2 ) = 8x 11/4 y y. x x = ( x x 1/2) 1/2 = (x /2 ) 1/2 = x /4 Rationalizing the Denominator For reasons beyond my realm of knowledge it is preferable to show algebraic expressions with no radicals in the denominator. If you have an expression with a radical in the denominator, find something to multiply the bottom by that will get rid of any roots, and then multiply the top and bottom by that thing = 1 a a 2 = 2 = 2 a a 2 =. Note that this expression is only valid if a > 0. a a

4 1. Algebraic Expressions Definition 1.4 A polynomial in the variable x is an expression of the form a n x n + a n 1 x n a 1 x 1 + a 0 where the a s are real numbers and n is a natural number. If a n 0, then we say that the polynomial has degree n. 1. P (x) = x + x + 2 is a polynomial with degree. Q(x) = 1 is a polynomial with degree 0.. F (x) = x 2 is a polynomial with degree 1. Definition 1.5 An algebraic expression is defined as the combination of polynomials by addition, subtraction, multiplication, division, powers, and roots. 1. 2x 2 + x + 4 x + 1. y 27 y x x 4. x x Adding and Subtracting Polynomials To add and subtract polynomials, group terms with like powers of x together. 1. Find the sum (x 6x 2 + 2x + 4) + (x + 5x 2 7x). (x 6x 2 + 2x + 4) + (x + 5x 2 7x) = (x + x ) + ( 6x 2 + 5x 2 ) + (2x 7x) + 4 = 2x x 2 5x + 4.

5 Find the difference (x 6x 2 + 2x + 4) (x + 5x 2 7x). (x 6x 2 + 2x + 4) (x + 5x 2 7x) = (x x ) + ( 6x 2 5x 2 ) + (2x ( 7x)) + 4 Multiplying Algebraic Expressions 1. Find the product (2x + 1)(x 5). = 0x 11x 2 + 9x + 4 = 11x 2 + 9x + 4. To multiply two polynomials together, you take each of the terms in the first one and multiply by each of the terms of the second one and add them. So in our case, we would take 2x x, and then add 2x ( 5), and do the same for 1. So we end up with (2x + 1)(x 5) = 2x x + 2x x + 1( 5) = 6x 2 10x + x 5 = 6x 2 7x 5 A teacher of mine called this process shooting the frogs into the buckets where the first terms were frogs (here 2x and 1) which were shot into the buckets (the second terms, here x and 5). Note that I do not condone shooting frogs into anything as that is animal cruelty. (1 + x)(2 x) = x 2 x x = 2 x x Factoring As above, if we are given (x 2)(x + 2), we can multiply it out to get (x 2)(x + 2) = x 2 4 Conversely, sometimes we are given an expression like x 2 4 and asked to find out if there are two smaller polynomials which multiply together to give it as a result. This is called factoring. Thus expanding (x 2)(x + 2) = x 2 4 factoring Another common example of factoring is finding a term common to all the terms in an expression and bringing it out front.

6 1. x 2 6x = x(x 2) 8x 4 y 2 + 6x y 2xy 4 = 2xy 2 (4x + x 2 y + y 2 ) Factoring Quadratics Factoring polynomials of degree 2 (i.e. quadratics) comes up frequently. In general, if we want to factor something like x 2 + bx + c, we want to find two numbers p and q that add up to b and that multiply together to give c. Then we will have x 2 + bx + c = (x + p)(x + q). Example: Suppose we want to factor x 2 5x + 6 into two smaller polynomials. Then we need to find two numbers that multiply together to give 6 and that add together to give 5. After a bit of guessing, the only numbers that do this are 2 and. Thus we have x 2 5x + 6 = (x 2)(x ) Multiplying the right hand side out shows that the two are indeed equal. Factoring with Radicals When factoring expressions involving rational exponents, factor out the lowest power present, even if it is negative. Example: Suppose we want to factor x /2 9x 1/2 + 6x 1/2. The lowest power present is x 1/2, so we bring it outside along with the common factor of : x /2 9x 1/2 + 6x 1/2 = x 1/2 (x 2 x + 2) = x 1/2 (x 2)(x 1) The last step is because 1 and 2 add up to and multiply to give Factoring by Grouping Higher-degree polynomials are a bit trickier to factor, but some can be done by some clever groupings. For example, consider x + x 2 + 4x + 4. The first two terms share a factor of x 2 while the last two share a factor of 4. Hence x + x 2 + 4x + 4 = (x + x 2 ) + (4x + 4) = x 2 (x + 1) + 4(x + 1) Now the first group and second group have a common factor of (x 1), so we can factor it out to get x + x 2 + 4x + 4 = (x + 1)(x 2 + 4)

7 Some Generally Useful Formulas Formula Example 1. (a b)(a + b) = a 2 b 2 (2x 5)(2x + 5) = 4x x 10x + 25 = 4x (a + b) 2 = a 2 + 2ab + b 2 (x + 5)(x + 5) = x 2 + 5x + 5x + 25 = x x (a b) 2 = a 2 2ab + b 2 (x 5)(x 5) = x 2 5x 5x + 25 = x 2 10x (a + b) = a + a 2 b + ab 2 + b (x + 5) = x + 5x x (a b) = a a 2 b + ab 2 b (x 5) = x 5x x a + b = (a + b)(a 2 ab + b 2 ) x + 8 = x + 2 = (x + 2)(x 2 2x + 4) 7. a b = (a b)(a 2 + ab + b 2 ) 27x 1 = (x) 1 = (x 1)(9x 2 + x + 1)