a b + c b = a+c a b c d = ac a b c d = a b d a does not exist

Transcription

1 Pre-precalculus Boot Camp: Arithmetic with fractions page Aug, 0 Arithmetic with fractions To add fractions with the same denominator, add the numerators: () a b + c b = a+c b To multiply fractions, multiply the numerators and denominators: () a b c d = ac bd To divide fractions, invert and multiply: () a b c d = a b d c Division by zero is not allowed: () a 0 does not eist (because a 0 =? would mean a = 0?, which has no solution if a 0 and infinitely many solutions if a = 0.) As a consequence of (), () ac bc = a b (because ac bc = a b c c = a b.) We use this either to simplify a fraction by canceling common factors,, = = or to obtain common denominators for adding fractions,, + = + = 0 + =. There s nothing inherently wrong with an improper fraction, but to write an improper fraction as a mied number, we use long divisio For eample, = with remainder means that = +.. Find the sum or differenc

2 Pre-precalculus Boot Camp: Arithmetic with fractions page Aug, 0. Find the sum or differenc t. + + w... Find the sum or differenc Hint: always use the least common demonimato t w Find the sum or differenc Hint: n = n Find the product. Hint: cancel common factors before multiplyin t. w... Find the quotient. Remember that division by zero is not allowe t. w.. y. z Answers t. 0 w t. 0 0 w DNE 0 DNE 0 t. w DNE DNE 0 DNE DNE t. w.. y. z

3 Pre-precalculus Boot Camp: Rationalizing the denominator page Sept, 0 Rationalizing the denominator It s sometimes handy to rewrite a fraction so that no radicals appear in the denominato To do this, we multiply the fraction by in an appropriate form, thus changing the appearance of the fraction but not its actual valu a Case : Rationalizing b c. In this case we can rationalize the denominator by multiplying by = c c. For eample, = = ( ) = = =. a Case : Rationalizing b±c d. When the last operation in the denominator is addition or subtraction, we rationalize the denominator by multiplying by = b c d b c and using d ( y)(+y) = y. (b c d is called the conjugate of b±c ) For instance, = + + = (+ ) ( ) = (+ ) = (+ ) = (+ ) Tip: look for common factors in the numerator and denominator and for perfect squares under the radical,, = =.. Rationalize the denominato 0 t. 0 w. 0.. Rationalize the denominato t w. +. Answers 0 0 t. w ( ) (+ ) ( ) (+ ) (+ ) ( 0) (+ ) ( ) 0 (+ ) (+ ) ( 0) (+ ) (+ ) 0 (+ ) t. (+ ) (+ 0) w. (+ ). (+ )

4 Pre-precalculus Boot Camp: Laws of eponents page Jan 0, 0 Laws of eponents We use eponents to denote repeated multiplication,, It s easiest to remember the laws a = a a a m a n = a m+n (a m ) n = a mn (ab) n = a n b n () ( a b) n = a n b n a m a n = am n = a n by thinking of simple eample For instance, and and a a = (a a a) (a a) = a, (a ) = (a a a) = (a a a) (a a a) = a, (ab) = (ab)(ab)(ab) = a a a b b b = a b, are eamples of the first three laws in (). The laws () are also true for eponents that are zero, negative or rational numbers after we define () a 0 = a n = a n a m/n = n (a m ) = ( n a) m, with some restriction For instance, we can t raise 0 to a negative eponent, since 0 is undefined, and the last two equations are true only if the corresponding roots are define (Remember that we can t take an even root of a negative numbe) The most common mistake students make with the laws of eponents is to confuse (ab) n with (a+b) n. In general, (a+b) n is not equal a n +b n. Radicals n a refers to a solution to the equation n = a, if one eist If n is odd, then every real number has eactly one nth root. For instance =, since is the unique solution to the equation =. If n is even, then negative numbers don t posses real nth roots, and every positive number has two real nth roots, one positive and one negativ In that case, the symbol n a stands for the positive nth root of The radical symbol with no inde is always assumed to mean the square root. Eample : Both and are solutions to = ; both and are square roots of, but stands for the positive square root, or. end Eample Eample : The equation = has no real solutions, so does not eist. One of the reasons that we can think of n a as a power of a is the fact that end Eample n ab = n a n b

5 Pre-precalculus Boot Camp: Laws of eponents page Jan 0, 0 provided all three roots eist. That is, (ab) /n = a /n b /n, as in (). This is useful when we have to simplify epressions of the form n a where n is a positive integer and a is an intege We look for the perfect nth powers that are factors of a, which we could always find by factoring a down to prime number Eample : = = =. end Eample Eample : = = =. If we hadn t noticed that is the largest square factor of, we still could have simplified this radical completely by finding that the prime factorization of is. Then = ( ) / = / / = + = / =. end Eample Eample : = = =. Since = / = /, the answer can be simplified further to. end Eample There s one important hitch in using the laws of eponents with fractional eponent If a 0, then both (a /n ) n and (a n ) /n correctly simplify to However, if a < 0 and n is an even integer, (a /n ) n doesn t eist, and, because n refers to the positive nth root, (a n ) /n = a, as in this eample: Eample : (( ) ) / = ( ) = =. In particular, the result is not. The rule to remember in general is end Eample If n is an even integer, then (a n ) /n = a.. Simplify the radical or state that it does not eist Simplify the radical or state that it does not eist , ,000 0,000 t w. 0. y. z...

6 Pre-precalculus Boot Camp: Laws of eponents page Jan 0, 0. Rewrite the epression in simplest radical form or state that it does not eist. / ( ) / ( ) / () / ( ) / ( ) / / / / / / / / / / / / / / t. 0 / () / () / w. 0 /. 0 / y. / z. ( 0) /. ( ) /. /. / 0. 0 /. ( 00) /. 00 /. Rewrite without parenthese Epress your answer without negative eponents, and again without fractions (other than eponents themselves). Assume all variables are positiv ( y ) (u / v / ) y ( y ) (a bc b a ) / (uv u v ) / (( y ) y ) / a (ab ) / b a b c (b) (cb ) (a c ) u (v ) wu 0 y( y) (y z) ( y) / ( / y ) / y ( / y / ) y ((+y)) (u v) (uv) ab (+ab) ( ab) (y y) ( y ) y ( y) ( y) y t. ( y ) (y ) v ( uv u v ) ( u v / ) / ( u v )(v u ) v v a(b a ) a ((b ) / ) / Answers DNE DNE DNE DNE t w. 0. 0y. 0z. 0. DNE 0. DNE w. t DNE.. y 0 = y 0 y. z. u v = u v y = y a / c v u = u v y/ = y / b / a / = a / b / a c b = a b c u v w / z = / y / z y / / y / y / + y + y uv v + u u v = u v v + u v u ab a b + a b = a b a b + a b y 0 = y 0 y = y v u = u v y y = y y t. u / v / b /

7 Pre-precalculus Boot Camp: Factoring polynomials page Jan 0, 0 Factoring polynomials To factor an algebraic epression is to rewrite it in an equivalent form in which the last operation performed is multiplicatio When factoring a polynomial,. Factor out any common factor. Then try to factor the polynomial with one of these method If the polynomial has two terms, try using one of the special product If the polynomial has three terms, try factoring it with backwards-foil. It also pays to be able to recognize the perfect square a +ab+b = (a+b). If the polynomial has four terms, try factoring it by groupin. Test whether you can factor the factor Eample : +0 Since the polynomial has four terms, we factor the first two terms and the last two and hope to find a common facto ( )+(0 ) = ( )+( ) = ( +)( ). end Eample Eample : y + y + y This polynomial has a common factor of y: y + y + y = y( ++). To factor the three-term polynomial, look for a factorization (a ± b)(c ± d) that, when multiplied using FOIL, gives First =, Last =, and Outside+Inside = : y( ++) = y(+)(+). The familiar difference of squares formula is just one of large number of special products: y = ( y)(+y) y = ( y)( +y +y ) y = ( y)( + y +y +y ) y = ( y)( + y + y +y +y ) y = ( y)( + y + y + y +y +y ). end Eample You can verify any of these yourself by carefully multiplying out the right sid When we replace y with y in the odd-lines, we also obtain +y = (+y)( y +y ) +y = (+y)( y + y y +y ).. Eample : u +v = (u) +v = (u+v)(u uv +v ) The three-term polynomial can t be factored further using integer end Eample Eample 0: = ( )( + ++)

8 Pre-precalculus Boot Camp: Factoring polynomials page Jan 0, 0 This polynomial hasn t been factored completely because the cubic polynomial factors further by grouping: ( + )+(+) = (+)+(+) = (+)( +), so = ( )(+)( +). We also could have arrived at this factorization by using square-minus-square: = ( ) = ( )( +) = ( )(+)( +). end Eample 0 The quadratic + is said to be irreducible, meaning that it has no real zeros and, consequently cannot be factored further without comple number The quadratic + is also irreducible, since b ac = ( ) = < 0, so u uv+v can t be factored without using comple numbers for coefficient In fact, the quadratic factor that appears in the difference-of-cubes formula is always irreducibl Every polynomial with real coefficients can be factored as a product of linear and irreducible quadratic factors with real coefficients (although there is no method that produces the eact factorization of all polynomials). If we allow ourselves to use comple coefficients, then every polynomials can be factored as a product of linear factors only. When we use only real coefficients in our factors, we re said to be factoring the polynomial over the real number When we allow comple coefficients in our factors, we re said to be factoring the polynomial over the comple number. Factor the polynomial over the real number u y y y + y y y u v +uv 0 +y +y y y t. u u v y w. y y. y y. +. Factor the polynomial over the real number y +y + +y+y y y 0 y +0 y +0y t. +y y y +y y +0y w. +y y. 0y+0y y. y +y z. + y +y. 0 y +y. + y +y

9 Pre-precalculus Boot Camp: Factoring polynomials page Jan 0, 0. Factor the polynomial over the real number y +y y 0u u v +uv v u u v uv +v u u v uv +v y+y y u v u u v + Answers (+)( ) ( )(+) ( )(+) ( )(+) ( u)(+u) ( )(+)) ( y)(+y) y( y)(+y) ( )(+ ) does not factor further ( y)( +y +y ) y ( y)( +y+y ) vu(u+v)(u uv+v ) (+y)( 0y+y ) ( +y)( y+y ) (y )( y +y+) ( )(+)( +) ( y)(+y)( +y ) ( )(+ )( +) t. ( u )(+u )(+u ) (u v)(u+v)(u +v ) ( y)(+y+ y ) w. y( y )(+y ). ( y)(+y)( y+y )( +y+y ) y. Has no real zeros; hence, does not factor without using comple coefficient (+)( ) ( )( ) (+)( ) ( )( ) (+) ( y) ( ) (+y) (+)( ) (+)( ) (+)( ) ( )(+) (+)( ) ( + )( ) ( + y)( y) 0y( + y) ( ) ( + ) ( ) ( + ) t. ( y)( + y) ( y)( y) ( y)( 0y) w. ( + y)( y). ( y)( y) y. ( y )( y ) z. ( +y )( +y ). ( y)( y). ( +y )( +y ) ( +)( ) ( + )( + ) ( + )( ), or, ( + )( ) ( + )( ) ( + y )( y) (u +v )(u v) ( )(+)( ) (+)( )(+) ( )(+)( ) (u v)(u+v)(u v) (u v)(u+v)(u v) (+ )( )( ) ( )(+ )(+), or ( )( +)(+) (+)( )( ++) ( y)(+y)( y+y ) ( )( )(+)( ++) (+)( +)( +) (uv )(uv +)(u )

10 Pre-precalculus Boot Camp: Solving polynomial equations page 0 Sept, 0 Solving polynomial equations One reason we factor polynomials is to find their zeros,, the -values that cause the polynomial to equal zer We rely heavily on a special property of the number zer For the product of two numbers to be zero, it is both necessary and sufficient that one of the two must be zer That is, ab = 0 if and only if a = 0 or b = 0. It is important to remember that no number other than zero has this property. For instance, ab = does not mean that a or b has to equal. To solve a polynomial equation, try to Get zero on one side and factor the othe Eample : Solve for in the equation 0 = + Factor the right sid + = (+)( ). So, for this product to equal zero, one factor or the other must be zero: + = 0 = 0 = = / = = / That is, 0 = + has two solutions: = / and = /. Eample : Solve for in the equation + =. Get a zero on one side by subtracting to both sides: + = 0. end Eample Now factor the left sid First factor out the common factor of, and then factor the three term polynomial using backwards FOIL: ( + ) = ( )(+) = 0. For the product of, ( ), and (+) to equal zero, one of these factors has to be zero: = 0 = 0 = 0 = / + = 0 = That is, the solutions to + = 0. are = 0, /, and. end Eample The problem of find y-intercepts usually boils down to solving an equation, as the net eample show Eample : Find the - and y-intercepts along the curve y =. To find the y-intercepts, we set = 0 and calculate y = 0 0 =. That it, (0, ) is the y-intercept. To find the -intercepts, set y = 0 and solve for by factoring: The solutions are 0 = = ( )(+). = 0 = = / so this curve has two -intercepts: (/,0) and ( /,0) + = 0 = = / end Eample

11 Pre-precalculus Boot Camp: Solving polynomial equations page Sept, 0. Find the zeros of the polynomia u. Solve for. = 0 + = 0 0 = 0 + = 0 + = = = + = + = 0 = + = + = 0 Answers = ± = ± = ± = ± u = ± = ±/ = ±/ = 0 or ± / = ± / =, / = /, =, = /,/ = / = / = 0,/ = / = /, / =, / =, / =,

12 Pre-precalculus Boot Camp: Rational functions page Jan, 0 Rational functions A rational function is a fraction whose numerator and denominator are both polynomial We perform arithmetic on rational functions just as with any fraction Simplification isn t something you do in your last step to satisfy your math professor; it s something you do at every step to make your work easie Be on the lookout for common factors of the numerator and denominator and cancel thes Eample : + = + ( )(+) = Notice that is considerably simpler than +. Canceling the + was a good mov end Eample Eample : Can I cancel the in +? + has no common factors, at least as it is written at present. You can cancel the from top and bottom if you factor it out first, but the result might not be very useful: + = ( ) + = +. end Eample. Simplify the rational functio t

13 Pre-precalculus Boot Camp: Rational functions page Jan, 0. Perform the operation indicated and simplify the resulting rational functio y t. w. z ( )(+) Perform the operation indicated and simplify the resulting rational functio Answers does not simplify further + + t ( )(+) ( )( ) ( )( ) + ( +)(+) ++ (+)(+)( ) 0 ( )(+) ( )(+) ( )( ) + ( )(+)( ) (+)(+) ( +)(+) + ( )( +) ++ ( )(+)( ++) ( )(+) + ( )( +)( +) + ( +)( ) + ( ) (+) t. y. (+)( ) z.. (+)( ) 0 + ( ) ( ) ( ) (+)(+) ( )( ) 0 w. + ( +)(+) ( )(+) (+)( +) 0 (+)( ) ( ) (+)( ) 0 ( )(+)