Algebraic Fractions Dave Saha 15 Jan 99 Dave

Algebraic Fractions Dave Saha 15 Jan 99 Dave Saha

Algebraic Expression DEFINITION An algebraic expression is f(x) = anxn + amxm + + a 0 , where an 0, x , where the exponents need not be positive integers and the coefficients are real numbers. Examples: x – , x¾ – 5 x½ + 2 x¼ + 1, 3 x– 5 – 7 x– 4 (Technically, an expression is algebraic if it can be constructed using a finite number of algebraic operations: addition, subtraction, multiplication, division, exponentiation, and taking of roots. This allows a much wider set of possible forms than just the one shown above. The form offered above is for comparison with the form of a polynomial. ) stands for the real numbers 2/25/2021 Dave Saha 2

Polynomial and Rational Functions DEFINITIONS A polynomial function is f(x) = anxn + an– 1 xn– 1 + + a 0, where an 0, x , where the exponents must be non-negative integers and the coefficients must be rational. Examples: x – 4, x 4 – 5 x 3 + 2 x 2 + 1, 1 x 5 – 7 x 4 3 A polynomial expression is referred to more often simply as a polynomial. An algebraic fraction is the quotient of two algebraic expressions. For example: – 12 x 52 + 7 x¾ – ¼x – 1 + 2 , x– 4 3 x– 5 – 7 x– 4 , x 2 – ½ x 4 – 3 – 5 x 3 + 2 x 2 + 1 A rational function is an algebraic fraction whose numerator and denominator are polynomials, as the last example shown. We will restrict all our examples to rational expressions, although the techniques shown apply to all algebraic fractions. 2/25/2021 Dave Saha 3

Operations with Rational Fractions Key Points to Remember • All operations with rational fractions – addition, subtraction, multiplication, and division – work the same way as with rational numbers. • In addition and subtraction the difficulty is developing the least common denominator. • In multiplication and division the difficulty is recognizing common factors in the numerator and denominator. • You may have to factor polynomials. EXAMPLE x– 2 – x+1 = x+1 x+2 (x – 2)(x + 2) – (x + 1)(x + 2)(x + 1) = x 2 – 4 x 2 + 2 x + 1 – (x + 1)(x + 2) = 2/25/2021 – 2 x – 5 x 2 + 3 x + 2 Dave Saha 4

EXAMPLE 2 2 – 1 2(k + 2) – 1(k + 1) + 2 = + 2 k + 4 k + 3 k + 5 k + 6 (k + 1)(k + 3) (k + 2)(k + 3) (k + 1)(k + 2)(k + 3) 2(k + 2) – 1(k + 1) 2 k + 4 – k – 1 = = (k + 1)(k + 2)(k + 3) = 1 k+3 = (k + 1)(k + 2)(k + 3) (k + 1)(k + 2) STEP factor denominators. STEP multiply each fraction by a form of 1, introducing the missing factor(s) into the denominators. Note the highlighting. STEP put over common denominator. STEPS clean up the numerator with algebra and arithmetic. STEP notice any numerator result which might match a factor in the denominator. (This example was selected to show this concept. ) STEP cancel common factors in numerator and denominator. 2/25/2021 Dave Saha 5

EXAMPLE 3 5 – 8 + = t 2 – 1 t 2 + 2 t + 1 5 – 8 + (t + 1)(t – 1) (t + 1) = 5(t + 1) – 8(t – 1) + (t + 1)(t – 1)(t + 1)(t – 1) = 5 t + 5 – 8 t + 8 – 3 t + 13 = (t + 1)(t – 1) STEP factor denominators. STEP multiply each fraction by a form of 1, introducing the missing factor(s) into the denominators. Note the highlighting. STEP put over common denominator. STEP clean up the numerator with algebra and arithmetic. STEP (notice any numerator result which might match a factor in the denominator) is unnecessary for this example. 2/25/2021 Dave Saha 6

EXAMPLE 4 2 z 2 + z – 3 z 2 + 2 z – 3 (z – 1)(2 z + 3) (z – 1)(z + 3) 2 = 2 z + 4 z + 3 2 z – 6 (z + 1)(z + 3) (z – 2)(2 z + 3) 2 (z – 1) z – 2 z + 1 = = (z + 1)(z – 2) z 2 – z – 2 STEP factor numerators and denominators. STEP (convert any division to multiplication by reciprocals) is unnecessary for this example. STEP cancel any common factors in numerator and denominator. Note the highlighting in the previous line. STEP clean up the numerator with algebra and arithmetic. 2/25/2021 Dave Saha 7

EXAMPLE 5 3 y + 5 2 y 2 – y – 3 2 y – 3 3 y + 5 (2 y – 3)(y + 1) 2 y – 3 = y 2 – 9 y + 20 3 y 2 + 11 y + 10 y 2 – 2 y – 8 (y – 5)(y – 4) (y + 2)(3 y + 5) (y – 4)(y + 2) (3 y + 5)(2 y – 3)(y + 1)(y – 4)(y + 2) = (y – 5)(y – 4)(y + 2)(3 y + 5)(2 y – 3) y + 1 = y– 5 STEP factor numerators and denominators. STEP convert any division to multiplication by reciprocals. Combine all factors as one algebraic fraction. STEP cancel any common factors in numerator and denominator. Note the highlighting in the previous line. STEP (clean up the numerator with algebra and arithmetic) is unnecessary for this example. 2/25/2021 Dave Saha 8

Relations Between Rational Functions a c Two fractions, b and d , are equivalent if and only if (a)(d) = (b)(c), cross multiply. a(x) c(x) By analogy two rational functions, and , are equivalent if and only if b(x) d(x) a(x) • d(x) = b(x) • c(x) But since we are dealing with functions, one other condition must apply: • the domains of a(x) and c(x) must be the same. b(x) d(x) Remember, the domain of a function, y = f(x), is the set of values for x for which f(x) has a value. When restricted to rational functions – when a(x), b(x), c(x), and d(x) are polynomials – the condition on domains becomes: • the zeros of b(x) must be the same as the zeros of d(x). Two fractions, a and c , have a definite order: a < c , or a = c , or a > c. b d b d a(x) c(x) Two rational functions, and , don’t work the same way because we are b(x) dealing with functions! One rational function may be greater than the other on part of their common domain and less than the other on another part. They may also be equal on still other parts of their common domain. 2/25/2021 Dave Saha 9

Practice Problems 1. – 4 x x+1 x+3 + – 2 x – 16 x– 4 x+4 t + 1 6. Simplify: t + 1 t 1 1_ + t t+1 2. r – 2 r + 5 r + 16 r + 2 r – 3 7. x 2 – y 2 x 2 – xy + y 2 x 3 + y 3 (x – y)2 x 2 – 2 xy + y 2 (x – y)4 3. 2 x 3 + 2 x – 3 3 – 2 x 8. 4. 6 a 2 – 3 a 3 a + 9 a 2 – 2 a + 1 4 a 2 – 1 2 a 2 – a – 1 a+3 9. Does x 2 – 2 x – 3 x 2 – x – 6 5. Simplify: 2/25/2021 1 2 4 – + x + 1 (x + 1)2 (x + 1)3 equal x 2 + 4 x + 3 x 2 + 5 x + 6 (x + y)– 1 x – 1 + y – 1 Dave Saha ? 10