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Simplifying algebraic fractions Contents Simplifying algebraic fractions Adding and subtracting algebraic fractions Multiplying and dividing algebraic fractions Improper fractions and polynomial division Examination-style question 1 of 37 © Boardworks Ltd 2006 Rational expressions Remember, a rational number is any number that can be written in the form a , where a and b are integers and b ≠ 0. b Numbers written in this form are often called fractions. In algebra, a rational expression is an algebraic fraction that can be written in the form f ( x ) , where f(x) and g(x) are g( x ) polynomials and g(x) ≠ 0. For example, 3 x2 3 x +1 x2 2 x3 2 x2 + 3 x 4 For which values of x are each of the above expressions undefined? 2 of 37 © Boardworks Ltd 2006 Rational expressions An algebraic fraction is undefined when the denominator is 0. So, 3 is undefined when x + 2 = 0. That is, when x = –2. x+2 3 x +1 2 – 2 = 0. That is, when x = ±√2. is undefined when x x2 2 x3 2 2 + 3x – 4 = 0. is undefined when x x2 + 3 x 4 We can factorize this to give (x + 4)(x – 1) = 0. x3 2 So 2 is undefined when x = –4 or x = 1. x + 3x 4 3 of 37 © Boardworks Ltd 2006 Simplifying fractions by cancelling When the numerator and the denominator of a numerical fraction contain a common factor, the fraction can be simplified by cancelling. For example, consider the fraction 2 2 28 = 42 3 3 The highest common factor of 28 and 42 is ___. 14 This fraction can therefore be written in its simplest terms by dividing both the numerator and the denominator by 14. 4 of 37 © Boardworks Ltd 2006 Simplifying algebraic fractions by cancelling Algebraic fractions can be cancelled in the same way. For example, 3 6a 6aa 3 = = 3 8a 8 a a a 4a 2 4 When the numerator or the denominator contains more than one term, we have to factorize before cancelling. For example, 3 pq Simplify 15 p 9 p 2 q 3 pq 3 pq = = 2 15 p 9 p 3 p(5 3 p ) 5 3 p 5 of 37 © Boardworks Ltd 2006 Simplifying algebraic fractions by cancelling 3b 6 Simplify 2 b 2b 3b 6 3(b 2) 3 = = 2 b 2b b(b 2) b x2 + x 2 Simplify x2 1 x 2 + x 2 ( x 1)( x 2) x 2 = = 2 x 1 ( x 1)( x 1) x 1 6 of 37 © Boardworks Ltd 2006 Using additive inverses When manipulating algebraic fractions it is helpful to remember that an expression of the form a – b is the additive inverse of the expression b – a. i.e. a – b = –(b – a) and b – a = –(a – b ) For example, 3 y 2 ( y 2 3) = 2 = –1 2 y 3 y 3 When cancelling, look for situations where a factor of –1 can be taken out of a pair of brackets. 7 of 37 © Boardworks Ltd 2006 Using additive inverses 14 7 x Simplify 2 x 5x + 6 7(2 x ) 14 7 x = x 2 5 x + 6 ( x 2)( x 3) 7(2 x ) = (2 x )( x 3) 7 = ( x 3) 7 = 3x 8 of 37 © Boardworks Ltd 2006 Simplifying complex fractions Sometimes the numerator or the denominator of an algebraic fraction contains another fraction. For example, 3x 1 x Simplify x2 This can be simplified by multiplying the numerator and the denominator by x. 3 x2 1 3 x 1x = 2 x3 x 1 2 a Simplify a a3 Multiply the numerator and the denominator by 3a: 1 2a 3a 6 3a 6 = 2 = 2 a a 3 3a a 4a 2 9 of 37 © Boardworks Ltd 2006 Simplifying complex fractions Simplify 3 42 x x 1 34x To simplify this algebraic fraction we multiply the numerator and the denominator by the lowest common multiple of x, x2 3x2 and 3x. That is ___. 3 42 9 x 12 x x = 2 4 1 3x 3 x 4 x 3(3 x + 4) = x(3 x + 4) = 10 of 37 3 x © Boardworks Ltd 2006 Contents Adding and subtracting algebraic fractions Simplifying algebraic fractions Adding and subtracting algebraic fractions Multiplying and dividing algebraic fractions Improper fractions and polynomial division Examination-style question 11 of 37 © Boardworks Ltd 2006 Adding and subtracting fractions Before looking at the addition and subtraction of algebraic fractions, let’s recall the method used for numerical fractions. 5 3 What is + ? 6 4 Before we can add these two fractions we have to write them as equivalent fractions over a common denominator. It is best to use the lowest common denominator of the two fractions. This is the lowest common multiple (LCM) of their denominators. 12 of 37 © Boardworks Ltd 2006 Adding and subtracting fractions The LCM of 6 and 4 is ___. 12 So we write, 5 3 10 + 9 + = 6 4 12 19 = 12 7 =1 12 We apply the same method to add or subtract algebraic fractions. 13 of 37 © Boardworks Ltd 2006 Adding and subtracting fractions 2 3 Write 2 + as a single fraction in its lowest terms. x x The LCM of x2 and x is ___. x2 2 3 2+3 3xx + = x2 x x2 y x as a single fraction in its lowest terms. Write 3x y The LCM of 3x and y is ___. 3xy y x y 22 3 3xx22 = 3x y 3 xy 14 of 37 © Boardworks Ltd 2006 Adding and subtracting fractions 2x 1 + Write as a single fraction in its lowest terms. x + 3 2x + 6 2x 1 + Start by factorizing where possible: x + 3 2( x + 3) The LCM of x + 3 and 2(x + 3) is ______. 2(x + 3) 2x 1 4x 1 + = + x + 3 2( x + 3) 2( x + 3) 2( x + 3) 4 x +1 = 2( x + 3) 15 of 37 © Boardworks Ltd 2006 Adding and subtracting fractions 2x + 8 Write 3 2 as a single fraction in its lowest terms. x +5 2 x + 8 3( x 2 + 5) 2 x + 8 3 2 = 2 2 x +5 x +5 x +5 3 x 2 +15 2 x 8 = x2 + 5 Notice that this becomes – 8. 3 x2 2 x 7 = x2 + 5 16 of 37 © Boardworks Ltd 2006 Contents Multiplying and dividing algebraic fractions Simplifying algebraic fractions Adding and subtracting algebraic fractions Multiplying and dividing algebraic fractions Improper fractions and polynomial division Examination-style question 17 of 37 © Boardworks Ltd 2006 Multiplying and dividing fractions Before looking at the multiplication and division of algebraic fractions, let’s recall the methods used for numerical fractions. 3 12 What is × ? 8 21 When multiplying two fractions, start by cancelling any common factors in the numerators and denominators: 1 3 2 7 3 12 × 8 21 Then multiply the numerators and multiply the denominators: 1 3 3 × = 2 7 14 18 of 37 © Boardworks Ltd 2006 Multiplying and dividing fractions To divide by a fraction we multiply by its reciprocal. 7 14 What is ÷ ? 9 15 This is equivalent to 1 5 3 2 7 15 1 5 × = × 9 14 3 2 5 = 6 19 of 37 © Boardworks Ltd 2006 Multiplying and dividing algebraic fractions We can apply the same methods to the multiplication and division of algebraic fractions. For example, 3 x 12 2 × Simplify 2x + 4 x 4 Start by factorizing where possible: 3 x 12 2 3( x 4) 2 × = × 2 x + 4 x 4 2( x + 2) x 4 = 20 of 37 3 x+2 © Boardworks Ltd 2006 Multiplying and dividing algebraic fractions x2 4 3x × Simplify x + 2 2x 4 x2 4 3x ( x 2)( x 2) 3x × = × x + 2 2x 4 x+2 2( x 2) 3x = 2 5 15 ÷ Simplify 2p p 1 5 15 5 p ÷ = × = 2p p 2 p 15 6 3 21 of 37 © Boardworks Ltd 2006 Multiplying and dividing algebraic fractions 14 7 ÷ 2 Simplify a+2 a a 6 14 a 2 a 6 14 7 × ÷ 2 = 7 a+2 a a 6 a+2 2 14 ( a + 2)( a 3) = × a+2 7 = 2(a – 3) 22 of 37 © Boardworks Ltd 2006 Contents Improper fractions and polynomial division Simplifying algebraic fractions Adding and subtracting algebraic fractions Multiplying and dividing algebraic fractions Improper fractions and polynomial division Examination-style question 23 of 37 © Boardworks Ltd 2006 Improper fractions and mixed numbers Remember, a numerical fraction is called an improper fraction if the numerator is larger than the denominator. Improper fractions are usually simplified by writing them as a whole number plus a proper fraction. This is called a mixed number. For example, the improper fraction 29 can be converted to a 6 mixed number as follows: 29 24 + 5 24 5 = = + = 6 6 6 6 5 4 6 When 29 is divided by 6, 4 is the quotient and 5 is the remainder. 24 of 37 © Boardworks Ltd 2006 Improper algebraic fractions An algebraic fraction is called an improper fraction when the numerator is a polynomial of degree greater than, or equal to, the degree of the denominator. For example, x3 x4 x3 x2 5 and 4 x2 ( x 4)( x 2) are improper algebraic fractions. Suppose we have an improper fraction f ( x) . g( x ) Dividing f(x) by g(x) will give us a quotient q(x) and a remainder r(x), which gives us the identity: f ( x) r( x ) q( x ) + g( x ) g( x) where the degree of f(x) ≥ the degree of g(x). 25 of 37 © Boardworks Ltd 2006 Writing improper fractions in proper form r( x ) We can think of the form q( x ) + as being the algebraic g( x ) equivalent of mixed number form. It is a polynomial plus a proper fraction. If the degree of f(x) is n and the degree of g(x) is m then: The degree of the quotient q(x) will be equal to n – m. The degree of the remainder r(x) will be less than m. An improper algebraic fraction can be written in proper form by: rewriting the numerator. writing an appropriate identity to equate the coefficients. using long division to divide the numerator by the denominator. 26 of 37 © Boardworks Ltd 2006 Writing improper fractions in proper form 27 of 37 © Boardworks Ltd 2006 Rewriting the numerator A useful technique for writing improper fractions in proper form is to look for ways to rewrite the numerator so that part of it can be divided by the denominator. For example, B x+3 Write in the form A + . x 1 x 1 x + 3 x 1+ 4 = x 1 x 1 x 1 4 = + x 1 x 1 4 =1+ x 1 28 of 37 © Boardworks Ltd 2006 Rewriting the numerator What is the quotient and the remainder when 3x2 + 2x is divided by x2 + 1? We can write this in fraction form as: 3 x2 + 2 x 3( x 2 +1) + 2 x 3 = 2 x +1 x 2 +1 3( x 2 +1) 2 x 3 = + 2 2 x +1 x +1 2x 3 =3 + 2 x +1 So when 3x2 + 2x is divided by x2 + 1 the quotient is 3 and the remainder is 2x – 3. 29 of 37 © Boardworks Ltd 2006 Rewriting the numerator Find the remainder when x3 is divided by x2 – 3. We can write this in fraction form as: x3 x3 3 x + 3 x = 2 x 3 x2 3 x( x 2 3) + 3 x = x2 3 x( x 2 3) 3x = 2 + 2 x 3 x 3 3x = x+ 2 x 3 The remainder is 3x. 30 of 37 © Boardworks Ltd 2006 Constructing an identity When the numerator cannot easily be manipulated to give an expression of the required form, we can write an identity using: f ( x) r( x ) q( x ) + g( x ) g( x) where q(x) is the quotient and r(x) is the remainder when f(x) is divided by g(x). What is x3 – 4x2 + 5 divided by x2 – 3? Let the quotient be Ax + B. (It must be linear because the degree of the dividend minus the degree of the divisor is 1). Let the remainder be Cx + D. (Its degree must be less than 2.) 31 of 37 © Boardworks Ltd 2006 Constructing an identity This gives us the identity x3 4 x 2 + 5 Cx + D Ax + B + x2 3 x2 3 Multiply through by x2 – 3: x3 4 x 2 + 5 ( Ax + B )( x 2 3) + Cx + D Ax3 3 Ax + Bx2 3 B + Cx + D Equate the coefficients: x 3: A=1 B = –4 x 2: x: constants: 32 of 37 C – 3A = 0 C = 3 D – 3B = 5 D + 12 = 5 D = –7 © Boardworks Ltd 2006 Constructing an identity We can now substitute these values into the original identity to give: x3 4 x 2 + 5 3x 7 x 4+ 2 2 x 3 x 3 Alternatively, use long division: x–4 x2 – 3 x3 – 4x2 + 0x + 5 x3 + 0x2 – 3x – 4x2 + 3x + 5 – 4x2 +0x + 12 3x – 7 The quotient is x – 4 and the remainder is 3x – 7 so, as before: x3 4 x 2 + 5 3x 7 x 4+ 2 2 x 3 x 3 33 of 37 © Boardworks Ltd 2006 Contents Examination-style question Simplifying algebraic fractions Adding and subtracting algebraic fractions Multiplying and dividing algebraic fractions Improper fractions and polynomial division Examination-style question 34 of 37 © Boardworks Ltd 2006 Examination-style question Given that f ( x ) x 3 9 + 2 { x , x 1, x 2}, x2 x x2 x2 + x 3 . show that f ( x ) x +1 3 9 3 9 x + 2 =x + x2 x x2 x 2 ( x 2)( x +1) x( x 2)( x +1) 3( x +1) 9 = + ( x 2)( x +1) ( x 2)( x +1) ( x 2)( x +1) x3 x 2 2 x 3 x 3 + 9 = ( x 2)( x +1) x3 x 2 5 x + 6 = ( x 2)( x +1) 35 of 37 © Boardworks Ltd 2006 Examination-style question Now divide x3 – x2 – 5x + 6 by x – 2: x–2 So, 36 of 37 x2 + x – 3 x3 – x2 – 5x + 6 x3 – 2x2 x2 – 5x x2 – 2x –3x + 6 –3x + 6 0 x3 x 2 5 x + 6 ( x 2)( x 2 + x 3) = ( x 2)( x +1) ( x 2)( x +1) x2 + x 3 as required. = x +1 © Boardworks Ltd 2006