#### Transcript A1 Algebraic manipulation

KS4 Mathematics A1 Algebraic manipulation 1 of 73 © Boardworks Ltd 2005 Contents A1 Algebraic manipulation A A1.1 Using index laws A A1.2 Multiplying out brackets A A1.3 Factorization A A1.4 Factorizing quadratic expressions A A1.5 Algebraic fractions 2 of 73 © Boardworks Ltd 2005 Multiplying terms Simplify: x + x + x + x + x = 5x x to the power of 5 Simplify: x × x × x × x × x = x5 x5 as been written using index notation. xn The number n is called the index or power. The number x is called the base. 3 of 73 © Boardworks Ltd 2005 Multiplying terms involving indices We can use index notation to simplify expressions. For example, 3p × 2p = 3 × p × 2 × p = 6p2 q2 × q3 = q × q × q × q × q = q5 3r × r2 = 3 × r × r × r = 3r3 3t × 3t = (3t)2 4 of 73 or 9t2 © Boardworks Ltd 2005 Multiplying terms with the same base When we multiply two terms with the same base the indices are added. For example, a4 × a2 = (a × a × a × a) × (a × a) =a×a×a×a×a×a = a6 = a (4 + 2) In general, xm × xn = x(m + n) 5 of 73 © Boardworks Ltd 2005 Dividing terms Remember, in algebra we do not usually use the division sign, ÷. Instead, we write the number or term we are dividing by underneath like a fraction. For example, (a + b) ÷ c 6 of 73 is written as a+b c © Boardworks Ltd 2005 Dividing terms Like a fraction, we can often simplify expressions by cancelling. For example, 3 n n3 ÷ n2 = 2 n 7 of 73 2 6p 6p2 ÷ 3p = 3p 2 n×n×n = n×n 6×p×p = 3×p =n = 2p © Boardworks Ltd 2005 Dividing terms with the same base When we divide two terms with the same base the indices are subtracted. For example, a5 ÷ a2 a×a×a×a×a = a×a = a × a × a = a3 = a (5 – 2) 2 4p6 ÷ 2p4 = 4×p×p×p×p×p×p = 2 × p × p = 2p2 = 2p(6 – 4) 2×p×p×p×p In general, xm ÷ xn = x(m – n) 8 of 73 © Boardworks Ltd 2005 Hexagon puzzle 9 of 73 © Boardworks Ltd 2005 Expressions of the form (xm)n Sometimes terms can be raised to a power and the result raised to another power. For example, (y3)2 = y3 × y3 (pq2)4 = pq2 × pq2 × pq2 × pq2 = (y × y × y) × (y × y × y) = p4 × q (2 + 2 + 2 + 2) = y6 = p4 × q8 = p4q8 10 of 73 © Boardworks Ltd 2005 Expressions of the form (xm)n When a term is raised to a power and the result raised to another power, the powers are multiplied. For example, (a5)3 = a5 × a5 × a5 = a(5 + 5 + 5) = a15 = a(3 × 5) In general, (xm)n = xmn 11 of 73 © Boardworks Ltd 2005 Expressions of the form (xm)n Rewrite the following without brackets. 1) (2a2)3 = 8a6 2) (m3n)4 = m12n4 3) (t–4)2 = t–8 4) (3g5)3 = 27g15 5) (ab–2)–2 = a–2b4 6) (p2q–5)–1 = p–2q5 7) (h½)2 = h 8) (7a4b–3)0 = 1 12 of 73 © Boardworks Ltd 2005 The zero index Look at the following division: Any number or term divided by itself is equal to 1. y4 ÷ y4 = 1 But using the rule that xm ÷ xn = x(m – n) y4 ÷ y4 = y(4 – 4) = y0 That means that y0 = 1 In general, for all x 0, x0 = 1 13 of 73 © Boardworks Ltd 2005 Negative indices Look at the following division: b2 ÷ b4 = b×b b×b×b×b = 1 1 = 2 b×b b But using the rule that xm ÷ xn = x(m – n) b2 ÷ b4 = b(2 – 4) = b–2 That means that b–2 = 1 b2 x–n = 1 xn In general, 14 of 73 © Boardworks Ltd 2005 Negative indices Write the following using fraction notation: u–1 = 2b–4 = 2 b4 x2y–3 = x2 y3 2a(3 – 15 of 73 1 u b)–2 = This is the reciprocal of u. 2a (3 – b)2 © Boardworks Ltd 2005 Negative indices Write the following using negative indices: 2 = 2a–1 a x3 = x3y–4 4 y p2 = p2(q + 2)–1 q+2 3m 2 + 2)–3 3m(n = (n2 + 2)3 16 of 73 © Boardworks Ltd 2005 Fractional indices Indices can also be fractional. 1 2 1 2 x ×x = x 1 2 1 +2 = x1 = x But, x × x = x So, x = x Similarly, But, So, 17 of 73 The square root of x. 1 2 1 3 1 3 1 3 x ×x ×x = x 1 3 + 1 3 + 1 3 = x1 = x x × x × x = x 3 3 3 x = x 1 3 3 The cube root of x. © Boardworks Ltd 2005 Fractional indices In general, x = x 1 n n 1 m Also, we can write x n as x n × m . Using the rule that (xm)n = xmn, we can write 1 n x m n We can also write x as xm × 1 n ×m n 1 n = (x )m = (x)m . 1 x m× n = 1 (xm)n n = xm In general, m n x = 18 of 73 n m x or m n n x = (x)m © Boardworks Ltd 2005 Index laws Here is a summary of the index laws. x–1 = 1 x x–n = 1n x xm × xn = x(m + n) xm ÷ xn = x(m – n) x0 19 of 73 (xm)n = xmn x = x x1 = x x = x = 1 (for x = 0) 1 2 1 n m n x = n xm n n or (x)m © Boardworks Ltd 2005 Contents A1 Algebraic manipulation A A1.1 Using index laws A A1.2 Multiplying out brackets A A1.3 Factorization A A1.4 Factorizing quadratic expressions A A1.5 Algebraic fractions 20 of 73 © Boardworks Ltd 2005 Expanding expressions with brackets Look at this algebraic expression: 3y(4 – 2y) This means 3y × (4 – 2y), but we do not usually write × in algebra. To expand or multiply out this expression we multiply every term inside the bracket by the term outside the bracket. 3y(4 – 2y) = 12y – 6y2 21 of 73 © Boardworks Ltd 2005 Expanding expressions with brackets Look at this algebraic expression: –a(2a2 – 2a + 3) When there is a negative term outside the bracket, the signs of the multiplied terms change. –a(2a2 – 3a + 1) = –2a3 + 3a2 – a In general, –x(y + z) = –xy – xz –x(y – z) = –xy + xz –(y + z) = –y – z –(y – z) = –y + z 22 of 73 © Boardworks Ltd 2005 Expanding brackets and simplifying Sometimes we need to multiply out brackets and then simplify. For example, 3x + 2x(5 – x) We need to multiply the bracket by 2x and collect together like terms. 3x + 2x(5 – x) = 3x + 10x – 2x2 = 13x – 2x2 23 of 73 © Boardworks Ltd 2005 Expanding brackets and simplifying Expand and simplify: 4 – (5n – 3) We need to multiply the bracket by –1 and collect together like terms. 4 – (5n – 3) = 4 – 5n + 3 = 4 + 3 – 5n = 7 – 5n 24 of 73 © Boardworks Ltd 2005 Expanding brackets and simplifying Expand and simplify: 2(3n – 4) + 3(3n + 5) We need to multiply out both brackets and collect together like terms. 2(3n – 4) + 3(3n + 5) = 6n – 8 + 9n + 15 = 6n + 9n – 8 + 15 = 15n + 7 25 of 73 © Boardworks Ltd 2005 Expanding brackets then simplifying Expand and simplify: 5(3a + 2b) – a(2 + 5b) We need to multiply out both brackets and collect together like terms. 5(3a + 2b) – a(2 + 5b) = 15a + 10b – 2a – 5ab = 15a – 2a + 10b – 5ab = 13a + 10b – 5ab 26 of 73 © Boardworks Ltd 2005 Find the area of the rectangle 27 of 73 © Boardworks Ltd 2005 Find the area of the rectangle What is the area of a rectangle of length (a + b) and width (c + d)? a b c ac bc d ad bd In general, (a + b)(c + d) = ac + ad + bc + bd 28 of 73 © Boardworks Ltd 2005 Expanding two brackets Look at this algebraic expression: (3 + t)(4 – 2t) This means (3 + t) × (4 – 2t), but we do not usually write × in algebra. To expand or multiply out this expression we multiply every term in the second bracket by every term in the first bracket. (3 + t)(4 – 2t) = 3(4 – 2t) + t(4 – 2t) = 12 – 6t + 4t – 2t2 = 12 – 2t – 2t2 29 of 73 This is a quadratic expression. © Boardworks Ltd 2005 Using the grid method to expand brackets 30 of 73 © Boardworks Ltd 2005 Expanding two brackets With practice we can expand the product of two linear expressions in fewer steps. For example, (x – 5)(x + 2) = x2 + 2x – 5x – 10 = x2 – 3x – 10 Notice that –3 is the sum of –5 and 2 … 31 of 73 … and that –10 is the product of –5 and 2. © Boardworks Ltd 2005 Matching quadratic expressions 1 32 of 73 © Boardworks Ltd 2005 Matching quadratic expressions 2 33 of 73 © Boardworks Ltd 2005 Squaring expressions Expand and simplify: (2 – 3a)2 We can write this as, (2 – 3a)2 = (2 – 3a)(2 – 3a) Expanding, (2 – 3a)(2 – 3a) = 2(2 – 3a) – 3a(2 – 3a) = 4 – 6a – 6a + 9a2 = 4 – 12a + 9a2 34 of 73 © Boardworks Ltd 2005 Squaring expressions In general, (a + b)2 = a2 + 2ab + b2 The first term squared … … plus 2 × the product of the two terms … … plus the second term squared. For example, (3m + 2n)2 = 9m2 + 12mn + 4n2 35 of 73 © Boardworks Ltd 2005 Squaring expressions 36 of 73 © Boardworks Ltd 2005 The difference between two squares Expand and simplify (2a + 7)(2a – 7) Expanding, (2a + 7)(2a – 7) = 2a(2a – 7) + 7(2a – 7) = 4a2 – 14a + 14a – 49 = 4a2 – 49 When we simplify, the two middle terms cancel out. This is the difference In general, between two squares. (a + b)(a – b) = a2 – b2 37 of 73 © Boardworks Ltd 2005 The difference between two squares 38 of 73 © Boardworks Ltd 2005 Matching the difference between two squares 39 of 73 © Boardworks Ltd 2005 Contents A1 Algebraic manipulation A A1.1 Using index laws A A1.2 Multiplying out brackets A A1.3 Factorization A A1.4 Factorizing quadratic expressions A A1.5 Algebraic fractions 40 of 73 © Boardworks Ltd 2005 Factorizing expressions Factorizing an expression is the opposite of expanding it. Expanding or multiplying out a(b + c) ab + ac Factorizing Often: When we expand an expression we remove the brackets. When we factorize an expression we write it with brackets. 41 of 73 © Boardworks Ltd 2005 Factorizing expressions Expressions can be factorized by dividing each term by a common factor and writing this outside of a pair of brackets. For example, in the expression 5x + 10 the terms 5x and 10 have a common factor, 5. We can write the 5 outside of a set of brackets and mentally divide 5x + 10 by 5. (5x + 10) ÷ 5 = x + 2 This is written inside the bracket. 5(x + 2) 42 of 73 © Boardworks Ltd 2005 Factorizing expressions Writing 5x + 10 as 5(x + 2) is called factorizing the expression. Factorize 6a + 8 43 of 73 Factorize 12n – 9n2 The highest common factor of 6a and 8 is 2. The highest common factor of 12n and 9n2 is 3n. (6a + 8) ÷ 2 = 3a + 4 (12n – 9n2) ÷ 3n = 4 – 3n 6a + 8 = 2(3a + 4) 12n – 9n2 = 3n(4 – 3n) © Boardworks Ltd 2005 Factorizing expressions Writing 5x + 10 as 5(x + 2) is called factorizing the expression. Factorize 3x + x2 The highest common factor of 3x and x2 is x. (3x + x 2) ÷x=3+x Factorize 2p + 6p2 – 4p3 The highest common factor of 2p, 6p2 and 4p3 is 2p. (2p + 6p2 – 4p3) ÷ 2p = 1 + 3p – 2p2 3x + x2 = x(3 + x) 2p + 6p2 – 4p3 = 2p(1 + 3p – 2p2) 44 of 73 © Boardworks Ltd 2005 Factorization 45 of 73 © Boardworks Ltd 2005 Factorization by pairing Some expressions containing four terms can be factorized by regrouping the terms into pairs that share a common factor. For example, Factorize 4a + ab + 4 + b Two terms share a common factor of 4 and the remaining two terms share a common factor of b. 4a + ab + 4 + b = 4a + 4 + ab + b = 4(a + 1) + b(a + 1) 4(a + 1) and + b(a + 1) share a common factor of (a + 1) so we can write this as (a + 1)(4 + b) 46 of 73 © Boardworks Ltd 2005 Factorization by pairing Factorize xy – 6 + 2y – 3x We can regroup the terms in this expression into two pairs of terms that share a common factor. When we take xy – 6 + 2y – 3x = xy + 2y – 3x – 6 out a factor of –3, – 6 = y(x + 2) – 3(x + 2) becomes + 2 y(x + 2) and – 3(x + 2) share a common factor of (x + 2) so we can write this as (x + 2)(y – 3) 47 of 73 © Boardworks Ltd 2005 Contents A1 Algebraic manipulation A A1.1 Using index laws A A1.2 Multiplying out brackets A A1.3 Factorization A A1.4 Factorizing quadratic expressions A A1.5 Algebraic fractions 48 of 73 © Boardworks Ltd 2005 Quadratic expressions A quadratic expression is an expression in which the highest power of the variable is 2. For example, 2 t x2 – 2, w2 + 3w + 1, 4 – 5g2 , 2 The general form of a quadratic expression in x is: ax2 + bx + c (where a = 0) x is a variable. a is a fixed number and is the coefficient of x2. b is a fixed number and is the coefficient of x. c is a fixed number and is a constant term. 49 of 73 © Boardworks Ltd 2005 Factorizing expressions Remember: factorizing an expression is the opposite of expanding it. Expanding or multiplying out a2 + 3a + 2 (a + 1)(a + 2) Factorizing Often: When we expand an expression we remove the brackets. When we factorize an expression we write it with brackets. 50 of 73 © Boardworks Ltd 2005 Factorizing quadratic expressions Quadratic expressions of the form x2 + bx + c can be factorized if they can be written using brackets as (x + d)(x + e) where d and e are integers. If we expand (x + d)(x + e) we have, (x + d)(x + e) = x2 + dx + ex + de = x2 + (d + e)x + de Comparing this to x2 + bx + c we can see that: The sum of d and e must be equal to b, the coefficient of x. The product of d and e must be equal to c, the constant term. 51 of 73 © Boardworks Ltd 2005 Factorizing quadratic expressions 1 52 of 73 © Boardworks Ltd 2005 Matching quadratic expressions 1 53 of 73 © Boardworks Ltd 2005 Factorizing quadratic expressions Quadratic expressions of the form ax2 + bx + c can be factorized if they can be written using brackets as (dx + e)(fx + g) where d, e, f and g are integers. If we expand (dx + e)(fx + g)we have, (dx + e)(fx + g)= dfx2 + dgx + efx + eg = dfx2 + (dg + ef)x + eg Comparing this to ax2 + bx + c we can see that we must choose d, e, f and g such that: a = df, b = (dg + ef) c = eg 54 of 73 © Boardworks Ltd 2005 Factorizing quadratic expressions 2 55 of 73 © Boardworks Ltd 2005 Matching quadratic expressions 2 56 of 73 © Boardworks Ltd 2005 Factorizing the difference between two squares A quadratic expression in the form x2 – a2 is called the difference between two squares. The difference between two squares can be factorized as follows: x2 – a2 = (x + a)(x – a) For example, 9x2 – 16 = (3x + 4)(3x – 4) 25a2 – 1 = (5a + 1)(5a – 1) m4 – 49n2 = (m2 + 7n)(m2 – 7n) 57 of 73 © Boardworks Ltd 2005 Factorizing the difference between two squares 58 of 73 © Boardworks Ltd 2005 Matching the difference between two squares 59 of 73 © Boardworks Ltd 2005 Contents A1 Algebraic manipulation A A1.1 Using index laws A A1.2 Multiplying out brackets A A1.3 Factorization A A1.4 Factorizing quadratic expressions A A1.5 Algebraic fractions 60 of 73 © Boardworks Ltd 2005 Algebraic fractions 3x 2a and are examples of algebraic fractions. 2 4x 3a + 2 The rules that apply to numerical fractions also apply to algebraic fractions. For example, if we multiply or divide the numerator and the denominator of a fraction by the same number or term we produce an equivalent fraction. For example, 3x 3 6 3y 3(a + 2) = = = = 2 4x 4x 8x 4xy 4x(a + 2) 61 of 73 © Boardworks Ltd 2005 Equivalent algebraic fractions 62 of 73 © Boardworks Ltd 2005 Simplifying algebraic fractions We simplify or cancel algebraic fractions in the same way as numerical fractions, by dividing the numerator and the denominator by common factors. For example, 6ab Simplify 3ab2 2 6ab 6×a×b = 2 3ab 3×a×b×b 2 = b 63 of 73 © Boardworks Ltd 2005 Simplifying algebraic fractions Sometimes we need to factorize the numerator and the denominator before we can simplify an algebraic fraction. For example, 2a + a2 Simplify 8 + 4a 2a + a2 8 + 4a a (2 + a) = 4(2 + a) a = 4 64 of 73 © Boardworks Ltd 2005 Simplifying algebraic fractions b2 – 36 is the difference between two squares. b2 – 36 Simplify 3b – 18 b2 – 36 (b + 6)(b – 6) = 3b – 18 3(b – 6) b+6 3 If required, we can write this as = b b 6 + = + 2 3 3 3 65 of 73 © Boardworks Ltd 2005 Manipulating algebraic fractions Remember, a fraction written in the form a+b c can be written as a b + c c However, a fraction written in the form c a+b cannot be written as c c + a b For example, 1+2 3 66 of 73 1 2 = + 3 3 but 3 1+2 3 3 = + 1 2 © Boardworks Ltd 2005 Multiplying and dividing algebraic fractions We can multiply and divide algebraic fractions using the same rules that we use for numerical fractions. In general, and, For example, a c ac × = b d bd a c a d ad ÷ = × = b d b c bc 3 3p 6p 3p 2 × = = (1 – p) 4(1 – p) 2(1 – p) 4 2 67 of 73 © Boardworks Ltd 2005 Multiplying and dividing algebraic fractions 2 4 ? What is ÷ 3y – 6 y–2 2 3y – 6 4 2 ÷ = y–2 3y – 6 This is the reciprocal 4 of y–2 y–2 × 4 y–2 2 = × 3(y – 2) 4 2 1 = 6 68 of 73 © Boardworks Ltd 2005 Adding algebraic fractions We can add algebraic fractions using the same method that we use for numerical fractions. For example, 1 2 What is + ? a b We need to write the fractions over a common denominator before we can add them. b 2a b + 2a 1 2 + = + = a b ab ab ab In general, a c ad + bc + = b d bd 69 of 73 © Boardworks Ltd 2005 Adding algebraic fractions y 3 What is + ? x 2 We need to write the fractions over a common denominator before we can add them. y y×x 3 3×2 + = + x x×2 2×x 2 xy 6 + = 2x 2x 6 + xy = 2x 70 of 73 © Boardworks Ltd 2005 Subtracting algebraic fractions We can also subtract algebraic fractions using the same method as we use for numerical fractions. For example, p q What is – ? 3 2 We need to write the fractions over a common denominator before we can subtract them. p q 2p 3q 2p – 3q – = – = 6 6 6 3 2 In general, a c ad – bc – = b d bd 71 of 73 © Boardworks Ltd 2005 Subtracting algebraic fractions 2+p 3 What is – ? 2q 4 2+p 3 (2 + p) × 2q 3×4 – = – 2q 4 4 × 2q 2q × 4 2q(2 + p) 12 = – 8q 8q 2q(2 + p) – 12 = 8q 4 6 q(2 + p) – 6 = 4q 72 of 73 © Boardworks Ltd 2005 Addition pyramid – algebraic fractions 73 of 73 © Boardworks Ltd 2005