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KS4 Mathematics. A1 Algebraic manipulation. A1 Algebraic manipulation. Contents. A. A1.2 Multiplying out brackets. A. A1.1 Using index laws. A1.3 Factorization. A. A1.4 Factorizing quadratic expressions. A. A1.5 Algebraic fractions. A. Multiplying terms. Simplify:.
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KS4 Mathematics A1 Algebraic manipulation
A1 Algebraic manipulation Contents • A A1.2 Multiplying out brackets • A A1.1 Using index laws A1.3 Factorization • A A1.4 Factorizing quadratic expressions • A A1.5 Algebraic fractions • A
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. The number n is called the index or power. xn The number x is called the base.
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 or 9t2
When we multiply two terms with the same base the indices are added. Multiplying terms with the same base 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)
Dividing terms a + b c 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, is written as (a + b) ÷ c
Dividing terms n3 6p2 n2 3p 6 ×p×p n×n×n n×n 3 ×p Like a fraction, we can often simplify expressions by cancelling. For example, n3÷ n2 = 6p2÷ 3p = 2 = = = n = 2p
a × a × a × a × a = a × a 4 × p × p × p × p × p × p = 2 × p × p = 2 × p × p × p × p When we divide two terms with the same base the indices are subtracted. Dividing terms with the same base For example, a5 ÷ a2 = a × a × a = a3 = a(5 – 2) 2 4p6 ÷ 2p4 = 2p2 = 2p(6 – 4) In general, xm÷xn=x(m – n)
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
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
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 1 7) (h½)2 = h 8) (7a4b–3)0 =
Any number or term divided by itself is equal to 1. Look at the following division: The zero index 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
= = 1 1 1 1 b × b b × b × b × b b × b xn b2 b2 x–n= Look at the following division: Negative indices b2 ÷ b4 = But using the rule that xm÷xn=x(m– n) b2 ÷ b4 = b(2 – 4) = b–2 That means that b–2 = In general,
2 x2 b4 y3 1 2a u (3 –b)2 Write the following using fraction notation: Negative indices This is the reciprocal of u. u–1 = 2b–4 = x2y–3 = 2a(3 –b)–2 =
= x3 = y4 p2 = q + 2 2 3m = a (n2 + 2)3 Write the following using negative indices: Negative indices 2a–1 x3y–4 p2(q + 2)–1 3m(n2 + 2)–3
x × x = x+ = 1 1 1 1 1 1 1 1 1 1 1 1 2 3 3 2 3 2 2 3 3 3 3 2 But, x × x = x x = x So, Similarly, x × x × x = x+ + = But, x × x × x = x 3 3 3 x = x So, 3 Indices can also be fractional. Fractional indices x1 = x The square root of x. x1 = x The cube root of x.
1 1 1 1 1 1 1 m m m m n n n n n n n n n n n Also, we can write x as x . ×m x × m= (x)m = (x)m n We can also write x as xm× . n x = (xm) = xm m× x = xm n or n x = (x)m In general, Fractional indices x = x n Using the rule that (xm)n=xmn,we can write In general,
1 xm × xn = x(m + n) x–1 = x 1 xm÷xn=x(m – n) x–n = xn 1 2 (xm)n=xmn x = x 1 m n n x = x n x1=x x0=1 (for x = 0) x = xm or (x)m n n Here is a summary of the index laws. Index laws
A1 Algebraic manipulation Contents A1.1 Using index laws • A • A A1.2 Multiplying out brackets A1.3 Factorization • A A1.4 Factorizing quadratic expressions • A A1.5 Algebraic fractions • A
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
Expanding expressions with brackets In general, –x(y + z) = –xy – xz –x(y – z) = –xy + xz –(y + z) = –y – z –(y – z) = –y + z 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
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
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 + 3 – 5n = 4 + 3 – 5n = 7 – 5n
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) = – 8 + 15 6n + 9n = 6n + 9n– 8 + 15 = 15n + 7
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
Find the area of the rectangle b a c d What is the area of a rectangle of length (a + b) and width (c + d)? ac bc ad bd In general, ac + ad + bc + bd (a + b)(c + d) =
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) This is a quadratic expression. = 12 – 6t + 4t – 2t2 = 12 – 2t – 2t2
Expanding two brackets With practice we can expand the product of two linear expressions in fewer steps. For example, – 10 (x – 5)(x + 2) = + 2x – 5x x2 = x2 – 3x – 10 Notice that –3 is the sum of –5 and 2 … … and that –10 is the product of –5 and 2.
Squaring expressions Expand and simplify: (2 – 3a)2 We can write this as, (2 – 3a)2 = (2 – 3a)(2 – 3a) Expanding, 2(2 – 3a) – 3a(2 – 3a) (2 – 3a)(2 – 3a) = = 4 – 6a – 6a + 9a2 = 4 – 12a + 9a2
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
The difference between two squares Expand and simplify (2a + 7)(2a – 7) Expanding, 2a(2a – 7) + 7(2a – 7) (2a + 7)(2a – 7) = – 49 = – 14a + 14a 4a2 = 4a2 – 49 When we simplify, the two middle terms cancel out. This is the difference between two squares. In general, (a + b)(a – b)= a2 – b2
A1 Algebraic manipulation Contents 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 • A
Expanding or multiplying out a(b + c) ab + ac Factorizing Factorizing an expression is the opposite of expanding it. Factorizing expressions Often: When we expand an expression we remove the brackets. When we factorize an expression we write it with brackets.
Expressions can be factorized by dividing each term by a common factor and writing this outside of a pair of brackets. Factorizing expressions 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 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) 5(x+ 2)
Writing 5x + 10 as 5(x + 2) is called factorizing the expression. Factorizing expressions Factorize 6a + 8 Factorize 12n – 9n2 The highest common factor of 6a and 8 is The highest common factor of 12n and 9n2 is 2. 3n. (6a + 8) ÷ 2 = 3a + 4 (12n – 9n2) ÷ 3n = 4 – 3n 6a + 8 = 2(3a + 4) 12n – 9n2 = 3n(4 – 3n)
Writing 5x + 10 as 5(x + 2) is called factorizing the expression. Factorizing expressions Factorize 3x + x2 Factorize 2p + 6p2 – 4p3 The highest common factor of 3x and x2 is The highest common factor of 2p, 6p2 and 4p3 is x. 2p. (2p + 6p2 – 4p3) ÷ 2p = (3x + x2) ÷ x = 3 + x 1 + 3p– 2p2 3x + x2 = x(3 + x) 2p + 6p2 – 4p3 = 2p(1 + 3p– 2p2)
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)
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 out a factor of –3, – 6 becomes + 2 xy – 6+ 2y – 3x =xy + 2y– 3x – 6 = y(x + 2)– 3(x + 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)
A1 Algebraic manipulation Contents A1.1 Using index laws • A A1.2 Multiplying out brackets • A A1.4 Factorizing quadratic expressions A1.3 Factorization • A • A A1.5 Algebraic fractions • A
Quadratic expressions t2 ax2 + bx + c (where a = 0) 2 A quadratic expression is an expression in which the highest power of the variable is 2. For example, x2 – 2, w2 + 3w + 1, 4 – 5g2 , The general form of a quadratic expression in x is: 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.
Expanding or multiplying out a2 + 3a + 2 (a + 1)(a + 2) Factorizing Remember: factorizing an expression is the opposite of expanding it. Factorizing expressions Often: When we expand an expression we remove the brackets. When we factorize an expression we write it with brackets.