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Learn about algebraic manipulation, index laws, multiplying terms, and simplifying expressions using index notation in mathematics.
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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.