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**A1 Algebraic manipulation**

KS4 Mathematics A1 Algebraic manipulation

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**A1 Algebraic manipulation**

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

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**xn 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. Start by asking pupils to simplify x + x + x + x + x. This is 5 lots of x, which is written as 5x. Next ask pupils how we could simplify x × x × x × x × x. Make sure there is no confusion between this repeated multiplication and the previous example of repeated addition. If x is equal to 2, for example, x + x + x + x + x equals 10, while x × x × x × x × x equals 32. Some pupils may suggest writing xxxxx. While this is not incorrect, neither has it moved us on very far. Point out the problems of readability, especially with high powers. When we write a number or term to the power of another number it is called index notation. The power, or index (plural indices), is the superscript number, in this case 5. The number or letter that we are multiplying successive times, in this case, x, is called the base. Practice the relevant vocabulary: x2 is read as ‘x squared’ or ‘x to the power of 2’; x3 is read as ‘x cubed’ or ‘x to the power of 3’; x4 is read as ‘x to the power of 4’. xn The number x is called the base.

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**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 Discuss each example briefly. Remind pupils to multiply any numbers together first followed by letters in alphabetical order. In the last example 3t × 3t the use of brackets may need further clarification. We must put a bracket around the 3t since both the 3 and the t are squared. If we wrote 3t2, then only the t would be squared. Give a numerical example, if necessary. If t was 2 then 3t would be equal to 6. We would then have 62, 36. If we wrote 3t2, that would mean 3 × 22 or 3 × 4 which is 12. Remember the order of operations - BIDMAS. Brackets are worked out before indices, but indices are worked out before multiplication. 3r × r2 = 3 × r × r × r = 3r3 3t × 3t = (3t)2 or 9t2

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**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, Stress that the indices can only be added when the base is the same. xm × xn = x(m + n)

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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, Point out that we do not need to write the brackets when we write a + b all over c. Since both letters are above the dividing line we know that it is the sum of a and b that is divided by c. The dividing line effectively acts as a bracket. (a + b) ÷ c is written as a + b c

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Dividing terms Like a fraction, we can often simplify expressions by cancelling. For example, n3 n2 6p2 3p n3 ÷ n2 = 6p2 ÷ 3p = 2 n × n × n n × n 6 × p × p 3 × p = = In the first example, we can divide both the numerator and the denominator by n. n ÷ n is 1. We can divide the numerator and the denominator by n again to leave n. (n/1 is n). If necessary, demonstrate this by substitution. For example, 3 cubed, 27, divided by 3 squared, 9, is 3. Similarly 5 cubed, 125, divided by 5 squared, 25, is 5. In the second example, we can divide the numerator and the denominator by 3 and then by p to get 2p. Again, demonstrate the truth of this expression by substitution, if necessary. For example, if p was 5 we would have 6 × 5 squared, 6 × 25, which is 150, divided by 3 × 5, divided by 15 is 10, which is 2 times 5. Stress that this will work for any number we choose for p. Pupils usually find multiplying easier than dividing. Encourage pupils to check their answers by multiplying (using inverse operations). For example, n × n2 = n3. And 2p × 3p = 6p2 = n = 2p

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**Dividing terms with the same base**

When we divide two terms with the same base the indices are subtracted. For example, a × a × a × a × a a × a = a5 ÷ a2 = a × a × a = a3 = a (5 – 2) 2 4 × p × p × p × p × p × p 2 × p × p × p × p 4p6 ÷ 2p4 = = 2 × p × p = 2p2 = 2p(6 – 4) Stress that the indices can only be subtracted when the base is the same. In general, xm ÷ xn = x(m – n)

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Hexagon puzzle The numbers in the squares are found by multiplying the terms in the circles on either side. Reveal each term until the puzzle is complete. The order can be modified to practice both multiplication and division of indices. Discuss briefly the fact that when we multiply two terms together with the same base (the letter n, in this case) we add the powers. When we divide two terms with the same base, we subtract the powers.

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**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 ( ) = y6 = p4 × q8 = p4q8

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**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( ) = a15 = a(3 × 5) In general, (xm)n = xmn

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**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 You may wish to ask pupils to complete this exercise individually before talking through the answers. The zero index is introduced in the last question and discussed on the next slide. 7) (h½)2 = h 8) (7a4b–3)0 = 1

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**Any number or term divided by itself is equal to 1.**

The zero index Any number or term divided by itself is equal to 1. Look at the following division: y4 ÷ y4 = 1 But using the rule that xm ÷ xn = x(m – n) y4 ÷ y4 = y(4 – 4) = y0 That means that y0 = 1 Stress that this rule is only true for non-zero values of x. 00 is undefined. In general, for all x 0, x0 = 1

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**Negative indices Look at the following division: b × b b × b × b × b =**

1 b × b = 1 b2 b2 ÷ b4 = But using the rule that xm ÷ xn = x(m – n) b2 ÷ b4 = b(2 – 4) = b–2 That means that 1 b2 Discuss the general form of each result where x is any number and m and n are integers. b–2 = In general, x–n = 1 xn

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**This is the reciprocal of u.**

Negative indices Write the following using fraction notation: This is the reciprocal of u. 1 u u–1 = 2 b4 2b–4 = x2 y3 x2y–3 = A number or term raised to the power of –1 is the reciprocal of the number or term. See N2.4 Reciprocals 2a (3 – b)2 2a(3 – b)–2 =

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**Negative indices Write the following using negative indices: 2 a =**

x3 y4 = x3y–4 p2 q + 2 = p2(q + 2)–1 Use the last two examples to explain that when we have terms linked by + and – in the denominator, or bracketed expressions, the whole of that expression must be raised to the negative power and not the individual terms. 3m (n2 + 2)3 = 3m(n2 + 2)–3

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**Fractional indices Indices can also be fractional. x × x = x + = x1 =**

2 x = 1 2 x1 = x But, x × x = x The square root of x. So, x = x 1 2 Similarly, x × x × x = 1 3 x = 1 3 x1 = x Discuss the fact that when we square the square root of a number we end up with the original number. Similarly, we get back to where we started when we cube the cube root of a number. But, x × x × x = x 3 The cube root of x. So, x = x 1 3

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**Fractional indices In general, x = x Also, we can write x as x . × m**

1 n n Also, we can write x as x m n 1 × m Using the rule that (xm)n = xmn, we can write x × m = (x )m = (x)m 1 n We can also write x as xm × . m n 1 x = (xm) = xm 1 n m× In general, x = xm x = (x)m m n or

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**1 xm × xn = x(m + n) x–1 = x 1 xm ÷ xn = x(m – n) x–n = xn (xm)n = xmn**

Index laws Here is a summary of the index laws. x–1 = 1 x xm × xn = x(m + n) xm ÷ xn = x(m – n) x–n = 1 xn (xm)n = xmn x = x 1 2 x1 = x Discuss the general form of each result where x is any number and m and n are integers. x = x 1 n x0 = 1 (for x = 0) x = xm or (x)m n m

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**A1.2 Multiplying out brackets**

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

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**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. The lines shown in orange show which terms we are multiplying together. 3y(4 – 2y) = 12y – 6y2

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**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

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**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

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**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 In this example, we have 4 and then –(5n – 3). This is equivalent to –1 × (5n – 3). Repeat that a minus sign in front of a bracket has the effect of changing the sign of every term inside the bracket. 5n becomes – 5n and – 3 becomes + 3. Let’s write the like terms next to each other. When we collect the like terms together we have 7 – 5n. = – 5n = 7 – 5n

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**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 In this example, we have two sets of brackets. The first set is multiplied by 2 and the second set is multiplied by 3. We don’t need to use a grid as long as we remember to multiply every term inside the bracket by every term outside it. Talk through the multiplication of (3n – 4) by 2 and (3n + 5) by 3. Let’s write the like terms next to each other. When we collect the like terms together we have 6n + 9n which is 15n and – = 7. = 6n + 9n – = 15n + 7

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**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 Here is another example with two sets of brackets. The first set is multiplied by 5 and the second set is multiplied by minus a. Talk through the multiplication of (3a + 2b) by 5. Next we need to multiply (2 + 5b) by –a. Talk through this. = 15a – 2a + 10b – 5ab = 13a + 10b – 5ab

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**Find the area of the rectangle**

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**Find the area of the rectangle**

What is the area of a rectangle of length (a + b) and width (c + d)? a b c d ac bc ad bd In general, (a + b)(c + d) = ac + ad + bc + bd

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**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. In this example, we need to multiply everything in the second bracket by 3 and then everything in the second bracket by t. We can write this as 3(4 – 2t) + t(4 – 2t). As pupils become more confident, they can leave this intermediate step out. (3 + t)(4 – 2t) = 3(4 – 2t) + t(4 – 2t) This is a quadratic expression. = 12 – 6t + 4t – 2t2 = 12 – 2t – 2t2

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**Using the grid method to expand brackets**

Demonstrate as many examples as is necessary for pupils to understand the method. Once pupils are happy with the method they may be able to multiply out brackets without using a grid.

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**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 … … and that –10 is the product of –5 and 2. Point out that for any expression in the form (x + a)(x + b), where a and b are fixed numbers, the expanded expression will have an x with a coefficient of a + b and the number at the end will be a × b.

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**Matching quadratic expressions 1**

Select a bracketed expression and ask a volunteer to find its corresponding expansion.

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**Matching quadratic expressions 2**

Select a bracketed expression and ask a volunteer to find its corresponding expansion.

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**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

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**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

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Squaring expressions Any of the terms in the expansion can be hidden or revealed to practice squaring expressions.

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**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 between two squares. In general, (a + b)(a – b) = a2 – b2

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**The difference between two squares**

This animation demonstrates geometrically why a2 – b2 = (a – b)(a + b). The first two steps in the animation demonstrate a square of area b2 being subtracted from a square of area a2 to give a shape of area a2 – b2. The shape of area a2 – b2 is then rearranged in to a rectangle of width a + b and height a – b to show that its area is equal to (a – b)(a + b). a2 – b2 is therefore equal to (a – b)(a + b).

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**Matching the difference between two squares**

Select a bracketed expression and ask a volunteer to find its corresponding expansion.

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**A1 Algebraic manipulation**

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

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**Factorizing expressions**

Factorizing an expression is the opposite of expanding it. Expanding or multiplying out Factorizing a(b + c) ab + ac Often: When we expand an expression we remove the brackets. When we factorize an expression we write it with brackets.

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**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. We can write the 5 outside of a set of brackets Encourage pupils to check this by multiplying the expression out to 5x + 10. (5x + 10) ÷ 5 = x + 2 This is written inside the bracket. 5(x + 2) 5(x + 2)

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**Factorizing expressions**

Writing 5x + 10 as 5(x + 2) is called factorizing the expression. 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 Point out that we do not normally show the line involving division. This is done mentally. We can check the answer by multiplying out the bracket. 6a + 8 = 2(3a + 4) 12n – 9n2 = 3n(4 – 3n)

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**Factorizing expressions**

Writing 5x + 10 as 5(x + 2) is called factorizing the expression. 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)

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Factorization Start by asking pupils to give you the value of the highest common factor of the two terms. Reveal this and then ask pupils to give you the values of the terms inside the brackets.

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**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 b Two terms share a common factor of 4 and the remaining two terms share a common factor of b. 4a + ab b = 4a 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)

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**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) This expression could also be written as xy – 3x + 2y – 6 to give x(y – 3) + 2(y – 3) = (y – 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)

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**A1.4 Factorizing quadratic expressions**

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

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**Quadratic expressions**

A quadratic expression is an expression in which the highest power of the variable is 2. For example, t2 2 x2 – 2, w2 + 3w + 1, 4 – 5g2 , The general form of a quadratic expression in x is: ax2 + bx + c (where a = 0) x is a variable. As well as the highest power being two, no power in a quadratic expression can be negative or fractional. Compare each of the quadratic expressions given with the general form. In x2 – 2, a = 1, b = 0 and c = –2. In w2 + 3w + 1, a = 1, b = 3 and c = 1. This is a quadratic in w. In 4 – 5g2, a = –5, b = 0 and c = 4. This is a quadratic in g. In t2/2, a = ½, b = 0 and c = 0. This is a quadratic in t. 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.

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**Factorizing expressions**

Remember: factorizing an expression is the opposite of expanding it. Expanding or multiplying out Factorizing (a + 1)(a + 2) a2 + 3a + 2 Often: When we expand an expression we remove the brackets. When we factorize an expression we write it with brackets.

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**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 Pupils will require lots of practice to factorize quadratics effectively. This slide explains why when we factorize an expression in the form x2 + bx + c to the form (x + d)(x + e) the values of d and e must be chosen so that d + e = b and de = c. (x + d)(x + e) = x2 + (d + e)x + de is an identity. This means that the coefficients and constant on the left-hand side are equal to the coefficients and constant on the right-hand side. 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.

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**Factorizing quadratic expressions 1**

Factorize the given expression by finding two integers that add together to give the coefficient of x and multiply together to give the constant. It may be a good idea to practice adding and multiplying negative numbers before attempting this activity. Use slide 31 in N1.2 Calculating with integers to do this if required. The lower of the two hidden integers will be given first in each case.

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**Matching quadratic expressions 1**

Select a quadratic expression and ask a volunteer to find its corresponding factorization.

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**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 Discuss the factorization of quadratics where the coefficient of x2 is not 1. Most examples at this level will have a as a prime number so that there are only two factors, 1 and the number itself. 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

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**Factorizing quadratic expressions 2**

Factorize each given expression using trial and improvement and the relationships shown on the previous slide. For each expression in the form ax2 + bx + c, start by using the pen tool to write down pairs of integers that multiply together to make a and pairs of integers that multiply together to make c. Use these to complete the factorization.

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**Matching quadratic expressions 2**

Select a quadratic expression and ask a volunteer to find its corresponding factorization.

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**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, See slide 37 to demonstrate the expansion of expressions of the form (x + a)(x – a). Pupils should be encouraged to spot the difference between two squares whenever possible. 9x2 – 16 = (3x + 4)(3x – 4) 25a2 – 1 = (5a + 1)(5a – 1) m4 – 49n2 = (m2 + 7n)(m2 – 7n)

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**Factorizing the difference between two squares**

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**Matching the difference between two squares**

Select an expression involving the difference between two squares and ask a volunteer to find the corresponding factorization.

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**A1 Algebraic manipulation**

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

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**Algebraic fractions 3x 4x2 2a 3a + 2**

and are examples of algebraic fractions. 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. It is important to realize that, like numerical fractions, multiplying or dividing the numerator and the denominator of an algebraic fraction by the same number, term or expression does not change the value of the fraction. This fact is used both when simplifying algebraic fractions and when writing algebraic fractions over a common denominator to add or subtract them. For example, 3x 4x2 = 3 4x = 6 8x = 3y 4xy = 3(a + 2) 4x(a + 2)

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**Equivalent algebraic fractions**

It is important to realize that, like numerical fractions, multiplying or dividing the numerator and the denominator of an algebraic fraction by the same number, term or expression does not change the value of the fraction. This fact is used both when simplifying algebraic fractions and when writing algebraic fractions over a common denominator to add or subtract them.

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**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, Simplify 6ab 3ab2 2 6ab 3ab2 = 6 × a × b 3 × a × b × b = 2 b

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**Simplifying algebraic fractions**

Sometimes we need to factorize the numerator and the denominator before we can simplify an algebraic fraction. For example, Simplify 2a + a2 8 + 4a 2a + a2 8 + 4a = a (2 + a) 4(2 + a) = a 4

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**Simplifying algebraic fractions**

b2 – 36 is the difference between two squares. Simplify b2 – 36 3b – 18 b2 – 36 3b – 18 = (b + 6)(b – 6) 3(b – 6) b + 6 3 = Pupils should be encouraged to spot the difference between two squares whenever possible. If required, we can write this as 6 3 = b + b 3 + 2

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**Manipulating algebraic fractions**

Remember, a fraction written in the form a + b c can be written as b a + However, a fraction written in the form c a + b cannot be written as b a + Stress that if two terms are added or subtracted in the numerator of a fraction, we can split the fraction into two fractions written over a common denominator. The converse is also true. However, if two terms are added or subtracted in the denominator of a fraction, we cannot split the fraction into two. Verify these rules using the numerical example. For example, 1 + 2 3 = 2 1 + 3 1 + 2 = 2 1 + but

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**Multiplying and dividing algebraic fractions**

We can multiply and divide algebraic fractions using the same rules that we use for numerical fractions. In general, a b × = c d ac bd a b ÷ = c d × = ad bc and, Point out to pupils that in the example we could multiply out the brackets in the denominator. However, it is usually preferable to leave expressions in a factorized form. For example, 3p 4 × = 2 (1 – p) 3 6p 4(1 – p) = 3p 2(1 – p) 2

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**Multiplying and dividing algebraic fractions**

What is 2 3y – 6 ÷ 4 y – 2 ? This is the reciprocal of 4 y – 2 2 3y – 6 ÷ = 4 y – 2 2 3y – 6 × 4 y – 2 2 3(y – 2) × = 4 y – 2 2 1 6 =

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**Adding algebraic fractions**

We can add algebraic fractions using the same method that we use for numerical fractions. For example, What is 1 a + 2 b ? We need to write the fractions over a common denominator before we can add them. 1 a + 2 b = b ab + 2a = b + 2a ab If necessary review the method for adding numerical fractions. In general, + = a b c d ad + bc bd

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**Adding algebraic fractions**

3 y What is + ? x 2 We need to write the fractions over a common denominator before we can add them. 3 x + y 2 = + 3 × 2 x × 2 y × x 2 × x + 6 2x xy = = 6 + xy 2x

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**Subtracting algebraic fractions**

We can also subtract algebraic fractions using the same method as we use for numerical fractions. For example, What is – ? p 3 q 2 We need to write the fractions over a common denominator before we can subtract them. – = p 3 q 2 – = 2p 6 3q 2p – 3q 6 In general, – = a b c d ad – bc bd

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**Subtracting algebraic fractions**

2 + p 4 3 2q What is – ? = – 2 + p 4 3 2q – (2 + p) × 2q 4 × 2q 3 × 4 2q × 4 = – 2q(2 + p) 8q 12 = 2q(2 + p) – 12 8q 6 The denominators in this example share a common factor. That means that we will either have to cancel at the end of the calculation (as shown) or use a common denominator of 4q in the first step. 4 = q(2 + p) – 6 4q

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**Addition pyramid – algebraic fractions**

Start by revealing the three fractions on the bottom row of the wall. Add the fractions together to find the missing values in the bocks above. Each block is the sum of the two fractions below it. Ask pupils if it is true to say that the fraction in the top row is the sum of the three fractions in the bottom row. Conclude that if the three factions on the bottom row are a, b and c then the fraction on the top row is a + 2b + c. The activity can be varied by revealing one fraction in each row and using subtraction to find those that are missing.

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Methods and Solving Equations

Methods and Solving Equations

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