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 exponent notation. The number n is called the exponent 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.

Multiplying terms involving exponents We can use exponent 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

Multiplying terms with the same base When we multiply two terms with the same base the exponents 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)

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

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, 15. 150 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

Dividing terms with the same base When we divide two terms with the same base the exponents 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)

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.

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 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