MEI C1 Study Resources Core1 Polynomials 5 Binomial Expansion

MEI Core 1 Polynomials Section 5 : Binomial expansions Study plan Background The binomial expansion and binomial coefficients crop up in many areas of life particularly when dealing with statistics. You may have met Pascal’s triangle before and this is very useful for expansions of expressions to relatively low powers but the binomial expansion is a powerful tool for expansions required to higher powers particularly when you only need the coefficient of one term. It is important that you are systematic and neat when doing this section as mistakes are easily made if work is muddled.

Detailed work plan 1. Read pages 108 – 111 and study the examples on pages 112 – 114 carefully. There are some further examples in the Notes and Examples. 2. You may find the Flash resources Pascal’s triangle and binomial expansions and Finding terms in binomial expansions helpful. There is also a Binomial expansions video. 3. Exercise 3F 1i), 1iv), 1v), 3i), 3iii), 3v)*, 5, 6*, 8, 9* 4. For further practice you can try the interactive questions Evaluating binomial coefficients and Finding coefficients in binomial expansions.

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MEI Core 1 Polynomials Section 5: Binomial expansions Notes and Examples These notes contain subsections on:  Using Pascal’s triangle  The formula for binomial coefficients  Binomial expansion using the formula  Approximations In this section you will learn to multiply out expressions of the form (a + b)n, where n is a positive integer.

Using Pascal’s triangle Expanding an expression of the form (a + b)n using Pascal’s triangle is quite straightforward.   

Find the appropriate row of binomial coefficients from Pascal’s triangle (the first two numbers in the row are 1 and n, and the row has n + 1 numbers in it in total) The first term is an and the powers of a go down by 1 in each term, so that the last term has no a in it. The first term has no b in it and the powers of b go up by 1 in each term, so that the last term is bn.

Example 1 Expand (p + q)5 . Solution

The row you need from Pascal’s triangle starts with 1, 5 and has 6 numbers in it. The row is 1, 5, 10, 10, 5, 1.

(p + q)5 = p5 + 5p4q + 10p3q2 + 10p2q3 + 5pq4 + q5 The Flash resource Pascal’s triangle and binomial coefficients shows how the expansion of (a + b)n is related to Pascal’s triangle. You expand (a + b)n in the same way whether a and b are numbers or letters. If they are numbers a little more work is involved as you need to work out the actual powers and multiply by the binomial coefficients. This is shown in the next example.

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MEI C1 Polynomials Section 5 Notes and Examples Example 2 Expand (x + 2)6 Solution Using Pascal’s triangle, the binomial coefficients are 1, 6, 15, 20, 15, 6, 1. (x + 2)6 = 1x6 + 6x5(2) + 15x4(22) + 20x3(23) + 15x2(24) + 6x(25) + 1(26) = x6 + 12x5 + 60x4 + 160x3 + 240x2 + 192x + 64

If a negative is involved, remember that even powers of a negative number are positive, e.g. (-2)4 = 16, but odd powers of a negative number are negative, e.g. (-2) 3 = -8.

Example 3 Expand (2x – 3y)4 Solution From Pascal’s triangle, the binomial coefficients are 1, 4, 6, 4, 1 (2x – 3y)4 = (2x)4 + 4(2x)3(-3y) + 6(2x)2(-3y)2 + 4(2x)(-3y)3 + (-3y)4 = 16x4 + 4(8x3)(-3y) + 6(4x2)(9y) + 4(2x)(-27y3) + 81y4 = 16x4 – 96x3y + 216x2y2 – 216xy3 + 81y4

The formula for binomial coefficients It is possible to calculate the numbers in Pascal’s triangle directly using a formula. If you are studying Statistics, you may have already met the factorial notation n! and the expression nCr.

 n n! The definition of the binomial coefficient nCr, or   , is . r !(n  r )! r n(n  1)....(n  r  1) However, it is usually easier to use the form . 1 2  3  ...  r This looks complicated in algebra but it is much easier when working with numbers. The denominator is the product of the integers from 1 up to r, and the numerator is the product of the integers from n down until you have r numbers altogether. So there must be an equal number of integers in the numerator and the denominator. If you have to work this out without a calculator, there will be a lot of cancelling you can do, so the calculation usually ends up quite simple. The example below demonstrates working out without a calculator.

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MEI C1 Polynomials Section 5 Notes and Examples Note that the binomial coefficients are symmetrical, so that n Cr  n Cnr . So if you want to work out is much easier.

Example 4 Find (i) Solution (i)

(ii)

6

C3 =

10

6

C3

15

(ii)

C12 , for example, you can work out

10

5

C3 instead, which

(iii) 12C4

C8

6  5 4 = 20 1 2  3

C8 = 10C2 =

15

By symmetry 10C8 = 10C2

10  9 = 45 1 2 5

12  11 10  9 (iii) C4 = 1 2  3  4 = 11  5  9 = 495 12

3 and 4 cancel with 12, and 2 cancels with 10 leaving 5.

For practice in examples like the one above, try the interactive questions Evaluating binomial coefficients.

Binomial expansion using the formula Although you can use Pascal’s triangle from tables for most of the expansions you are likely to meet at this stage, there are times when you will have to use the formula for the binomial coefficients. If the value of n is large, the table may not show the row that you need. In the A2 part of your ‘A’ level course you will learn to expand expressions involving a value of n that is not a positive whole number, such as –3 or ½. For values of n like these, you cannot use Pascal’s triangle to find the binomial coefficients, so you have to use the formula. The work you are doing now is good preparation for this.

Example 5 In the expansion of  3  2x  , find the coefficient of 8

(i)

x2

(ii)

x5

(iii)

x7

Solution (i)

8 7  729  4 x 2 1 2  81648 x 2

The term in x 2 is 8 C2  36   2 x   2

The coefficient of x 2 is 81648.

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MEI C1 Polynomials Section 5 Notes and Examples (ii)

8 7  6  27  32 x 2 1 2  3  48384 x 2

The term in x 5 is 8 C5  33   2 x   5

The coefficient of x 5 is –48384. (iii)

The term in x 7 is 8 C7  31   2 x   8  3  128 x 2 7

 3072 x 2 The coefficient of x is –3072. 7

You can see more examples like this using the Flash resource Finding terms in binomial expansions. You can also look at the Binomial expansions video. For practice in examples like the one above, try the interactive questions Finding coefficients in binomial expansions.

Approximations If the value of x is small, then the first few terms of a binomial expansion make a good approximation. The example below shows this.

Example 6 (i) Find the first four terms in the expansion of (1 + 2x)30 (ii) By substituting x = 0.01, find an approximate value for 1.02 30 (iii) Compare this approximate value with the answer you get on a calculator. Solution (i) (1 + 2x)30 = 1 + 30C1(1)29(2x) + 30C2(1)28(2x)2 + 30 C3(1)27(2x)3 + … 30  29 30  29  28 = 1 + 30(2x) + (4x2) + (8x3) + … 1 2 1 2  3 = 1 + 60x + 1740x2 + 32480x3 + … (ii)

1.0230  1 + 600.01 + 1740(0.01)2 + 32480(0.01)3  1 + 0.6 + 17400.0001 + 324800.000001  1.6 + 0.174 + 0.03248  1.80648

(iii)

Using a calculator 1.02 30 = 1.811 The approximation is correct to two decimal places.

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Core 1 Polynomials Section 5: Binomial expansions Crucial points 1. When using the binomial expansion, make sure that you remember to raise the whole term to the appropriate power e.g. in the expansion of (1 + 2x)n, remember that (2x)r = 2rxr.

8

Wrong:

9

Right:

( 2 + 3x )

3

= 23 + 3 × 2 2 × 3 x + 3 × 2 × 3 x 2 + 3 x 3 = 8 + 36 x + 18 x 2 + 3x3

( 2 + 3x )

3

8

= 23 + 3 × 22 × (3x) + 3 × 2 × (3 x) 2 + (3 x)3 = 8 + 36 x + 54 x 2 + 27 x3

9

2. Make sure that you can use the formula for binomial coefficients confidently You need to know what is meant by nCr - this could be tested in your examination, and you may need to show that you know this formula rather than just using Pascal’s triangle. 3. Make sure you can find specific binomial coefficients efficiently If asked to find a particular term in a binomial expansion, don’t do the full expansion (which would waste a lot of time), just find the coefficient you need, making sure you use the right binomial coefficient. Also, remember that n C r = n C n − r . Example: Solution:

Find the coefficient of x 5 in the expansion of ( 3x + 2 ) . 7

⎛7⎞ The required binomial coefficient is 7 C5 or ⎜ ⎟ . ⎝ 5⎠ 7×6 7 C5 = = 21 , so the coefficient of x 5 is 2 ×1 21× (3)5 × 22 = 20412

9

Be careful not to make an error like in 1 above! A very common incorrect answer would be: 21× (3) × 22 = 252

8

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MEI Core 1 Polynomials Section 5: Binomial expansions Exercise Do not use a calculator in this exercise. 1. Write out the following binomial expansions. (i) (ii) ( x  1)6 ( x  2)5 (iii) (iv) (3x  2 y)3 (2 x  1)4 2. Calculate the following binomial coefficients. 8 9 (i) (ii) C3 C5 (iii) 3. Find (i) (ii)

12

(iv)

C4

20

C18

the coefficient of x4 in the expansion of (1 + 2x)15 the coefficient of x23 in the expansion of (3 – x)25 10

(iii)

1  the coefficient of x in the expansion of  x   . x  2

4. Expand the expression (1 + x)4(1 – x)7 up to and including the term in x². 5. (i) (ii) (iii)

Write down the first three terms in the binomial expansion of (1 – x)15. By substituting x = 0.01, find an approximate value for 0.9915. Find the percentage error in using this approximate value instead of the true value. 9

6. (i) (ii) (iii)

 x Write down the first four terms in the binomial expansion of 1   .  2 9 By substituting x = 0.1, find an approximate value for 1.05 . Find the percentage error in using this approximate value instead of the true value.

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MEI Core 1 Polynomials Section 5: Binomial expansions Solutions to Exercise 1. Pascal’s triangle:

1 1 1 1 1

1

2 3

4

1

5 6

1 1 3 6

10 15

1 4

10 20

1 5

15

1 6

1

(i) ( x  1)6  x 6  6 x 5  15 x 4  20 x 3  15 x 2  6 x  1 (ii) ( x  2)5  x 5  5 x 4( 2)1  10 x 3( 2)2  10 x 2( 2)3  5 x( 2)4  ( 2)5

 x 5  5 x 4  2  10 x 3  4  10 x 2  8  5 x  16  32  x 5  10 x 4  40 x 3  80 x 2  80 x  32 (iii) (2 x  1)4  (2 x )4  4(2 x )3  6(2 x )2  4(2 x )  1

 16 x 4  4  8 x 3  6  4 x 2  8 x  1  16 x 4  32 x 3  24 x 2  8 x  1 (iv) (3 x  2 y )3  (3 x )3  3(3 x )2( 2 y )  3(3 x )( 2 y )2  ( 2 y )3

 27 x 3  3  9 x 2  2 y  3  3 x  4 y 2  8 y 3  27 x 3  54 x 2 y  36 xy 2  8 y 3

2. (i)

8

C3 

8 7 6  8  7  56 1 2  3 2

C 5  9C 4 

(ii)

9

(iii)

12

(iv)

20

9 8  7  6  9  2  7  126 1 2  3  4 5

12  11  10  9 C4   11  9  5  495 1 2  3  4 10

C 18 

20

20  19 C2   10  19  190 1 2

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MEI C1 Polynomials Section 5 Exercise solutions 3. (i) Term in x4  15C 4(2 x )4 

15  14  13  12  16 x 4  2180 x 4 12 34

Coefficient of x4 is 2180 (ii) Term in x23 

25

C 23(3)2(  x )23  25C 2(3)2(  x )23

25  24  9   x 23  2700 x 23 12 Coefficient of x23 is -2700. 

4

10  9  8  7 x 6 1 (iii) Term in x²  C 4 x     210 x 2 4 12 34 x x 2 Coefficient of x is 210. 10

6

4. (1  x )4  1  4 x  6 x 2  ... (1  x )7  1  7 x  21 x 2 ...

(1  x )4(1  x )7   1  4 x  6 x 2  ...  1  7 x  21 x 2  ...   (1  7 x  21 x 2 )  4 x(1  7 x )  6 x 2  ...  1  7 x  21 x 2  4 x  28 x 2  6 x 2  ...  1  3 x  x 2  ...

5. (i) (1  x )15  1  15 x  (ii) Putting x = 0.01:

15  14 2 x  ...  1  15 x  105 x 2  ... 12

(1  0.01)15  1  15  0.01  105  0.01 2  ... 0.9915  1  0.15  0.0105  ...  0.8605

0.8605  0.9915  100  0.051% (iii) Percentage error  0.9915 6. (i)

9 2 3  1  x   1  9  x   9  8  x   9  8  7  x   ...         2 123  2    2  12  2  9  1  2 x  9 x 2  212 x 3  ... 9

0.1  9 2 3 21 (ii) Putting x = 0.1:  1    1  2  0.1  9  0.1  2  0.1  ... 2   9 1.05  1  0.45  0.09  0.0105  ...  1.5505 (iii) Percentage error 

1.5505  1.05 9  100  0.053% 1.05 9

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MEI Core 1 Polynomials Section 5: Binomial expansions Multiple Choice Test Do not use a calculator for this test. 1) The value of 12C3 is (a) 1320 (c) 440 (e) I don’t know

(b) 220 (d) 36

2) The value of 15C11 is (a) 32760 (c) 8196 (e) I don’t know

(b) 1365 (d) 165

Questions 3 and 4 refer to the expansion of (a – b)5 3) The term in a2 is (a) –5a2b3 (c) 5a2b3 (e) I don’t know

(b) 10a2b3 (d) –10a2b3

4) The term in b4 is (a) 5ab4 (c) 10ab4 (e) I don’t know

(b) –5ab4 (d) –10ab4

Questions 5, 6 and 7 refer to the expansion of (x + 3)8. 5) The coefficient of x3 is (a) 243 (c) 13608 (e) I don’t know

(b) 56 (d) 1512

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MEI C1 Polynomials Section 5 MC test 6) The coefficient of x4 is (a) 4536 (c) 81 (e) I don’t know

(b) 70 (d) 5670

7) The coefficient of x6 is (a) 20412 (c) 9 (e) I don’t know

(b) 28 (d) 252

Questions 8 and 9 refer to the expansion of (2x – 1)30 8) The coefficient of x2 is (a) 1740 (c) –1740 (e) I don’t know

(b) 435 (d) –435

9) The coefficient of x3 is (a) –4060 (c) –32480 (e) I don’t know

(b) 32480 (d) 4060

10) Using the first three terms of the expansion of (1 – 2x)12 and the substitution x = 0.01 gives an approximate value for 0.9812 of (a) 0.7847 (c) 0.7732 (e) I don’t know

(b) 0.7864 (d) 0.7336

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Core 1 Polynomials Chapter assessment Do not use a calculator for this test. 1. (a) (b) (c) (d)

Add (x³ + 2x² – 3x + 1) to (2x³ + 5x – 3) Subtract (2x³ – 3x² + x – 2) from (x4 + x³ – 2x² + 1) Multiply (x³ + 4x² – 2x + 3) by (2x – 1) Multiply (x² + 2x + 3) by (x² – x + 1)

[2] [2] [3] [3]

2. (x – 3) is a factor of the polynomial x³ + ax² – 5x + 6. Find the value of a.

[2]

3. Find the remainder when 2x³ – 3x² + x + 1 is divided by (x + 2).

[2]

4. (a) Solve the equation 2x³ – x² – 5x – 2 = 0 (b) Sketch the graph of y = 2x³ – x² – 5x – 2

[4] [2]

5. (a) Show that (x – 3) is a factor of 6x³ – 17x² – 5x + 6. (b) Hence solve the equation 6x³ – 17x² – 5x + 6 = 0. (b) Sketch the graph of y = 6x³ – 17x² – 5x + 6

[2] [2] [2]

6. (a) Write the quadratic expression x² + 2x + 5 in the form A(x + B)² + C. [3] (b) Hence write down the coordinates of the minimum point of the graph y = x² + 2x + 5. [1] 7. Sketch the following graphs. (a) y = x³ (b) y = (x + 1)³ (c) y = x³ – 2 (d) y = (x – 2)³ + 3

[1] [2] [2] [2]

8. f(x) = x³ + ax² + bx + 8 (a) When f(x) is divided by (x + 1), the remainder is 18. When f(x) is divided by (x – 2), the remainder is 0. Find the values of a and b. [4] (b) Factorise f(x) completely and hence solve the equation f(x) = 0. [4] (c) Sketch the graph of y = f(x). [2] (d) Sketch the graph of y = f(x – 1), showing the coordinates of the points where the graph cuts the x axis. [2] 9. Expand (2 – x)5.

[5]

10. (a) Find the first three terms in the expansion of (1 + 2x)10. (b) Use the substitution x = 0.01 and your answer to (a) to find an approximate value for 1.0210. Give your answer to an appropriate number of decimal places.

[3]

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