MEI C1 Study Resources Core1 Polynomials 2 The Factor and Remainder Theorems

MEI Core 1 Polynomials Section 2: The factor and remainder theorems Study plan Background You are familiar with the ideas of factors of numbers and this section extends this to algebraic expressions. If one expression divides into another without remainder then it must be a factor of the second expression. Factorising is an essential skill when solving equations so the more practice at this that you get, the better. Detailed work plan 1. Read pages 88 through to the top of page 92 carefully, paying particular attention to the examples. The Notes and Examples have further examples. 2. You may find the Flash resource Factorising a cubic, the PowerPoint presentation Factorising polynomials, and the Solving cubics video helpful. 3. Exercise 3C Try questions including 1, 3*, 5, 6, 7*, 10 and 11. 4. Read pages 92 – 93 carefully. The Notes and Examples have further examples. 5. You may find the Flash resource The remainder theorem helpful. 6. Exercise 3C Try questions 9, 12, 14, 15*, 17, 18* and 19. 7. For additional practice, try the interactive resources Finding a remainder and Finding a polynomial, given the remainder.

© MEI, 16/03/09

1/1

MEI Core 1 Polynomials Section 2: The factor and remainder theorems Notes and Examples These notes contain subsections on  The factor theorem  The remainder theorem

The factor theorem If  x  a  is a factor of f(x), then f(a) = 0 and x = a is a root of the equation f(x) = 0. Conversely, if f(a) = 0, then  x  a  is a factor of f(x). The factor theorem probably won’t come as any great surprise to you. After all, you already know that you can solve some quadratics by factorising them. e.g.

to solve the quadratic equation you factorise: and deduce the solutions

x² + 3x – 10 = 0 (x + 5)(x – 2) = 0 x = -5 and x = 2

Clearly, for f(x) = x² + 3x – 10, f(-5) = 0 and f(2) = 0. (x + 5) is a factor of f(x)  f(-5) = 0 (x – 2) is a factor of f(x)  f(2) = 0 The factor theorem simply extends this idea to other polynomials such as cubics, and provides a method for solving cubics and higher polynomials.

Example 1 (i) Solve the equation x³ + 2x² – 5x – 6 = 0 (ii) Sketch the graph of y = x³ + 2x² – 5x – 6

Solution (i)

The first step is to find one solution by trial and error. If there is an integer solution x = a, then by the factor theorem (x – a) must be a factor of x³ + 2x² – 5x – 6. So a must be a factor of 6. a could therefore be 1, -1, 2, -2, 3, -3, 6 or –6.

Let f(x) = x³ + 2x² – 5x – 6 f(1) = 1 + 2 – 5 – 6 = -8 f(-1) = -1 + 2 + 5 – 6 = 0

You need to find a value of x for which f(x) = 0.

f(-1) = 0 so by the factor theorem x + 1 is a factor of f(x).

© MEI, 16/03/09

1/4

MEI C1 Polynomials Section 2 Notes and Examples The next step is to factorise f(x) into the linear factor x + 1 and a quadratic factor.

x³ + 2x² – 5x – 6 = (x + 1)  quadratic factor. Let the quadratic factor be ax² + bx + c. x³ + 2x² – 5x – 6 = (x + 1)(ax² + bx + c) = ax³ + bx² + cx + ax² + bx + c = ax³ + (a + b)x² + (b + c)x + c

Multiply out the brackets

Equating coefficients of x³  a = 1 Equating constant term  c = -6 Equating coefficients of x²  a + b = 2  b = 1 Check coefficient of x: b + c = 1 – 6 = -5 x³ + 2x² – 5x – 6 = (x + 1)(x² + x – 6) = (x + 1)(x – 2)(x + 3)

Factorise the quadratic factor

The solutions of the equation are x = -1, x = 2 and x = -3. (ii) Part (i) shows that the graph of y = x³ + 2x² – 5x – 6 crosses the x-axis at (-3, 0), (-1, 0) and (2, 0). By putting x = 0 you can see that it crosses the y-axis at (0, -6). This information allows you to sketch the graph.

In the example above, the quadratic factor was found by equating coefficients. There are a number of other methods of finding this quadratic factor. You can do it by polynomial division, or by inspection (which means doing it in your head). Which one you use is really down to personal preference. The PowerPoint presentation Factorising polynomials shows these alternative methods.

© MEI, 16/03/09

2/4

MEI C1 Polynomials Section 2 Notes and Examples You can look at some more examples of factorising cubics using the Flash resource Factorising a cubic. You can also look at the Solving cubics video.

Example 2 f(x) = 2x³ + px² + 5x – 6 has a factor x – 2. Find the value of p and hence factorise f(x) as far as possible. Solution x – 2 is a factor of f(x)  f(2) = 0 f(2) = 16 + 4p + 10 – 6 = 20 + 4p 20 + 4p = 0  p = -5 f(x) = 2x³ - 5x² + 5x – 6 2x³ - 5x² + 5x – 6

= (x – 2)(ax² + bx + c) = ax³ + bx² + cx – 2ax² - 2bx – 2c = ax³ + (b – 2a)x² + (c – 2b)x – 2c

Equating coefficients of x³  a = 2 Equating constant terms  -2c = -6  c = 3 Equating coefficients of x²  b – 2a = -5  b – 4 = -5  b = -1 Check coefficient of x: c – 2b = 3 + 2 = 5 2x³ - 5x² + 5x – 6 = (x – 2)(2x² – x + 3) The discriminant of the quadratic factor is (-1)² - 4 2  3 = 1 – 24 = -23 As this is negative, the quadratic factor cannot be factorised further.

The remainder theorem The factor theorem is a special case of the remainder theorem. The remainder theorem says: For a polynomial f(x), f(a) is the remainder when f(x) is divided by  x  a  .

f ( x)   x  a  g( x)  f (a)

If the remainder is zero, then x – a is a factor of f(x). So if x – a is a factor of f(x), then f(a) = 0. This is the factor theorem.

© MEI, 16/03/09

3/4

MEI C1 Polynomials Section 2 Notes and Examples Don’t forget about the remainder theorem! When a remainder is asked for in examination questions, there are usually quite a number of candidates who divide out to find it. This takes longer and you are much more likely to make an error. The remainder theorem is quick and straightforward to use.

Example 3 Find the remainder when f(x) = x4 – 3x³ + x² – 4 is divided by (i) x – 2 (ii) x + 1 Solution (i) By the remainder theorem the remainder is f(2) f(2) = 24 – 32³ + 2² - 4 = 16 – 24 + 4 – 4 = -8 Remainder = -8 (ii) By the remainder theorem the remainder is f(-1) f(-1) = (-1)4 – 3(-1)³ + (-1)² – 4 = 1 + 3 + 1 – 4 = 1 Remainder = 1

You can look at similar examples using the Flash resource The remainder theorem. For extra practice in examples like the one above, try the interactive questions Finding a remainder.

Example 4 f(x) = 2x³ + ax² + bx + 1 When f(x) is divided by x –1 the remainder is 7 When f(x) is divided by x + 3 the remainder is -5 Find the values of a and b. Solution f(1) = 7 f(-3) = -5

Adding:

2+a+b+1=7 a+b=4  2(-3)³ + (-3)²a – 3b + 1 = -5  -54 + 9a – 3b + 1 = -5  9a – 3b = 48  3a – b = 16 a +b=4 3a – b = 16 4a = 20 a = 5, b = -1

For extra practice in examples like the one above, try the interactive questions Finding a polynomial, given the remainder.

© MEI, 16/03/09

4/4

Core 1 Polynomials Section 2: The factor and remainder theorems Crucial points 1. Look out for chances to use the remainder theorem It is quite common in examination questions for candidates to divide out to find a remainder, forgetting that the remainder theorem can be used. Dividing out usually takes longer and is more susceptible to errors – using the remainder theorem is quick and easy! 2. Don’t lose easy marks when sketching polynomials You will often be asked to sketch a polynomial which you have already factorised. Remember you are being asked for a sketch, so you should do this in your answer booklet and NOT on graph paper. You should certainly not be wasting time plotting points. Make sure that your graph does not stop at the axes, that you have shown all the points at which the graph cuts the axes (including the y-axis), and that your graph is the correct way up. 3. Take care with signs Be careful about signs when using the factor and remainder theorems:

8

Wrong

f(2) = 0 ⇒ (x + 2) is a factor

9

Right

f(2) = 0 ⇒ (x – 2) is a factor

8

9

4. When dividing out a polynomial, remember that you can check by multiplying You can also check the solutions of an equation by substitution.

© MEI, 18/01/07

1/1

MEI Core 1 Polynomials Section 2: The factor and remainder theorems Exercise Do not use a calculator for this exercise. 1. (i) (ii) (iii)

Show that x + 1 is a factor of 2 x 3 − 5 x 2 − x + 6. Hence factorise 2 x3 − 5 x 2 − x + 6 completely. Sketch the graph of y = 2 x3 − 5 x 2 − x + 6

2. x – 2 is a factor of the polynomial x3 + ax 2 − 4 x + 12 . (i) Find the value of a. (ii) Factorise the polynomial completely. 3. Find the remainder when 2 x3 − 5 x 2 + x − 2 is divided by (i) x+2 (ii) x – 1. 4. (i) (ii) (iii)

Show that x – 2 is a factor of the polynomial f(x) = x3 − x 2 − x − 2 . Hence factorise f(x) as far as possible. What can you say about the graph of y = f(x)?

5. When the polynomial x3 + kx 2 − 5 x + 4 is divided by x – 3 the remainder is –2. Find the value of k. 6. When the polynomial 3x3 + ax 2 + bx − 1 is divided by x – 2 the remainder is 5. When it is divided by x + 1 the remainder is –1. Find the values of a and b. 7. Solve the equation 3x3 − 2 x 2 − 11x + 10 = 0 . 8. Solve the equation 2 x3 + 5 x 2 − 14 x − 8 = 0 . 9. Solve the equation 4 x3 + 12 x 2 − 7 x − 30 = 0 .

© MEI, 28/07/08

1/1

MEI Core 1 Polynomials Section 2: The factor and remainder theorems Solutions to Exercise 1. (i)

f( x )  2 x 3  5 x 2  x  6 f( 1)  2( 1)3  5( 1)2 ( 1)  6  2  5  1  6  0 so by the factor theorem, x + 1 is a factor.

(ii) 2 x 3  5 x 2  x  6  ( x  1)(2 x 2  7 x  6)

 ( x  1)(2 x  3)( x  2) (iii) y  2 x 3  5 x 2  x  6  ( x  1)(2 x  3)( x  2) When x = 0, y = 6 When y = 0, x = -1 or x  23 or x = 2 6

-1 3 2

2. (i)

2

f( x )  x 3  ax 2  4 x  12 f(2)  2 3  a  2 2  4  2  12  8  4a  8  12  4a  12 x – 2 is a factor so by the factor theorem f(2) = 0 4a  12  0 a  3

(ii) x 3  3 x 2  4 x  12  ( x  2)(ax 2  bx  c ) By inspection x 3  3 x 2  4 x  12  ( x  2)( x 2  x  6)  ( x  2)( x  2)( x  3)

3. f( x )  2 x 3  5 x 2  x  2 (i) By the remainder theorem, the remainder when f(x) is divided by x + 2 is

© MEI, 11/10/10

1/3

MEI C1 Polynomials Section 2 Exercise solutions f(-2). Remainder  f( 2)  2( 2)3  5( 2)2  2  2  16  20  2  2  40 (ii) By the remainder theorem, the remainder when f(x) is divided by x – 1 is f(1). Remainder  f(1)  2  1 3  5  1 2  1  2  2  5  1  2  4

4. (i)

f( x )  x 3  x 2  x  2 f(2)  2 3  2 2  2  2  8  4  2  2  0 so by the factor theorem, x – 2 is a factor.

(ii) x 3  x 2  x  2  ( x  2)( x 2  x  1) The quadratic expression x 2  x  1 cannot be factorised, so the expression has been factorised as far as possible. (iii) The discriminant of x 2  x  1 is 1 2  4  1  1  3 , so the quadratic equation x 2  x  1  0 has no real roots. Therefore the graph of y  x 3  x 2  x  2 crosses the x-axis once only.

5. f( x )  x 3  kx 2  5 x  4 By the remainder theorem, f(3)  2

3 3  k  3 2  5  3  4  2 27  9k  15  4  2 16  9k  2 9k  18

k  2 6. f( x )  3 x 3  ax 2  bx  1 By the remainder theorem, f(2)  5

3  23  a  22  b  2  1  5 24  4a  2b  1  5 4a  2b  18 2a  b  9 By the remainder theorem, f( 1)  1 3( 1)3  a( 1)2  b( 1)  1  1 3  a  b  1  1

a b  3

© MEI, 11/10/10

2/3

MEI C1 Polynomials Section 2 Exercise solutions Adding:

3a  6

a  2,b  5 7. 3 x 3  2 x 2  11 x  10  0 f( x )  3 x 3  2 x 2  11 x  10 f(1)  3  2  11  10  0 so (x – 1) is a factor

( x  1)(3 x 2  x  10)  0 ( x  1)(3 x  5 )( x  2)  0

x  1 or x 

5 3

or x  2

8. 2 x 3  5 x 2  14 x  8  0 f( x )  2 x 3  5 x 2  14 x  8 f(1)  2  5  14  8  15 f( 1)  2  5  14  8  9 f(2)  16  20  28  8  0 so (x – 2) is a factor

( x  2)(2 x 2  9 x  4)  0 ( x  2)(2 x  1)( x  4)  0

x  2 or x   21 or x  4

9. 4 x 3  12 x 2  7 x  30  0 f( x )  4 x 3  12 x 2  7 x  30 f(1)  4  12  7  30  21 f( 1)  4  12  7  30  15 f(2)  32  48  14  30  36 f( 2)  32  48  14  30  0 so (x + 2) is a factor

( x  2)(4 x 2  4 x  15 )  0 ( x  2)(2 x  5 )(2 x  3)  0

x  2 or x   52 or x 

3 2

© MEI, 11/10/10

3/3

MEI Core 1 Polynomials Section 2: The factor and remainder theorems Multiple Choice Test Do not use a calculator for this test. 1) Which of the following is a factor of x³ + x² + 2x + 8? (a) x + 2 (c) x + 1 (e) I don’t know

(b) x – 2 (d) x – 1

2) Which of the following is NOT a factor of 2x³ + x² – 2x – 1? (a) 2x + 1 (c) x + 1 (e) I don’t know

(b) 2x – 1 (d) x – 1

3) x – 2 is a factor of 3x³ – 5x² + ax + 2. The value of a is (a) 21 (c) -21 (e) I don’t know

(b) 3 (d) -3

4) (x – 1) is a factor of 2x³ + x² – 5x + 2. This expression can be written in the form (a) (x – 1)(2x² + x – 2) (c) (x – 1)(x² + 2x + 2) (e) I don’t know

(b) (x – 1)(2x² + 3x – 2) (d) (x – 1)(2x² - 3x + 2)

5) The factorisation of 6x³ + 7x² – x – 2 is (a) (x + 1)(2x – 1)(3x + 2) (c) (x – 1)(2x – 1)(3x – 2) (e) I don’t know

(b) (x + 1)(2x + 1)(3x – 2) (d) (x – 2)(2x – 1)(3x – 1)

6) When x³ – 3x² + 2x – 4 is divided by x – 3, the remainder is (a) -64 (c) -2 (e) I don’t know

(b) 2 (d) 64

© MEI, 01/12/08

1/3

MEI C1 Polynomials Section 2 MC test 7) When x4 + 2x – 5 is divided by x + 1, the remainder is (a) 6 (c) 2 (e) I don’t know

(b) -6 (d) -2

8) When x³ – 3x² + ax + 1 is divided by x – 2, the remainder is 3. The value of a is (a) 9 (c) 3 (e) I don’t know

(b) 0 (d) 2

9) When 2x4 + px² + qx – 1 is divided by x + 2, the remainder is 15. When it is divided by x – 1, the remainder is 0. The values of p and q are (a) p = -3, q = 2 (c) p = 3, q = 2 (e) I don’t know

(b) p = -3, q = -2 (d) p = 3, q = -2

10) Which of the graphs below represents y = (x – a)(x – b)(x + c), where a, b and c are all positive constants? y y

P

Q

x

x

y

y

S

R

x

© MEI, 01/12/08

x

2/3

MEI C1 Polynomials Section 2 MC test (a) P (c) R (e) I don’t know

(b) Q (d) S

© MEI, 01/12/08

3/3