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the factor & remainder theorems - solutions

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AQA Core 1 Polynomials 1 of 4 10/01/13 © MEI  Section 2: The factor and remainder theorems Solutions to Exercise 1.  i) 3 2 3 2 7 1 1 1 1 1 2 2 2 2 4 4 2 f ) 2 7 4 f ) 2 7 4 4 0 x x x x  By the factor theorem, 2  + 1 is a factor, so there is no remainder. 3 2 2 3 2 2 3 2 7 4 2 1) ) 2 2 2 2 2) 2) x x x x x xc  x x x x cx c  x x cx c  Equating coefficients of x³  1  Equating constant terms  4  Equating coefficients of x 7 1 c Check: coefficient of x² 1 2 1   2 2  2 7 4) 2 1) 4 x x x x x  ii) 3 2 f ) 2 4 f 1) 1 2 4 1 x x x   3 2 2 3 2 2 3 2 2 4 1) ) 1  ) ) 1 x x x x xc  x x x x xc  x x c xc  Equating coefficients of x³  1  Equating constant terms 4 3 c  Equating coefficients of x² 3  Check: coefficient of x 3 0   2 2  2 4) 1) 3 3 x x x x  remainder -1. 2.  i) 2 f ) 2 5 6 x x x   2 f 1) 2 1) 5 1) 1) 6 2 5 1 6 0  so by the factor theorem, x + 1 is a factor. ii) 3 2 2 2 5 6 1) 2 7 6)  1) 2 3) 2) x x x x x x x x x  iii) 2 2 5 6 1) 2 3) 2) x x x x x x  When x = 0, y = 6 When y = 0, x = -1 or  3 2  or x = 2

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  • AQA Core 1 Polynomials

    1 of 4 10/01/13 MEI

    Section 2: The factor and remainder theorems

    Solutions to Exercise

    1. (i)

    3 2

    3 2 71 1 1 1 1 12 2 2 2 4 4 2

    f( ) 2 7 4

    f( ) 2 7 4 4 0

    x x x x

    By the factor theorem, 2x + 1 is a factor, so there is no remainder.

    3 2 2

    3 2 2

    3 2

    2 7 4 (2 1)( )

    2 2 2

    2 ( 2 ) ( 2 )

    x x x x ax bx c

    ax ax bx bx cx c

    ax a b x b c x c

    Equating coefficients of x 1a Equating constant terms 4c

    Equating coefficients of x 2 7 1b c b

    Check: coefficient of x 2 1 2 1a b

    3 2 2(2 7 4) (2 1) 4x x x x x x

    (ii)

    3 2f( ) 2 4

    f(1) 1 2 4 1

    x x x

    The remainder is therefore -1.

    3 2 2

    3 2 2

    3 2

    2 4 ( 1)( ) 1

    1

    ( ) ( ) 1

    x x x ax bx c

    ax ax bx bx cx c

    ax b a x c b x c

    Equating coefficients of x 1a Equating constant terms 1 4 3c c

    Equating coefficients of x 2 3b a b

    Check: coefficient of x 3 3 0c b

    3 2 2( 2 4) ( 1) 3 3x x x x x remainder -1.

    2. (i) 3 2f( ) 2 5 6x x x x

    3 2f( 1) 2( 1) 5( 1) ( 1) 6 2 5 1 6 0

    so by the factor theorem, x + 1 is a factor.

    (ii)

    3 2 22 5 6 ( 1)(2 7 6)

    ( 1)(2 3)( 2)

    x x x x x x

    x x x

    (iii) 3 22 5 6 ( 1)(2 3)( 2)y x x x x x x

    When x = 0, y = 6

    When y = 0, x = -1 or 32x or x = 2

  • AQA C1 Polynomials 2 Exercise solutions

    2 of 4 10/01/13 MEI

    3. (i) 3 2f( ) 4 12x x ax x

    3 2f(2) 2 2 4 2 12

    8 4 8 12

    4 12

    a

    a

    a

    x 2 is a factor so by the factor theorem f(2) = 0

    4 12 0

    3

    a

    a

    (ii) 3 2 23 4 12 ( 2)( )x x x x ax bx c

    By inspection

    3 2 23 4 12 ( 2)( 6)

    ( 2)( 2)( 3)

    x x x x x x

    x x x

    4. 3 2f( ) 2 5 2x x x x

    (i) By the remainder theorem, the remainder when f(x) is divided by x + 2 is

    f(-2).

    Remainder 3 2f( 2) 2( 2) 5( 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 3 2f(1) 2 1 5 1 1 2 2 5 1 2 4

    5. (i) 3 2f( ) 2x x x x

    3 2f(2) 2 2 2 2 8 4 2 2 0

    so by the factor theorem, x 2 is a factor.

    (ii) 3 2 22 ( 2)( 1)x x x x x x

    The quadratic expression 2 1x x cannot be factorised, so the expression has been factorised as far as possible.

    (iii) The discriminant of 2 1x x is 21 4 1 1 3 , so the quadratic

    equation 2 1 0x x has no real roots.

    -1

    2

    6

    32

  • AQA C1 Polynomials 2 Exercise solutions

    3 of 4 10/01/13 MEI

    Therefore the graph of 3 2 2y x x x crosses the x-axis once only.

    6. 3 2f( ) 5 4x x kx x

    By the remainder theorem,

    3 2

    f(3) 2

    3 3 5 3 4 2

    27 9 15 4 2

    16 9 2

    9 18

    2

    k

    k

    k

    k

    k

    7. 3 2f( ) 3 1x x ax bx

    By the remainder theorem,

    3 2

    f(2) 5

    3 2 2 2 1 5

    24 4 2 1 5

    4 2 18

    2 9

    a b

    a b

    a b

    a b

    By the remainder theorem,

    3 2

    f( 1) 1

    3( 1) ( 1) ( 1) 1 1

    3 1 1

    3

    a b

    a b

    a b

    Adding:

    3 6

    2, 5

    a

    a b

    8. 3 23 2 11 10 0x x x 3 2f( ) 3 2 11 10x x x x

    f(1) 3 2 11 10 0 so (x 1) is a factor

    2

    53

    ( 1)(3 10) 0

    ( 1)(3 5)( 2) 0

    1 or or 2

    x x x

    x x x

    x x x

    9. 3 22 5 14 8 0x x x 3 2f( ) 2 5 14 8x x x x

    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

  • AQA C1 Polynomials 2 Exercise solutions

    4 of 4 10/01/13 MEI

    2

    12

    ( 2)(2 9 4) 0

    ( 2)(2 1)( 4) 0

    2 or or 4

    x x x

    x x x

    x x x

    10. 3 24 12 7 30 0x x x

    3 2f( ) 4 12 7 30x x x x

    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

    2

    5 32 2

    ( 2)(4 4 15) 0

    ( 2)(2 5)(2 3) 0

    2 or or

    x x x

    x x x

    x x x


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