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 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
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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
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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
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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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