Polynomial
Past Paper Questions
Polynomial Past Paper Questions
2001 P2 Q1.
Given (x + 2) is a factor of 2x3 + x2 + kx + 2, find the value of k.
Hence solve the equation 2x3 + x2 + kx + 2= 0, when k takes this value.
3
2
2x3 + x2 + kx + 2
2
1
k
2
-2
2
-4
6
-2k – 12
–3
K+6
-2k – 10
If (x + 2) is a factor the remainder, R = 0
 - 2k – 10 = 0
- 2k = 10
k=-5
3
2001 P2 Q1.
Given (x + 2) is a factor of 2x3 + x2 + kx + 2, find the value of k.
Hence solve the equation 2x3 + x2 + kx + 2= 0, when k takes this value.
3
2
2x3 + x2 – 5x + 2
If k = -5
2
-2
2
1
-5
2
-4
6
-2
–3
1
0
Solving
(x + 2)(2x2 – 3x + 1) = 0
(x + 2)(2x – 1)(x – 1) = 0
x = - 2; x = ½ ; x = 1
2
Polynomial Past Paper Questions
2002 WD P1 Q5
Given (x – 2) and (x + 3) are factors of f(x) = 3x3 + 2x2 + cx + d
Find the values of c and d.
5
f(x) = 3x3 + 2x2 + cx + d
3
2
2
6
3
8
If this is a factor then:
c
16
d
32+2c
16+c 32+2c+d
32 + 2c + d = 0
1
Polynomial Past Paper Questions
2002 WD P1 Q5
Given (x – 2) and (x + 3) are factors of f(x) = 3x3 + 2x2 + cx + d
Find the values of c and d.
5
f(x) = 3x3 + 2x2 + cx + d
3
-3
3
2
c
-9
21
-7
If this is a factor then:
d
-63-3c
21+c -63-3c+d
-63 – 3c + d = 0
1
Polynomial Past Paper Questions
2002 WD P1 Q5
Given (x – 2) and (x + 3) are factors of f(x) = 3x3 + 2x2 + cx + d
Find the values of c and d.
5
Using both equations we can solve for c and d simultaneously
2c + d = -32 ----(1)
– 3c + d = 63 ----(2)
(1) – (2) 
5c = - 95
c = -19
If c = -19 subst into (1) to find d:
2c + d = -32
2(-19) + d = -32
– 38 + d = -32
d=6
3
Polynomial Past Paper Questions
2002 WD P2 Q6: The graph of f(x) = 2x3 – 5x2 – 3x + 1 has a root between 0 and 1.
Find the value of this root to one decimal place.
3
f(x) = 2x3 – 5x2 – 3x + 1
x = 0:
x = 1:
f(0) = 2(0)3 – 5(0)2 – 3(0) + 1 = 1
f(1) = 2(1)3 – 5(1)2 – 3(1) + 1 = – 5
As a change of sign occurs a root exists between x = 0 & x = 1
(with the root being much closer to x = 0 as f(x) is closer to zero)
So try x = 0.1:
x = 0.2:
x = 0.3:
f(0.1) = 2(0.1)3 – 5(0.1)2 – 3(0.1) + 1 = 0.652
f(0.2) = 2(0.2)3 – 5(0.2)2 – 3(0.2) + 1 = 0.216
f(0.3) = 2(0.3)3 – 5(0.3)2 – 3(0.3) + 1 = -0.296
As f(0.2) = 0.216 is closest to zero solution is x = 0.2 to 1 d.p.
3
2003 P2 Q1 Show (x – 2) is a factor of f(x) = 6x3 – 5x2 – 17x + 6
Express f(x) in its fully factorised form.
4
f(x) = 6x3 – 5x2 – 17x + 6
6
2
6
-5
-17 6
12
14
7
-3
-6
0
As remainder, R = 0
 (x – 2) is a factor
f(x) = 6x3 – 5x2 – 17x + 6
= (x – 2)(6x2 + 7x – 3)
= (x – 2)(3x – 1)(2x + 3)
4
2004 P1 (i) Show (x + 1) is a factor of f(x) = x3 – x2 – 5x – 3
(ii) Hence or otherwise factorise f(x) fully.
One of the turning points of the graph lies on the x-axis.
Write down the coordinates of this turning point.
5
1
f(x) = x3 – x2 –5x – 3
1
-1
1
-1
-5
-3
-1
2
3
-2
-3
0
As remainder, R = 0
 (x – 2) is a factor
f(x) = x3 – x2 – 5x – 3
= (x + 1)(x2 – 2x – 3)
= (x + 1)(x + 1)(x – 3)
= (x + 1) 2 (x – 3)
Turning Point rests on axis when (x + a) 2  (-1, 0)
6
Polynomial Past Paper Questions
2005 P1 Show (x – 3) is a factor of f(x) = 2x3 – 7x2 + 9 and factorise f(x) fully. 5
f(x) = 2x3 – 7x2 + 9
2
3
2
-7
0
9
6
-3
-9
-1
-3
0
As remainder, R = 0
 (x – 2) is a factor
f(x) = 2x3 – 7x2 + 9
= (x – 3)(2x2 – x – 3)
= (x + 1)(2x – 3)(x + 1)
5
2005 P2
(a) Show that x = -1 is a solution of the cubic x3 + px2 + px + 1 = 0 1
(b) Hence find the range of values of p for which all the roots are real 7
x3 + px2 + px + 1 = 0
1
-1
p
-1
1 P–1
p
1
1–p -1
1
0
As Remainder, R = 0  x = -1 is a solution
1
2005 Alternatively use substitution
x3 + px2 + px + 1 = 0
LHS = (-1)3 + p(-1)2 + p(-1) + 1
= -1 + p – p + 1
=0
As LHS = RHS
RHS = 0
 x = -1 is a solution as satisfies the equation
1
2005 P2(b) Hence find the range of values of p for which all the roots are real 7
1
From (a)
-1
p
-1
1 P–1
a=1
b = (p – 1)
c=1
p
1
1–p -1
1
0
1x2 + (p – 1)x + 1 = 0
b2 – 4ac
(p – 1)2 – 4(1)(1) = 0
p2 – 2p + 1 – 4 = 0
p2 – 2p – 3 = 0
(p – 3)(p + 1) = 0
p=3 & p=–1
If REAL  b2 – 4ac ≥ 0
• = 0 & Solve roots
• Sketch Quadratic
• Determine range using sketch
for b2 – 4ac ≥ 0
y
-1
If real roots can be EQUAL or DISTINCT
i.e. b2 – 4ac ≥ 0  p ≤ -1 & p ≥ 3
3
x
7
Higher Polynomial
Past Paper Questions
Total = 36 Marks