f () = 4() 3 – () 2 – 16() + 4
= 0
By the Remainder Theorem, is a root and (4x-1) is a factor of
y = 4x3-x 2-16x+4
f (x) = 4x3-x 2-16x+4
= (4x-1) (x2-4)
= (4x-1) (x-2) (x+2)
Therefore, by the Factor Theorem, the equation in (ii) has roots at
x = -2, and 2.
Hence, the smallest positive root is .
1.(b)
From the graph, it seems that the smallest positive root is
f () =() 3 – () 2 – () + 4
= 0
Hence, the smallest positive root is.
1.(c)
From the graph, it seems that the smallest positive root is
f () =2() 3 – 3() 2 – 8() + 12
= 0
By the Remainder Theorem, is a root and (2x-3) is a factor of
y = 4x3-x 2-16x+4
f (x) = 2x3-3x 2-8x + 12
= (2x-3) (x2-4)
= (2x-3) (x-2) (x+2)
Therefore, by the Factor Theorem, the equation y = 2x3-3x 2-8x + 12 has roots at x = -2, and 2.
-
(a) (i) y = x3 –x2 –9x +9
When y=0, x3 – x2 – 9x + 9 = 0
Or, x2(x - 1) –9 (x - 1) = 0
Or, (x - 3) (x - 1) (x + 3) = 0
Therefore, x = 3 or x =1 or x = -3
Smallest positive root = 1 =
(ii) y = 3x3 –x2 –27x +9
When y = 0, 3x3 – x2 – 27x + 9 = 0
Or, x2 (3x - 1) –9 (3x - 1) = 0
Or, (x - 3) (3x - 1) (x + 3) = 0
Therefore, x = 3 or x=or x = -3
Smallest positive root =
(iii) y = x3 – x2 – x + 9
When y = 0, x3 – x2 – x + 9= 0
Or, x2(x - 1) –9(x – 1 ) = 0
Or, (x - 3)( x - 1) (x + 3) = 0
Hence, x = 3 or x = or x = -3
Smallest positive root =
(b) (i) y = 30x3 –20x2 –270x +180
When y = 0, 30x3 –20x2 –270x + 180 = 0
Or, 10 x2(3x - 2) –90(3x - 2) = 0
Or, (x - 3)(3x - 2)(x + 3) = 0
Hence, x = 3 or x= or x = -3
Smallest positive root =
(ii) y = 2x3 – 3x2 – 18x + 27
When y = 0, 2x3 –3x2 –18x + 27 = 0
Or, x2(2x - 3) –9(2x - 3) = 0
Or, (x - 3)(2x - 3)(x + 3) = 0
Hence, x = 3 or x = or x = -3
Smallest positive root =
(iii) y = 2x3 – 5x2 – 18x + 45
When y = 0, 2x3 – 5x2 – 18x + 45 = 0
Or, x2 (2x - 5) –9(2x - 5) = 0
Or, (x - 3)(2x - 5)(x + 3) = 0
Hence, x = 3 or x =or x = -3
Smallest positive root =
All the equations handled so far in this paper are put into the table below along with the smallest positive root of each equation.
3. (a)
(Shown)
(b)
It is given that q and p have no common factors. Therefore, q cannot be a factor of p. Thus, q3 cannot be a factor of p. However p multiplied by a constant, gives us . Therefore, p must be a factor of d.
(c)
(shown)
(d)
It is given that q and p have no common factors. Thus, q cannot be a factor of p. So, q3 cannot be a factor of p. However q multiplied by a constant,
-) gives us . Therefore, q must be a factor of a.
(e) f(x) = anxn + an-1xn-1 +……………..a1x + a0
f() = an + an-1 + … … … … a1 + a0 = 0
an pn + an-1 pn-1q + … … … … …a1 p qn-1 + a0 qn = 0
p (an pn-1 + an-1 pn-2q +… … … … a1 qn-1) = - a0 qn
It is given that a, p and q are all integers with an ≠ 0.
p and q have no common factors and so p is not a factor of qn
Therefore, p must be a factor of a0.
F() = an + an-1+… … … … a1 +a0 = 0
an pn + an-1pn-1q +… … … … a1pqn-1 +a0qn = 0
q (an-1pn-1 +… … … … a1pqn-2 +a0qn-1) = - an pn
It is given that a, p and q are all integers with an ≠ 0.
p and q have no common factors and so q is not a factor of pn
Therefore, q must be a factor of an.
4.
P (x) = 6x4 -7x3 + 8x2 -7x + 2
If p is a factor of 2, the constant term and q is a factor of 6, the leading co-efficient, the rational zero of P(x) = p/q
The factors of 2 are ± 1 and ± 2
The factors of 6 are ± 1, ± 2, ± 3 and ± 6.
The candidates for rational zeros are: ± 2, ± 1, ±, ± , ± and ± .
When P(x) has a rational zeros, P(x) = 0
Finding Rational Zeros by Trial and Error Method:
P(-2) = 6(-2)4 –7(-2)3 + 8(-2)2 –7(-2) + 2
= 96 + 56 + 32 + 14 + 2 = 200
P(2) = 6(2)4 –7(2)3 + 8(2)2 –7(2) + 2
= 96 – 56 + 32 - 14 + 2 = 60
P(-1) = 6(-1)4 –7(-1)3 + 8(-1)2 –7(-1)+2
= 6 + 7 + 8 + 7 + 2 = 30
P(1) = 6(1)4 –7(1)3 + 8(1)2 –7(1)+2
= 6 – 7 + 8 – 7 + 2 = 2
P(-) = 6(-)4 –7(-)3 + 8(-)2 –7(-)+2
=+ + + +2 =
P() = 6()4 –7()3 + 8()2 –7()+2
=- + - +2 = 0
P(-)= 6(-)4 –7(-)3 + 8(-)2 –7(-)+2
= =
P()= 6()4 –7()3 + 8()2 –7()+2
= = 0
P(-)= 6(-)4 –7(-)3 + 8(-)2 –7(-)+2
= =
P()= 6()4 –7()3 + 8()2 –7()+2
= =
P(-)= 6(-)4 –7(-)3 + 8(-)2 –7(-)+2
= =
P()= 6()4 –7()3 + 8()2 –7()+2
= =
Hence, P(x) = 0 when x = or x =
The rational zeros of P are and.
