- Level: AS and A Level
- Subject: Maths
- Word count: 1351
By the software Equation Grapher, we can find the smallest positive root of y=0 and then factorize the equation to find the other roots.
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Introduction
1.(a) By the software Equation Grapher, we can find the smallest positive root of y=0 and then factorize the equation to find the other roots.
(i) From the graph, we can see that the smallest positive root is 1
f (1) = 13 – 12 – 4(1) + 4
= 0
By the Remainder Theorem, 1 is a root and (x–1) is a factor of y = x3– x 2– 4x + 4
f (x) = x3– x 2– 4x + 4
= (x–1) (x2–4)
= (x–1) (x–2) (x+2)
Therefore, by the Factor Theorem, the equation y = x3– x 2– 4x + 4 has roots at
x = -2, 1 and 2.
Hence, the smallest positive root is 1 or
(ii) From the graph, it seems that the smallest positive root is
f () = 2() 3 – () 2 – 8() + 4
= 0
By the Remainder Theorem, is root and (2x-1) is a factor of y = 2x3-x 2-8x+4
f(x) = 2x3-x 2-8x+4
= (2x-1) (x2-4)
= (2x-1) (x-2) (x+2)
Therefore, by the Factor Theorem, the equation y = 2x3-x 2-8x+4 has roots at
x = -2, and 2.
Hence, the smallest positive root is .
(iii) From the graph, it seems that the smallest positive root is
f () = 3() 3 – () 2 – 12() + 4
= 0
By the Remainder Theorem, is a root and (3x-1) is a factor of
y = 3x3-x 2-12x+4
f (x) = 3x3-x 2-12x+4
= (3x-1) (x2-4)
= (3x-1) (x-2) (x+2)
Middle
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.
Question Number | Equation | Smallest positive root |
1 (a) (i) | y = x3 – x2 –4x + 4 | |
1 (a) (ii) | y = 2x3 –x2 – 8x + 4 | |
1 (a) (iii) | y = 3x3 –x2 – 12x + 4 | |
1 (a) (iv) | y = 4x3 – x2 –16x + 4 | |
1 (b) | ||
1 (c) | ||
2 (a) (i) | y = x3 –x2 –9x +9 | |
2 (a) (ii) | y = 3x3 –x2 –27x +9 | |
2 (a) (iii) | y = x3 – x2 – x + 9 | |
2 (b) (i) | y = 30x3 –20x2 –270x +180 | |
2 (b) (ii) | y = 2x3 – 3x2 – 18x + 27 | |
2 (b) (iii) | y = 2x3 – 5x2 – 18x + 45 |
Conclusion
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.
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