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# 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.

Extracts from this document...

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

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.

 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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