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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.

image00.png

(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 image01.png

image22.png

(ii)        From the graph, it seems that the smallest positive root is image02.png

f (image02.png) = 2(image02.png) 3 – (image02.png) 2 – 8(image02.png) + 4

= 0

By the Remainder Theorem, image02.png 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, image02.png and 2.

Hence, the smallest positive root is image02.png.

image09.png

(iii)        From the graph, it seems that the smallest positive root is image11.png

f (image11.png) = 3(image11.png) 3 – (image11.png) 2 – 12(image11.png) + 4

= 0

By the Remainder Theorem, image11.png 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)

...read more.

Middle

image26.pngx – 1 ) = 0

                           Or, (x - 3)( image26.pngx - 1) (x + 3) = 0

                 Hence, x = 3 or x = image04.png or x = -3

  Smallest positive root = image04.png

(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=image04.png or x = -3

Smallest positive root = image04.png

                (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 = image26.png or x = -3

Smallest positive root = image26.png

(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 =image31.pngor x = -3

Smallest positive root = image31.png

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

image01.png

1 (a) (ii)

y = 2x3 –x2 – 8x + 4

image32.png

1 (a) (iii)

y = 3x3 –x2 – 12x + 4

image33.png

1 (a) (iv)

y = 4x3 – x2 –16x + 4

image34.png

1 (b)

image35.png

image36.png

1 (c)

image37.png

image38.png

2 (a) (i)

y = x3 –x2 –9x +9

image01.png

2 (a) (ii)

y = 3x3 –x2 –27x +9

image11.png

2 (a) (iii)

y = image26.pngx3 – x2 – image30.pngx + 9

image38.png

2 (b) (i)

y = 30x3 –20x2 –270x +180

image38.png

2 (b) (ii)

y = 2x3 – 3x2 – 18x + 27

image36.png

2 (b) (iii)

y = 2x3 – 5x2 – 18x + 45

image39.png

...read more.

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, ±image04.png, ± image02.png, ± image11.pngand ± image03.png.

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(-image04.png) = 6(-image04.png)4 –7(-image04.png)3 + 8(-image04.png)2 –7(-image04.png)+2

            =image05.png+ image06.png+ image07.png + image08.png +2 = image10.png

P(image04.png) = 6(image04.png)4 –7(image04.png)3 + 8(image04.png)2 –7(image04.png)+2

            =image05.png- image06.png+ image07.png - image08.png +2 = 0

P(-image02.png)= 6(-image02.png)4 –7(-image02.png)3 + 8(-image02.png)2 –7(-image02.png)+2

           = image12.png = image13.png

P(image02.png)= 6(image02.png)4 –7(image02.png)3 + 8(image02.png)2 –7(image02.png)+2

          = image12.png  = 0

P(-image11.png)= 6(-image11.png)4 –7(-image11.png)3 + 8(-image11.png)2 –7(-image11.png)+2

           = image14.png = image15.png

P(image11.png)= 6(image11.png)4 –7(image11.png)3 + 8(image11.png)2 –7(image11.png)+2

           = image16.png = image17.png

P(-image03.png)= 6(-image03.png)4 –7(-image03.png)3 + 8(-image03.png)2 –7(-image03.png)+2

           = image18.png = image19.png

P(image03.png)= 6(image03.png)4 –7(image03.png)3 + 8(image03.png)2 –7(image03.png)+2

           = image20.png = image21.png

Hence, P(x) = 0 when x = image02.pngor x = image04.png

The rational zeros of P are image02.png andimage04.png.

...read more.

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