Chapter 1: Roots of polynomial equations We are working with Cambridge Assessment International Education towards endorsement of this title. So 8 + 12(y − 1) − (y − 1) 3 = 0 ⇒ S2 = ( Σ a ) 2 − 2 Σ ab = 1 + 9 = 289 64 2 64 3 2 This gives 8 + 12y − 12 − y + 3y − 3y + 1 = 0 Similarly, S−3 for [1] is equal to S−1 for [2] which then simplifies to y 3 − 3y 2 − 9y + 3 = 0 So S−3 = − 1 (a + 2)( b + 2)(g + 2) 2 a The expression is the product abg of the roots of our cubic equation 8 Let y = x 2 So
Rearranging the equation gives:
(a + 2)( b + 2)(g + 2) = −3 abg
x 3 − x = −4 Squaring both sides:
b g a + + is the a +2 b +2 g +2 reciprocal of the sum of the roots, so for our cubic Σ ab equation, this is Σ a1 = = −9 = 3 −3 Σ abg
b The expression
Substituting for x 2 in the equation:
y 3 − 2y 2 + y − 16 = 0.......................................................[2]
S6, S8, S10 for [1] are equal to S3, S4, S5 for [2].
Rearranging the equation gives: 2x = x + 6 4
⇒ x 6 − 2x 4 + x 2 = 16......................................................[1]
FT
7 Let y = x 3
(x 3 − x) 2 = (−4) 2
3
Cubing both sides: (2x 4) 3 = (x 3 + 6) 3 Expanding the brackets: 8x 12 = x 9 + 18x 6 + 108x 3 + 216
S1 = Σ a = 2, Σ ab = 1, S2 = ( Σ a ) 2 − 2 Σ ab = (2) 2 − 2 × 1 = 2
For y 3 − 2y 2 + y − 16 = 0, the roots are such that S3 − 2S2 + S1 − 48 = 0 ⇒ S3 = 2S2 − S1 + 48 = 50 Multiplying by y gives
y 4 − 2y 3 + y 2 − 16y = 0, which leads to
Substituting for x 3 in the equation:
S4 − 2S3 + S2 − 16S1 = 0
8y 4 − y 3 − 18y 2 − 108y − 216 = 0.....................................[2]
⇒ S4 = 2S3 − S2 + 16S1
So S6 for [1] is equal to S2 for [2]
So S4 = 100 − 2 + 32 = 130
Dividing [2] by 8:
Multiplying by y again leads to
A
⇒ 8x 12 − x 9 − 18x 6 − 108x 3 − 216 = 0.............................[1]
S5 − 2S4 + S3 − 16S2 = 0
Then Σ a = 1 , Σ ab = − 9 and 8 4 27 Σ abg 2 Σ a1 = = =−1 2 −27 Σ abgd
= 260 − 50 + 32 = 242
R
27 y4 − 1 y3 − 9 y2 − y − 27 = 0 8 2 4
⇒ S5 = 2S4 − S3 + 16S2
D
Hence S6 = 50, S8 = 130, S10 = 242
7
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