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secondary 3 | A Maths
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universe
Universe

secondary 3 chevron_right A Maths chevron_right Singapore

help!thankyou!

Date Posted: 4 years ago
Views: 232
snell
Snell
4 years ago
let f(x) = 2x³ + Ax² + Bx + C

f(1) = 2 + A + B + C = 48
A + B + C = 46 ... ... ... [1]



perform long division:

_____________ x + ½(A+6)
2x² - 6x - 7 √ 2x³ + Ax² + Bx + C
___________- ) 2x³ - 6x² - 7x
____________ (A+6)x²+ (B+7)x + C
________ - ) (A+6)x²- 3(A+6)x - 7(A+6)/2
_________ (B+7+3A+18)x + C+7(A+6)/2

remainder is (B+3A+25)x + (C+7A/2+21)

hence,
B+3A + 25 = -1
B + 3A = -26
B = -26 - 3A ... ... ... [2]


C + 7A/2 + 21 = 5
C = -7A/2 - 16 ... ... ... [3]

subst. [2] & [3] into [1],
A - 26 - 3A - 7A/2 - 16 = 46
A = -16

B = -26 - 3(-16) = 22

C = -7(-8) - 16 = 40

hence,
f(x) = 2x³ - 16x² + 22x + 40 (shown)



(ii)
f(x) = 0
2x³ - 16x² + 22x + 40 = 0
x³ - 8x² + 11x + 20 = 0


let g(x) = x³ - 8x² + 11x + 20
g(-1) = -1 - 8 - 11 + 20 = 0
(x+1) is a factor of g(x).


perform long division:

_______ x² - 9x + 20
x+1 √x³ - 8x² + 11x + 20
___ - ) x³ + x²
_____ -9x² + 11x + 20
____ - ) -9x²- 9x
___________ 20x + 20
________ - ) 20x + 20
__________________ 0


hence,
x³ - 8x² + 11x + 20 = 0
(x+1)(x² - 9x + 20) = 0
(x+1)(x-4)(x-5) = 0
x = -1, 4 or 5


...

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Tan Qiao Yi
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