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Alex_Xolod [135]
3 years ago
14

Factorise and solve : x^3-x^2-4x+4

Mathematics
1 answer:
Alika [10]3 years ago
6 0

(x - 1)(x - 2)(x + 2)

note that the sum of the coefficients 1 - 1 - 4 + 4 = 0

thus x = 1 is a root and (x - 1 ) is a factor

dividing x³ - x² - 4x + 4 by (x - 1)

x³ - x² - 4x + 4 = (x - 1)(x² - 4 ) (note (x² - 4 ) is a difference of squares )

x³ - x² - 4x + 4 = (x - 1)(x - 2)(x + 2)

(x - 1)(x - 2)(x + 2 ) =0

x - 1 = 0 ⇒ x = 1

x - 2 = 0 ⇒ x = 2

x + 2 = 0 ⇒ x = - 2

solutions are x = 1 or x = ± 2



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Answer:

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Step-by-step explanation:

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We know that s^2+3s+2=(s+1)(s+2), so we have

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By using the method of partial fraction we have:

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Using linearity of inverse transform we get:

y(t)=L^{-1}[\frac{11}{3(s+1)}](t) -L^{-1}[\frac{5}{2(s+2)}](t) -L^{-1}[\frac{1}{6(s+4)}](t)

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