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Fynjy0 [20]
3 years ago
5

A fourth degree polynomial has real roots at 0, 3, a double real root at -1, and passes through the point

Mathematics
1 answer:
REY [17]3 years ago
6 0

Answer:

Step-by-step explanation:

y = a * x(x - 3)(x + 1)^2

2,12 is used to find the value of a

y = a (2)(2 - 3)(2 + 1)^2

12 = a (2)(-1)(3)^2

12 = a (2) (-1) (9)

12 = - 18*a

12/-18 = a

a = - 2/3

y = (-2/3)*x*(x - 3)(x + 1)^2

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Since \alpha and \beta are roots of ax^2+bx+c, we can factorize the quadratic in terms of the roots as

ax^2+bx+c = a(x-\alpha)(x-\beta)

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(x-\alpha^2)(x-\beta^2)

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and to make things look cleaner, scale the whole expression by a^2 to get

\boxed{a^2x^2 + (2ac-b^2)x + c^2}

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