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If p (x) is a polynomial of degree 3 with p(0) = p(1) = p(-1) = 0 and p(2) = 6, then

a show that p(-x) = -p (x).
b find the interval in which p(x) is less than zero.​

User Laurencevs
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Answer:

p(x) is a function, and a polynomial of degree 3, then

By remainder theorem and factor theorem,

p(0) = p(1) = p (-1) = 0 if and only if

x = 0, 1, -1 are zeroes if and only if

p(x) = k(x - 0)(x - 1)(x + 1) = k(x)(x^2 - 1) = k(x^3 - x) = kx^3 - kx

p(2) = k(2)^3 - k(2) = 6

8k - 2k = 6

6k = 6

k = 1

Thus p(x) = x(x - 1)(x + 1) = x(x^2 - 1) = x^3 - x

p(-x) = (-x)^3 - (-x) = x^3 + x

-p(x) = -(x^3 - x) = -x^3 + x

As you can see, p(x) does not equal to -p(x), unless x = 0, 1, -1,

because p(0) = -p(0) = 0, p(-1) = - p(1) = 0, p(1) = -p(-1) = 0

p(x) = x(x - 1)(x + 1)

If x < -1, then x < 0, x - 1 < 0, x + 1 < 0

That means p(x) < 0

If -1 < x < 0, then x < 0, -2 < x - 1 < -1 < 0, 0 < x + 1 < 1

That means p(x) > 0

If 0 < x < 1, then x > 0, -1 < x - 1 < 0, 0 < 1 < x + 1 < 2

That means p(x) < 0

If x > 1, then x > 0, x - 1 > 0, x + 1 > 2 > 0

That means p(x) > 0

Thus p(x) < 0 if x < -1 or 0 < x < 1

User Christoph Winkler
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