Question 1061853: Find an equation for f(x), the polynomial of smallest degree with real coefficients such that f(x) bounces off of the x-axis at −1, bounces off of the x-axis at 2, has complex roots of −1+5i and 2+i and passes through the point (0,−14).
Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website! Find an equation for f(x), the polynomial of smallest degree with real coefficients such that f(x) bounces off of the x-axis at −1, bounces off of the x-axis at 2, has complex roots of −1+5i and 2+i and passes through the point (0,−14).
Sketch those points and sketch a curve that follows the
given instructions:
bounces at x = -1 implies factor of (x+1)^2
bounces at x = 2 implies factor of (x-2)^2
real coefficients implies roots at -1+5i and -1-5i,
implies factors ((x-1)+5i)((x-1)-5i) = (x-1)^2+25 = x^2-2x+26
real coefficients implies roots at 2+i and 2-i,
implies factors ((x-2)-i)((x-2)+i) = (x-2)^2+1 = x^2-4x+5
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Then:: f(x) = a(x+1)^2*(x-2)^2*(x^2-2x+26)*(x^2-4x+5)
Solve for "a" using (0,-14)
f(0) = a(1^2)(-2)^2*(26)*(+5) = -14
a[1*4*130] = -14
a = -14/520
a = -7/260
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Ans: f(x) = (-7/260)(x+1)^2*(x-2)^2*(x^2-2x+26)(x^2-4x+5)
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Cheers,
Stan H.
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