Question 920515: Please help! Ive tried several times and can not get the correct answer.
Construct a polynomial with the stated properties. Reduce all fractions to lowest terms.
Third-degree, with zeros of -2,-1, and 3,and passes through the point (2,10)
Answer p(x)=??
Answer by Theo(13342)   (Show Source):
You can put this solution on YOUR website! the zeroes are -2, -1, and 3
this means the factors are (x+2), (x+1), and (x-3)
(x+2) * (x+1) = x^2 + x + 2x + 2
combine like terms to get x^2 + 3x + 2
x^2 + 3x + 2 * (x-3) = x^3 + 3x^2 + 2x - 3x^2 - 9x - 6
combine like terms to get x^3 - 7x - 6
if the equation goes through the point (2,10), then f(2) must be equal to 10.
f(2), however, is equal to 2^3 - 7*2 - 6 which is equal to 8 - 14 - 6 which is equal to -12.
you can't just add 22 to it because that will change the roots.
you can, however, multiply it by a factor that will allow f(2) to be equal to 10 and not change the roots.
that factor is -5/6.
how did we find it?
we set a * (x^3 - 7x - 6) = y
replace y with 10 and x with 2 and you get:
a * (2^3 - 7*2 - 6) = 10
simplify to get:
a * -12 = 10
divide both sides of the equation by -12 to get:
a = 10/-12.
simplify to get a = -5/6
your equation becomes:
-5/6 * (x^3 - 7x - 6) = y
simplify to get:
-5/6 * x^3 + 35/6 * x + 30/6 = y
this simplifies to:
-5/6 * x^3 + 35/6 * x + 5 = y
f(2) now becomes:
f(2) = -5/6 * 2^3 + 35/6 * 2 + 5 which becomes:
f(2) = -5/6 * 8 + 70/6 + 5 which becomes:
f(2) = -40/6 + 70/6 + 5 which becomes:
f(2) = 30/6 + 5 which becomes:
f(2) = 5 + 5 which is equal to 10.
your equation now satisfies the requirements and has the same roots as it had before.
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