If P(x)=9x5−21x4+10x3+6x2−3x−1, then
a.) Find all zeros of P, and state their multiplicities.
b.) Sketch the graph of P.
a.) To find the zeros of P, we apply synthetic division with the possible rational zeros of the factor of 1 divided by the factor of 9 which are ±1,±13,±19. Then, by trial and error
Again, by applying Synthetic Division
Again,
Again,
Thus,
P(x)=9x5−21x4+10x3+6x2−3x−1=(x+13)(9x4−24x3+18x2−3)=(x+13)(x−1)(9x3−15x2+3x2+3)=(x+13)(x−1)(x−1)(9x2−6x−3)=(x+13)(x−1)(x−1)(x−1)(9x+3)=(x+13)(x−1)3(9x+3)
Therefore, rational zeros of P are −13 and 1. Then, the zeros have multiplicity of 1 and 3 respectively.
b.) To sketch the graph of P, we must know first the intercepts of the function. The values of the x intercepts are the zeros of the function, that is −13 and 1. Next, to determine the y intercept, we set x=0 so that
P(0)=(0+13)(0−1)3(9(0)+3)=(13)(−1)3(3)=−1
Since the function has an odd degree and a positive leading coefficient, then its end behavior is y→−∞ as x→−∞ and y→∞ as x→∞. Then, the graph is
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