Thursday, May 18, 2017

College Algebra, Chapter 4, Chapter Review, Section Review, Problem 46

If P(x)=9x521x4+10x3+6x23x1, 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)=9x521x4+10x3+6x23x1=(x+13)(9x424x3+18x23)=(x+13)(x1)(9x315x2+3x2+3)=(x+13)(x1)(x1)(9x26x3)=(x+13)(x1)(x1)(x1)(9x+3)=(x+13)(x1)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)(01)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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