4.2E: More on Finding The Zeros of Polynomial Functions (Exercises)

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Section 4.2 Exercises

In exercises #1-7, use polynomial long division to perform the indicated division. If a non-zero remainder exists, write the result of your division as a mixed fraction.

1. $\left(4x^{2} +3x-1\right)\div (x-3)$

2. $\left(3x^{2} -13x-10\right)\div \left(x-5\right)$

3. $\left(8x^{3} + 27\right)\div \left(2x+3\right)$

4. $\left(5x^{4} -3x^{3} +2x^{2} -1\right)\div \left(x^{2} +4\right)$

5. $\left(9x^{3} +5\right)\div \left(2x-3\right)$

6. $\left(x^{4} - 9\right)\div \left(x^{2} +3\right)$

7. $\left(-2x^{5} +3x^{4}-2x^{2}\right)\div \left(x^{2} +3\right)$

In exercises #8-14, given a zero(s) of the polynomial, find the other zeros.

8. $x^{3} +6x^{2} +11x+6,\; \; x=-1$

9. $x^{3} -24x^{2} +192x-512,\; x=8$

10. $3x^{3} +4x^{2} -x-2,\; \; x=-1$

11. $x^{3} +2x^{2} -3x-6,\; \; x=-2$

12. $2x^{3} -3x^{2} -11x+6,\; \; x=\dfrac{1}{2}$

13. $3x^{3} +7x^{2} -4,\; \; x=\dfrac{2}{3}$

14. $x^{4} -2x^{3}-11x^{2} -8x+4,\; \; x=5, x = -3$

In exercises 15-16, list all of the possible zeros using the rational zero theorem. Then find one zero.

15. $f(x)=x^{4} +3x^{2}-8x+4$

16. $f(x)=3x^{5} +2x^{4}-5x^{3}+x$

In exercises #17-22, use the rational zero theorem and long division of polynomials to find all the zeros.

17. $f(x)=x^{3} -6x^{2}+11x-6$

18. $f(x)=x^{3} -3x-2$

19. $f(x)=x^{3} -5x^{2}-4x+20$

20. $f(x)=6x^{3} +17x^{2}+x-10$

21. $f(x)=x^{4} +2x^{3}-9x^{2}-2x+8$

22. $f(x)=x^{4} +2x^{3}-4x^{2}-2x+3$

In exercises #23-24, use the given complex zero of the polynomial to find the other zeros.

23. $f(x)=x^{4} -2x^{3}-11x^{2}-8x-60,\; \; x=-2i$

24. $f(x)=x^{4} -x^{3}+7x^{2}-9x-18,\; \; x=3i$

In exercises #25-28, use the rational zero theorem and long division of polynomials to find all the zeros.

25. $f(x)=x^{3} -x^{2}+4x-4$

26. $f(x)=x^{3} -3x^{2}+5x-15$

27. $f(x)=x^{3} +x^{2}-2$

28. $f(x)=x^{4} -2x^{3}-2x^{2}-2x-3$

29. Explain the mistake(s) that was(were) made in the following division

$A long division problem with two mistakes.$

30. Explain the mistake(s) that was(were) made.

Given that $x = 2$ is a zero of $f(x)=x^{3} +2x^{2}-7x-2$, find the other zeros.

Solution:

Step 1: Since $x=2$ is a zero, then $x=-2$ is also a zero

Step 2: Since $x=2$ and $x=-2$ are zeros, $f(x)$ has factors $ (x-2) $ and $ (x+2) $

Step 3: Therefore $ (x-2)(x+2)=x^2-4 $ is also a factor.

Step 4: To find the other factor, we divide $f(x)$ by $x^2-4 $

\[\dfrac{x^{3} +2x^{2}-7x-2}{x^2-4}=x+2 + \dfrac{-3x+6}{x^2-4} \nonumber \]

This has a non-zero remainder. The division should have a remainder of zero if $x = 2$ is a zero. What went wrong?

31. Approximate a real zero of the polynomial of $f(x)=x^{3} +3x^{2} -5$ on the interval [0,2] using seven iterations of The Midpoint Algorithm.

32. Approximate a real zero of the polynomial of $f(x)=x^{4} -2x^{2} +x -5$ on the interval [1,3] using seven iterations of The Midpoint Algorithm.

33. Approximate a real zero of the polynomial of $f(x)=7x^{3} -x^{2} +3$ on the interval [-1,0] using seven iterations of The Midpoint Algorithm.

Answer

1. $\dfrac{4x^{2} +3x-1}{x-3} = 4x+15 +\dfrac{44}{x-3} $

5. $\dfrac{x^{4} - 9 }{x^2+3} = x^2-3 $

8. The zeros are $x=-1$, $x=-2$, and $x=-3$ all with multiplicity one.

13. The zeros are $x=\dfrac{2}{3}$, $x=-1$, and $x=-2$ all with multiplicity one.

17. The zeros are $x=1$, $x=2$, and $x=3$ all with multiplicity one.

21. The zeros are $x=-4$, $x=-1$, $x=1$,and $x=2$ all with multiplicity one.

23. The zeros are $x=2i$, $x=-2i$, $x=5$,and $x=-3$ all with multiplicity one.

25. The zeros are $x=2i$, $x=-2i$, and $x=1$ all with multiplicity one.

31.

The interval [a,b]	[0,2]	[1, 2]	[1, 1.5]	[1,1.25]	[1,1.125]	[1.0625,1.125]	[1.09375,1.125]
The midpoint $x_{mid}$	1	1.5	1.25	1.125	1.0625	1.09375	1.109375
The value at the midpoint $y=f(x_{mid})$	$-1 $	$ 5.125 $	$\approx 1.641 $	$ \approx 0.221 $	$ \approx -0.414 $	$ \approx -0.10269 $	$ \approx 0.057461 $
Compare the values at each endpoint	$f(0)= -5$, $f(1) =-1$, and $ f(2) =15 $	$ f(1)=-1 $, $ f(1.5)=5.125 $, and $ f(2) = 15 $	$ f(1)=-1 $,$f(1.25) \approx 1.641 $, and $ f(1.5)=5.125 $	$f(1)=-1 $, $ f(1.125) \approx $ $ 0.221 $, and $ f(1.25) \approx 1.641 $	$f(1)=-1 $, $ f(1.0625) \approx -0.414 $, and $ f(1.125) \approx $ $ 0.221 $	$ f(1.0625) \approx -0.414 $, $ f(1.09375) \approx -0.10269 $, and $ f(1.125) \approx $ $ 0.221 $

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