## anonymous 5 years ago How do you use the given root to assist in finding the remaining roots of the equation? 2x^4 - 15x^3 + 34x^2 - 19x - 20 = 0; 1/2

1. anonymous

divide. if $\frac{1}{2}$ is a root then you know your polynomial is $(x-\frac{1}{2})q(x)$ or more easily written as $p(x)=(2x-1)q(x)$

2. anonymous

you find $q(x)$ by dividing by $(x-\frac{1}{2})$ synthetic division is easiest

3. anonymous

do you know how to use synthetic division?

4. mathteacher1729

These are the synthetic division steps. You know to use synthetic division because you are given a factor of polynomial. It's like saying "458 is divisible by 2... what is the remainder when divisible by 2?) Did you copy your polynomial correctly? You are left with a third degree polynomial which has complex roots... Usually the problems give "Friendlier" answers. :)

5. anonymous

if this is correct there must be a typo somewhere because you (wolfram) got a remainder of -91/4

6. anonymous

Great

7. anonymous

remainder should be 0 of course

8. mathteacher1729

Jennifer -- I think you might have mis-copied the problem. The answer is ... too ugly for something which might be assigned for a practice/hw or even test problem.

9. anonymous

Let me check quick

10. anonymous

in any case 1/2 is not a zero according to wolfram

11. anonymous

Oh wow, after the equation, the 1/2 is a negative..I am so sorry..

12. anonymous

2x^4 - 15x^3 + 34x^2 - 19x - 20 = 0; -1/2

13. anonymous

ohhhhhhhhhhhhhhhhhhhh -1/2

14. anonymous

You sure are one smart cookie to notice that. My goodness; i sure wish I could get math as easy as you .

15. mathteacher1729

That makes more sense, but you're still going to be left with a 3rd degree polynomial... And that 3rd order polynomial will not be very easy to solve.

16. anonymous

then it factors as $(x+\frac{1}{2})q(x)$

17. mathteacher1729

18. anonymous

btw you can find the quotient relatively easily. very easy if you can synthetic division

19. anonymous

do you know how? long division is a pain.

20. anonymous

21. anonymous

list the coefficients they are 2 -15 34 -19 20

22. anonymous

since you root is -1/2 put that on the side. i will write -.5 2 -15 34 -19 20 -.5 ___________________________________

23. anonymous

drop the 2 2 -15 34 -19 20 -.5 ___________________________________ 2

24. anonymous

-.5*-1 2 -15 34 -19 20 -.5 -1 ___________________________________ 2

25. anonymous

-15-1=-16 2 -15 34 -19 20 -.5 -1 ___________________________________ 2 -16

26. anonymous

-16*-.5 = 8 2 -15 34 -19 20 -.5 -1 8 ___________________________________ 2 -16

27. anonymous

34 + 8 = 42 2 -15 34 -19 20 -.5 -1 8 ___________________________________ 2 -16 42

28. anonymous

-.5*42=-21 2 -15 34 -19 20 -.5 -1 8 -21 ___________________________________ 2 -16 42

29. anonymous

-19-21=-40 2 -15 34 -19 20 -.5 -1 8 -21 ___________________________________ 2 -16 42 -40

30. anonymous

-.5*-40= 20 and now i see i have a typo. the last number in the top line should be -20, not 20 easy to fix 2 -15 34 -19 - 20 -.5 -1 8 -21 20 ___________________________________ 2 -16 42 -40

31. anonymous

-20+20=0 so the remainder is 0 \ 2 -15 34 -19 - 20 -.5 -1 8 -21 20 ___________________________________ 2 -16 42 -40 0

32. anonymous

and your answer is $p(x)=(x+\frac{1}{2})(2x^3-16x^2+42x-40)$

33. anonymous

or $p(x)=(2x-1)(x^3-8x^2+21x-10)$

34. anonymous

done!