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

- anonymous

- jamiebookeater

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- 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
\[p(x)=(2x-1)q(x)\]

- anonymous

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

- anonymous

do you know how to use synthetic division?

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## More answers

- 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. :)

##### 1 Attachment

- anonymous

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

- anonymous

Great

- anonymous

remainder should be 0 of course

- 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.

- anonymous

Let me check quick

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

ohhhhhhhhhhhhhhhhhhhh -1/2

- anonymous

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

- 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.

- anonymous

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

- mathteacher1729

##### 1 Attachment

- anonymous

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

- anonymous

do you know how? long division is a pain.

- anonymous

a little bit..i'm bad..really

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

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

- 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

- anonymous

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

- anonymous

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

- anonymous

or \[p(x)=(2x-1)(x^3-8x^2+21x-10)\]

- anonymous

done!

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