Ask your own question, for FREE!
Mathematics 15 Online
OpenStudy (anonymous):

Use the Rational Zeros Theorem to write a list of all possible rational zeros of the function. f(x) = -2x4 + 4x3 + 3x2 + 18

OpenStudy (welshfella):

+18 * -2 = -36 so you need to list the factors of -36

OpenStudy (welshfella):

eg +/- 1 , +/- 2 , etc

OpenStudy (anonymous):

±6,±6 ±2,±18 ±3,±12 ±4,±9

OpenStudy (anonymous):

Right?

OpenStudy (anonymous):

@welshfella

Nnesha (nnesha):

hmm

OpenStudy (anonymous):

?

Nnesha (nnesha):

hmm ho would that work ? are we going to divide all factors by 2 maybe

Nnesha (nnesha):

how**

Nnesha (nnesha):

p/q method \[\rm \frac{ p }{ q }=\frac{ \pm~ list~factors~of~ constant~term }{\pm~ list ~factors~of~leading ~coefficient }\]

OpenStudy (welshfella):

ah yes - i'd forgotten that formula. So +/- 18 would not be included

OpenStudy (anonymous):

So I divide all those numbers by 2?

OpenStudy (welshfella):

- to be honest I'm not sure ....

Nnesha (nnesha):

no i was asking should we divide by 2..

OpenStudy (welshfella):

It's been a long time since i did these

Nnesha (nnesha):

p/q method \[\rm \frac{ p }{ q }=\frac{ \pm~ list~factors~of~ constant~term }{\pm~ list ~factors~of~leading ~coefficient }\] do this divide factors of constant term by factors leading coefficient

OpenStudy (welshfella):

i suppose 18 would be included as its 18 / 1

OpenStudy (welshfella):

???

Nnesha (nnesha):

what do you mean welshfell ? sorry i don't get it .. :o

OpenStudy (welshfella):

well 18 is a factor of the constant term and 1 is a factor of -2

OpenStudy (anonymous):

Yes, you would include 18 for precisely the reason you just described.

OpenStudy (welshfella):

I guess I should revise this stuff!!!

Nnesha (nnesha):

f(x) = -2x4 + 4x3 + 3x2 + 18 leading coefficient = -2 so factors are \[\pm 1 , \pm2 \] constant term is 18= so factors 1 ,2 ,3 , 6,9,18 so we should include all of them

OpenStudy (welshfella):

yea

Nnesha (nnesha):

\[\rm \frac{p}{q}= \frac{ \pm 1 , \pm 2 , \pm 3,\pm 6, \pm 9,\pm 18}{\pm 1 , \pm2}\] now just divide and remember each number at the numerator are dividing by1 and 2

Nnesha (nnesha):

ofc +/- 1 , +/- 2

OpenStudy (welshfella):

yes thats it

OpenStudy (anonymous):

±1, 2, 3, 6, 9, 18, 1/2, 3/2, 9/2

OpenStudy (welshfella):

Of course its possible that none of those are roots but that'f not relevant to the question.

Nnesha (nnesha):

^^ye that to find real you can use synthetic division or just replace all x's with each possible solution if you get zero as final answer then that would be the real zero

OpenStudy (welshfella):

Just out of interest I'll do this one on my calc...

OpenStudy (anonymous):

So what I posted should be the answer that's needed for this question?

Nnesha (nnesha):

yes

Nnesha (nnesha):

ye use calculator for real zeros ;P

OpenStudy (anonymous):

Thank you two

OpenStudy (welshfella):

real zeros are 2.89 and -1.524 the other 2 roots are complex

OpenStudy (welshfella):

yw

Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!
Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!