Ask your own question, for FREE!
Mathematics 13 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 Show work to receive credit.

ganeshie8 (ganeshie8):

1. find all factors of leading coefficient 2. find all factors of constant term

OpenStudy (anonymous):

how would I find the factors of leading coefficient and constant term?

ganeshie8 (ganeshie8):

wats the leading coefficent here ?

OpenStudy (anonymous):

-2?

OpenStudy (anonymous):

is it?

ganeshie8 (ganeshie8):

yes. so factors are just \(\pm 1, \pm 2\)

ganeshie8 (ganeshie8):

similarly find factors for constant term also

OpenStudy (anonymous):

ok so 1 and 2 are possible rational zeros of the function?

ganeshie8 (ganeshie8):

nope we're not finished yet.

ganeshie8 (ganeshie8):

Find factors of constant term, them we can figure out the possible rational roots

ganeshie8 (ganeshie8):

*then

OpenStudy (anonymous):

are the factors of the constant term 1,2, and 4?

ganeshie8 (ganeshie8):

you missed a few

ganeshie8 (ganeshie8):

and 4 is not a factor of 18. 4 doesnt divide 19

ganeshie8 (ganeshie8):

factors of a number are just all the numbers that divide it.

ganeshie8 (ganeshie8):

for ex, 9 is a factor of 18, cuz 9 divides 18

OpenStudy (anonymous):

so its 1 2 and 3 then in this case

ganeshie8 (ganeshie8):

thats all ?

ganeshie8 (ganeshie8):

factors of 18 : \(\pm 1 , \pm 2, \pm 3, \pm 6, \pm 9, \pm 18\)

OpenStudy (anonymous):

so since we have all the factors for the function, do we start finding the possible rational zeros for the function?

ganeshie8 (ganeshie8):

yes, possible rational zeroes are just the fraction of factors of 'constant term' and 'leading coeffecient'

OpenStudy (anonymous):

oh ok. can you draw it out so I can see the work better? if you can, that'd be nice

ganeshie8 (ganeshie8):

factors of constant term : \(\pm 1 , \pm 2, \pm 3, \pm 6, \pm 9, \pm 18 \) factors of leading coeff : \(\pm 1 , \pm 2\) possible rational zeroes : \(\pm\frac{1}{1},\pm\frac{2}{1},\pm\frac{3}{1},\pm\frac{6}{1},\pm\frac{9}{1},\pm\frac{18}{1} \) and, \(\pm\frac{1}{2},\pm\frac{2}{2},\pm\frac{3}{2},\pm\frac{6}{2},\pm\frac{9}{2},\pm\frac{18}{2} \)

ganeshie8 (ganeshie8):

simplify the fractions

OpenStudy (anonymous):

so the simplified fractions would be: 1, 2, 3, 6, 9, and 18 as well as 1/2, 1, 3/2, 3, 9/2, and 9. is this right?

ganeshie8 (ganeshie8):

thats right, arrange them in order

ganeshie8 (ganeshie8):

possible rational zeroes : \(\pm 1/2, \pm 1, \pm 3/2, \pm 3, \pm 9/2, \pm 6, \pm 9, \pm 18\)

ganeshie8 (ganeshie8):

dont forget the negative ones !

OpenStudy (anonymous):

oh ok and thank you so much! and I won't

ganeshie8 (ganeshie8):

np :)

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!