Question

Find a rational zero of the polynomial function and use it to find all the zeros of the function. f(x) = x^4 + 3x^3 - 5x^2 - 9x - 2

Answer 1

@chris00 can you help me with this last one

Answer 2

gimmi a sec

Answer 3

okay!

Answer 4

sythetic division?

Answer 5

to find a rational zero we usually have to pick any random number

Answer 6

so by inspection x=2 is a rational zero

Answer 7

huh?

Answer 8

it says find a rational zero...

Answer 9

then use it to find all the other zero's

Answer 10

but where'd you get 2 from?

Answer 11

i said by inspection...

Answer 12

they wouldn't give a nasty question...its either in the range of -3 to 3 usualy

Answer 13

@jim_thompson5910 can you help?

Answer 14

yummydum you can use the rational root theorem to find that the possible rational roots are:

-1, 1, -2, 2

Answer 15

I would say to do synthetic division:

Answer 16

ie the factors (both plus and minus) of 2, which is the last term

Answer 17

im confused..

Answer 18

to do so you find the \({p\over q}'s\) which would be \(\pm1,\pm2\)
\[x^3+3x^3-5x^2-9x-2\]
\[\underline{2}|~~~1~~~~~~3~~~-5~~~-9~~~-2\]\[~~~~~~\underline{~~~~~~~~~2~~~~~~10~~~~~10~~~~~~~2~}\]\[~~~~~~~1~~~~~~5~~~~~~~5~~~~~~~~1~~~~~~~0\]
so now you know that one of the roots are \(x=2\) and you have left \(x^3+5x^2+5x+1\)

so if you divide sythetically again:\[\underline{-1}|~~~1~~~~~~~5~~~~~~~~5~~~~~~~~1\]\[~~~~~~\underline{~~~~~~~-1~~~~-4~~~~-1~}\]\[~~~~~~~~~1~~~~~~~4~~~~~~~~1~~~~~~~~0\] and now you know that another root is \(x=-1\) and you have left \(x^2+4x+1\)

so that you can just plug into the quadratic equation:\[x={{b\pm\sqrt{b^2-4ac}}\over2a}\]\[x={{4\pm\sqrt{4^2-4(1)(1)}}\over2(1)}\]\[x={{4\pm\sqrt{16-4}}\over2}\]\[x={4\over2}\pm{\sqrt{12}\over2}\]\[x={2}\pm{2\sqrt{3}\over2}\]\[x=2\pm\sqrt{3}\]

and so the roots of \(f(x)= x^4 + 3x^3 - 5x^2 - 9x - 2\) are:
\[x={2,~~x=-1,2~~x=\sqrt{3},~~x=-2\sqrt{3}}\]

Answer 19

sorry, typo:
its \(x={2,~~x=-1,~~x=2\sqrt{3},~~x=-2\sqrt{3}}\)

Answer 20

@ErinWeeks