Question

PLEASE HELP
Use the rational zero theorem to find all the real zero of the polynomial function. Use the zeros to factor f over the real numbers.

f(x)=2x^4+x^3-15x^2-7x+7

x=_____
OR
There is no real zero

Use the real zeros to factor f
f(x)=________
(type answer in factored form)

Answer 1

Do you know how to use the rational root test?

Answer 2

no. not that I remember

Answer 3

Basically what it is is that if a polynomial has rational roots, they can be written in the form p/q where p is an integer factor of the constant term, and q is an integer factor of the leading coefficient

So in your case, if it has any rational roots, they will be one of
\[\pm \frac{1,7}{1,2}\]

which is basically shorthand for
\[-1, 1, -7, 7, -\frac{1}{2}, \frac{1}{2}, -\frac{7}{2}, \frac{7}{2}\]

So these are your possible roots, all you have to do is plug them in, and if you get zero it is a root :)

Answer 4

Then once you have found a root \(r\), you can use synthetic division to factor it out by dividing the polynomial by \(x - r\)

Answer 5

okay?

Answer 6

How you found any zero's in the above list?

Answer 7

still confused

Answer 8

Basically the rational root theorem tells us that if there are any rational roots, they have to be one of these:
\[-1, 1, -7, 7, -\frac{1}{2}, \frac{1}{2}, -\frac{7}{2}, \frac{7}{2}\]

You can check if these are roots by plugging them into the function and finding out if they equal zero, for example
\[\text{f}(x)=2x^4+x^3-15x^2-7x+7\]
\[\text{f}(-1) = 2(-1)^4 + (-1)^3 - 15(-1)^2 - 7(-1) + 7 =\]
\[2 - 1 - 15 + 7 + 7 = 0\]
So since f(-1) = 0, -1 is a root of the polynomial

There is another rational root in that list, x = \(1 \over 2\), and the others aren't rational roots

So we can use syntetic division to factor out \((x + 1)\) and \((x - {1 \over 2})\)

Answer 9

x= -1 and (1/2)

Answer 10

Yes, so use long division / synthetic division to factor them both out, then check if there are any other real zero's by solving the quadratic factor that's left

Answer 11

how would you start with the synthetic division?

Answer 12

It's a bit hard to type it here, and I'm terrible at drawing so the best way to explain it is just give you a link: http://www.youtube.com/watch?v=bZoMz1Cy1T4

If you don't understand you can ask here :)

Answer 13

so I know how to start it now but when I did it I got -4 as a remainder for using -1

Answer 14

ok let me draw it


There should be no remainder
\[2x^4 + x^3 - 15x^2 - 7x + 7 = (x + 1)(2x^3 - x^2 -14x + 7)\]

Answer 15

opps did my wrong! then I use the quadratic factor??
and those will be my x? how will I find the real zeros??

Answer 16

for the zero \(\frac{1}{2}\) you can repeat the process to get
\[(x + 1)(x - {1 \over 2})(2x^2 - 14)=(x + 1)(2x - 1)(x^2 - 7)\]

The other two roots are if \(x^2 - 7 = 0\), and we already know that these are irrational because if they were rational we would've found them using the rational root theorem :)

Answer 17

how will we find the real zero of the factor of f?

Answer 18

set it equal to zero:
\[\begin{eqnarray}
(x + 1) = 0 & \text{gives} & x = -1\\
(2x - 1) = 0 & \text{gives} & x = {1 \over 2}\\
(x^2 - 7) = 0 & \text{gives} & x = \pm\sqrt{7}
\end{eqnarray}\]

So the four real roots are -1, \(1 \over 2\), \(\sqrt 7\) and \(-\sqrt 7\).

Answer 19

okay so then it would be the same with the x on the first question.

Answer 20

yes, the four real zeros are x = -1, x = 1/2, x = sqrt(7) and x = -sqrt(7)

Answer 21

so basically its the same thing.

Answer 22

yeah, a zero is the same as a root

Answer 23

so I need my answer in factored from or radicals as needed. would squ. root of 7 still count. sorry asking a lot of questions

Answer 24

I would leave the factored form as (x + 1)(2x - 1)(x^2 - 7), because radicals don't look that pretty in factored form. They might want you to also factor (x^2 - 7) as (x - sqrt(7))(x + sqrt(7)), but I wouldn't do it like that myself.

Answer 25

okay thank you!!!!!!