Question

I need the explanations of how as well..thanks Given the polynomial f(x) = x^4 - 37x^2+36?a. Use Descartes Rule of Signs to determine the number of positive and negative roots.b. Use the Rational Zero Theorem (aka Rational Roots Theorem) to determine a list of possible zeros.c. Use the Intermediate Value Theorem to prove that the polynomial has a zero in the interval [0, 4].d. Solve for the zeros of f(x).

Answer 1

a good explanation of descartes rule of sign, as well as a couple worked out examples, is here
http://www.purplemath.com/modules/drofsign.htm

Answer 2

we can work this one out in detail if you like

Answer 3

then, we can work out the next part
Use the Rational Zero Theorem (aka Rational Roots Theorem) to determine a list of possible zeros

which is absolutely stated incorrectly

Answer 4

Yes please show me how to work this out in detail because I have no clue as yo where to start

Answer 5

ok for descares rule of sign, lets list the coefficients of your polynomial
\[f(x) = x^4 - 37x^2+36\]
they are 1, -37, 36

Answer 6

there are 2 changes in sign of the coefficients, from \(+1\) to \(-37\) and then from \(-37\) to \(+36\)

Answer 7

is that part clear? that is, it is clear how to count the changes in sign?

Answer 8

Yes it is.

Answer 9

ok so this tell you there are at most 2 POSITIVE zeros
it does not say there are 2, it says there are at most 2
and you count down by 2's so there could be 2 or there could be none

Answer 10

ok

Answer 11

now we compute \(f(-x)=(-x)^4-37(-x)^2+36=x^4-37x^2+36=f(x)\) because this is an even function (all exponents are even)
generally this is not the case, this is just a special example
in this case we see there are also at most two NEGATIVE zeros
so there are either 2 negative zeros , or no negative zeros

Answer 12

again this is a special example
because this is an even function it is symmetric with respect to the y axis, so if there are two positive zeros there are also two negative zeros.
this is not always the case, it is particular to this example

Answer 13

that takes care of descartes rule of sign
we note the changes in sign of the coefficients of \(f(x)\) to count the number of possible positive zeros, and the changes is sign of the coefficients of \(f(-x)\) to find the number of possible negative zeros

Answer 14

you with me so far?

Answer 15

if so we can move on to the rational root theorem

Answer 16

Yes I am

Answer 17

ok
before we continue, this problem has been cooked up to be very easy, because it is a quadratic in \(x^2\)and you can factor it as \((x^2-1)(x^2-36)\) so the zeros are easy to find, but lets put that on hold for a second

Answer 18

Alright

Answer 19

the rational root theorem tells you how to find the POSSIBLE RATIONAL roots
first off no one says the roots have to be rational
but IF they are rational (i.e. fractions) then they are of the form \(\frac{p}{q}\) and the rational root theorem tells you that the numerator \(p\) divides the constant and the denominator \(q\) divides the leading coefficient

Answer 20

in your case the constant is 36 and the leading coefficient is 1, so your "possible rational roots" must be in this case integers that divide 36
there are a lot of them

Answer 21

\[1,2,3,6,-,12,18,36\] all divide 36, as do their opposites
\[-1,-2,-3,-6,-9,-12,-18,-36\]

Answer 22

okay

Answer 23

it would be even worse if the leading coefficient was not one, because then you would have a large list of fractions as well

Answer 24

i see

Answer 25

and don't forget, no one says the roots have to be rational
they could be none of these

Answer 26

now for part 3
check that \(f(4)\) and \(f(0)\) have opposite sign
in this case \(f(0)=36\) i will let you compute \(f(4)\) for your self, but it will probably be negative
therefore, since the function is positive when \(x=0\) and negative when \(x=4\) and continuous (since it is a polynomial) it must have a zero (i.e. cross the \(x\) axis, go from being positive to negative) somewhere in the interval \((0,4)\)

Answer 27

now as i wrote above, finding the zeros in this case is very simple
it is a polynomial of degree 4, which should make it very hard, but there is no third degree term and no first degree term, so you can think of it as a quadratic in \(x^2\) and easily factor as
\[(x^2-1)(x^2-36)\] set each factor equal to zero and solve for \(x\)

Answer 28

hope all steps are clear

Answer 29

Thank you