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vijeya3 · Answer

http://www.sparknotes.com/math/algebra2/polynomials/section4.rhtml Check this site...try to understand the method....if u need further help..lemme know :)

whpalmer4 · Answer

I think it helps to see some examples of why the RZT/RRT theorem works.

say we multiply two binomials representing the zeroes of a polynomial:

$$P(x)=(x-a)(x-b)=x*x-b*x-a*x-(a)(-b)=x^2-ax-bx+ab$$
notice  the constant term $ab$ is simply the product of the constant terms of the factor binomials.

misty1212 · Answer

HI!!

misty1212 · Answer

$$f(x) = x^3 - 3x^2 - 25x - 21
$$ since the constant is 21 and the leading coefficient is 1, the possible rational zeros are $$\pm1,\pm3,\pm7\pm21$$ the divisors of $21$

misty1212 · Answer

after that, plug in and see which one works
1 does not for sure but $f(-1)=-1-3+25-21=0$

anonymous · Answer

You can verify your answer by http://www.wolframalpha.com/input/?i=factor+x%5E3+-+3x%5E2+-+25x+-+21

whpalmer4 · Answer

As I was saying, the constant term is built up from the product of the constant terms in the factors.  As a polynomial with a given rational zero $a$ will have at least one factor $(x-a)$ we can use that knowledge to guess at the possible factors/zeros.  If the coefficient of the highest order term is $1$, then the rational zeros (if any) of the polynomial will be found in the set of positive and negative factors of the constant term.

Notice that if the factors of the constant term are not prime numbers, there may be considerably more possible rational zeros than actual ones.

$$x^2+8x+7$$only has possible rational zeroes of $\pm1,\ \pm7$
$$x^2+5x+24$$has possible rational zeroes of $\pm1,\pm2,\pm3,\pm4,\pm6,\pm8,\pm12$

Once you have your list of possible rational zeroes, you eliminate the ones which are not actual zeroes by plugging each potential zero into the polynomial and seeing if the value of the polynomial is $0$.  

Horner's rule and synthetic substitution are good ways to make testing potential zeros in cumbersome polynomials easier.

thesmartone · Answer

@UMULAS Do you understand your question now or do you need more help? :)