Question

HELP ME PLEASE QUICK! How do I solve x^3 - 7x^2 +12x using Descartes Rule of Signs?

Answer 1

I guess I've put this off long enough :)
Are you trying to find how many positive and negative solutions this has?

Answer 2

No, I honestly dont know how to solve with descartes rule of signs, i have never heard of that before. If you know how to do it please help me :)

Answer 3

All I know about it is that the number of positive roots would be the number of times the sign of the term changes at every term..

Answer 4

From x^3 to -7x^2, it went from + to - so that's one...

Answer 5

And from -7x^2 it went to +12x, so from - to + so that's two

Answer 6

ohhh ok alright thank you I'll see what i can do with this

Answer 7

JACK I'M GOING TO KILL YOU , YOU TROLL

Answer 8

Descartes' Rule of Signs doesn't solve the equation for you, but it does give you some guidance about what the solution looks like.

Answer 9

Thank you

Answer 10

If you have some polynomial P(x) and you want to apply Descartes' Rule of Signs:

First write it in order of descending exponents (which you already have). This is crucial if you want to get the right answer!

Now count from left to right the number of sign changes you encounter in P(x). I like to write down a string of +- characters to indicate the sign of each term, which for this polynomial \(P(x) = x^3 - 7x^2 +12x\) would be +-+

+-+ has 2 sign changes (+ to -, - to +).

The number of sign changes in P(x) gives you the number of positive real roots, possibly adjusted downward by a multiple of 2. The adjustment comes because complex roots in a polynomial with real coefficients always come in conjugate pairs of the form \(a \pm bi\) where \(i = \sqrt{-1}\). We don't know how many such pairs there are.

Next, rewrite the polynomial as P(-x). The easy way to do this is simply copy the polynomial, changing the sign of any term with an odd exponent.

For our P(x), \(P(-x) = -x^3-7x^2-12x\)
Now we repeat the counting of sign changes: ---
We have 0 sign changes in P(-x), so we have 0 negative real roots. If that number had turned out to be 2 or great, again, it could be potentially adjusted downward by a multiple of 2 for complex roots.

Finally, look at the highest exponent of P(x). That is the total number of roots of P(x), combining positive real, negative real, and complex roots. Any difference between the sum of the counts of positive real roots and negative real roots and the total root count is made up by complex roots, or roots which are 0 (which is real, but neither positive nor negative).

We have 3 roots, and we have decided that 2 of them may be positive real roots. We have 1 "missing" root, but we know it cannot be a complex root. If we take a look at our equation, it becomes apparent that we could factor out \(x\) from each term, rewriting \[P(x) = x(x^2-7x+12)\]so when solving \[P(x) = x(x^2-7x+12) = 0\](which is what we do when we find the roots) one of those solutions is simply \(x=0\). The other two solutions will be the solutions to \[x^2-7x+12 =0\]and can be found by factoring, completing the square, or the quadratic formula.

By using the discriminant test on that remaining polynomial, we can clear up the question as to whether we have 2 positive real roots or a pair of complex conjugate roots. If the polynomial is written in the form \(ax^2+bx+c=0\) then the discriminant can be found by evaluating \(b^2-4ac\). If the result is positive, we have two real roots. If the result is 0, we have a perfect square with one real root with multiplicity 2. If the result is negative, we have two complex conjugate roots.

Answer 11

Thank you so much for your help!