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It's fairly clear that x = 0 is a root of P(x)
This is because P(x) = 0 when x = 0
P(x) = x^5+4x^3-x^2+8x
P(0) = (0)^5+4(0)^3-(0)^2+8(0)
P(0) = 0
Let's factor x from each term
P(x) = x^5+4x^3-x^2+8x
P(x) = x(x^4+4x^2-x+8)
We have something of the form
P(x) = x*Q(x)
where
Q(x) = x^4+4x^2-x+8
Let's use Descartes rule of signs to determine the possible number of roots for Q(x).
It will in turn help us determine the possible number of roots for P(x).
We need to count the number of sign changes in Q(x) to determine the possible number of positive roots for Q(x).
Think of x^4 as +x^4
The jump from +x^4 to +4x^2 has no sign change because each term is positive.
But the jump from +4x^2 to -x has the sign go from positive to negative. We have a sign change.
Another sign change happens when going from -x to +8.
Q(x) has 2 sign changes in total.
It means Q(x) has either 2 positive real roots or 0 positive real roots.
We step down by 2 because complex roots (of the form a+bi) come in conjugate pairs.
This is only when all coefficients of the polynomial are real numbers.
Now we need to look at the sign changes of Q(-x) to determine the possible number of negative roots for Q(x).
Replace each x with -x and simplify.
Q(x) = x^4 + 4x^2 - x + 8
Q(-x) = (-x)^4 + 4(-x)^2 - (-x) + 8
Q(-x) = x^4 + 4x^2 + x + 8
There are NO sign changes at all here.
Each term is positive.
Therefore, the number of possible negative real roots is 0.
In other words, Q(x) will not have x-intercepts in the negative territory.
Furthermore, P(x) doesn't have any x-intercepts in negative territory.
Here are all the possible outcomes for Q(x)
Positive
Negative
Imaginary
Total
2
0
2
4
0
0
4
4
The first row represents having 2 positive real number roots. Notice the total is always 4 due to the Fundamental Theorem of Algebra.
This theorem says the total number of roots (real+imaginary) must be the degree of the polynomial.
So if we know there are 2 positive roots and 0 negative roots. Then so far we have 2+0 = 2 real roots. The remaining 4-2 = 2 roots must be in the form a+bi where i = sqrt(-1)
We use the same logic for the second row as well. All four roots for that row are complex.
This is what the table looks like for Q(x) when we list the possible number of real roots and imaginary roots.
Real
Imaginary
Total
2
2
4
0
4
4
This is what the table looks like for P(x) when we list the possible number of real roots and imaginary roots.
Real
Imaginary
Total
3
2
5
1
4
5
I've added 1 to each real count for Q(x) since we determined earlier x = 0 was a root of P(x).
Each total also increases by 1.
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