Use Descartes' rules of signs to discuss the possibilities for the roots of each equation. Do not solve equation. $$p(x)= x^3+5x^2+7x+1=0$$
$p(x)$ I saw no sign change
$p(-x)$ I saw 2 sign changes because $-x^3+5x^2-7x+1=0$. I am stuck how is the answer $3$ negative roots for $p(x)$ and how do I get $1$ negative and $2$ imaginary roots for $p(x)$?
On
A very late answer, hoping it will benefit someone in future:
The word you missed is "at most" - 3 sign changes means "it has at most 3 negative roots", and not that "it has 3 negative roots"
Positive roots:The rule states that if the terms of a single-variable polynomial with real coefficients are ordered by descending variable exponent, then the number of positive roots of the polynomial is either equal to the number of sign differences between consecutive nonzero coefficients, or is less than it by an even number. Multiple roots of the same value are counted separately.
Negative roots:As a corollary of the rule, the number of negative roots is the number of sign changes after multiplying the coefficients of odd-power terms by −1, or fewer than it by an even number. This procedure is equivalent to substituting the negation of the variable for the variable itself. - Wikipedia
As you have seen, $p(-x)= -x^3+5x^2-7x+1=0$ has three sign changes: and that assures that there are at most 3 negative roots or less than it by an even number.
Since cubic equations has 3 roots and one is essentially a real root, and since $p(x)$ has no positive roots, it has either all 3 negative roots or 1 negative root and two complex conjugates as roots. This is all we can infer from Descartes' rule of signs.
Now, solving we see that it has one real negative root, and two complex roots.
There are sign changes from $-x^3$ to $+5x^2$, from $+5x^2$ to $-7x$, and from $-7x$ to $+1$. So that is three sign changes.