Finding the Smallest Positive Root of the Polynomial

We are asked to find the smallest positive root of the equation:

$x^5 - 5x^4 - 10 x^3 + 50 x^2 + 9 x - 45 = 0$

Let the polynomial be $P(x) = x^5 - 5x^4 - 10 x^3 + 50 x^2 + 9 x - 45$. We can try to factor this polynomial by grouping terms.

Polynomial Factorization

Group the terms as follows:

$ (x^5 - 5x^4) + (-10 x^3 + 50 x^2) + (9 x - 45) = 0 $

Factor out common terms from each group:

$ x^4(x - 5) - 10x^2(x - 5) + 9(x - 5) = 0 $

Notice that $(x - 5)$ is a common factor:

$ (x - 5)(x^4 - 10x^2 + 9) = 0 $

Solving the Factors

This equation holds if either factor is zero:

1. $x - 5 = 0 \implies x = 5$
2. $x^4 - 10x^2 + 9 = 0$

For the second factor, let $y = x^2$. The equation becomes a quadratic in $y$:

$ y^2 - 10y + 9 = 0 $

Factor the quadratic equation:

$ (y - 1)(y - 9) = 0 $

This gives $y = 1$ or $y = 9$. Substitute back $x^2 = y$:

• If $y = 1$, then $x^2 = 1 \implies x = \pm 1$.
• If $y = 9$, then $x^2 = 9 \implies x = \pm 3$.

Identifying Roots and the Smallest Positive Root

The roots of the original equation are $5, 1, -1, 3, -3$.

The positive roots are $1, 3, 5$.

The smallest positive root is $x=1$.

Determining the Correct Interval

We need to find the interval that contains the smallest positive root, $x=1$. Let's check the given options:

• Option 1: $0 < x \le 2$. Since $0 < 1 \le 2$, this interval contains the root $x=1$.
• Option 2: $2 < x \le 4$. $x=1$ is not in this interval.
• Option 3: $6 \le x \le 8$. $x=1$ is not in this interval.
• Option 4: $10 \le x \le 100$. $x=1$ is not in this interval.

Therefore, the smallest positive root lies in the interval $0 < x \le 2$.