To find the roots of the polynomial \( p(x) = x^3 + 4x^2 - 7x - 10 \), we can use methods such as the Rational Root Theorem, synthetic division, or numerical methods.
First, let's test for potential rational roots using the Rational Root Theorem. The possible rational roots could be the factors of the constant term (-10) divided by the factors of the leading coefficient (1). The candidates are:
\[ \pm 1, \pm 2, \pm 5, \pm 10 \]
We can test these candidates by substituting them into the polynomial.
-
Testing \( x = 1 \): \[ p(1) = 1^3 + 4(1^2) - 7(1) - 10 = 1 + 4 - 7 - 10 = -12 \quad (\text{not a root}) \]
-
Testing \( x = -1 \): \[ p(-1) = (-1)^3 + 4(-1)^2 - 7(-1) - 10 = -1 + 4 + 7 - 10 = 0 \quad (\text{is a root}) \]
Since \( x = -1 \) is a root, we can use synthetic division to divide \( p(x) \) by \( x + 1 \).
Using synthetic division: \[ \begin{array}{r|rrrr} -1 & 1 & 4 & -7 & -10 \ & & -1 & -3 & 10 \ \hline & 1 & 3 & -10 & 0 \ \end{array} \]
The quotient is \( x^2 + 3x - 10 \).
Now we need to find the roots of \( x^2 + 3x - 10 = 0 \). We can use the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
- Here, \( a = 1 \), \( b = 3 \), and \( c = -10 \).
Calculating the discriminant: \[ b^2 - 4ac = 3^2 - 4(1)(-10) = 9 + 40 = 49 \]
Now applying the quadratic formula: \[ x = \frac{-3 \pm \sqrt{49}}{2 \cdot 1} = \frac{-3 \pm 7}{2} \]
This gives us the two roots:
- \( x = \frac{4}{2} = 2 \)
- \( x = \frac{-10}{2} = -5 \)
Thus, the roots of the polynomial \( p(x) = x^3 + 4x^2 - 7x - 10 \) are:
- \( x = -5 \)
- \( x = -1 \)
- \( x = 2 \)
Now listing the roots in ascending order: \[ \text{The roots are } -5, -1, \text{ and } 2. \]