Question

What are the roots of the polynomial function p(x)=x3+4x2−15x−18
𝑝
(
𝑥
)
=
𝑥
3
+
4
𝑥
2
−
15
𝑥
−
18
?(1 point)
Responses

x=−6
𝑥
=
−
6
, x=−1
𝑥
=
−
1
and x=3
𝑥
=
3
x equals negative 6 , x equals negative 1 and x equals 3

x=3
𝑥
=
3
, x=1
𝑥
=
1
and x=6
𝑥
=
6
x equals 3 , x equals 1 and x equals 6

x=−6
𝑥
=
−
6
, x=1
𝑥
=
1
and x=3
𝑥
=
3
x equals negative 6 , x equals 1 and x equals 3

x=−3
𝑥
=
−
3
, x=−1
𝑥
=
−
1
and x=6

Answer 1

To find the roots of the polynomial function \( p(x) = x^3 + 4x^2 - 15x - 18 \), we can use the Rational Root Theorem or synthetic division to test possible rational roots.

Let's start by checking the possible rational roots derived from the factors of the constant term (-18) and the leading coefficient (1).

The possible rational roots are \( \pm 1, \pm 2, \pm 3, \pm 6, \pm 9, \pm 18 \).

Let's evaluate the polynomial at these values:

1. **Testing \( x = -6 \)**:
\[
p(-6) = (-6)^3 + 4(-6)^2 - 15(-6) - 18 = -216 + 144 + 90 - 18 = 0
\]
So, \( x = -6 \) is a root.

2. **Testing \( x = -1 \)**:
\[
p(-1) = (-1)^3 + 4(-1)^2 - 15(-1) - 18 = -1 + 4 + 15 - 18 = 0
\]
So, \( x = -1 \) is also a root.

3. **Now we have found two roots: \( x = -6 \) and \( x = -1 \)**. We can now divide the polynomial by \( (x + 6)(x + 1) = x^2 + 7x + 6 \) to find the remaining root.

Using synthetic division for \( p(x) \) divided by \( (x + 6)(x + 1) \):
1. Divide \( p(x) \) by \( x + 6 \):
- The quotient should give us \( x^2 + 7x + 6 \).

We can now factor \( x^2 + 7x + 6 \):
\[
x^2 + 7x + 6 = (x + 1)(x + 6)
\]
So the full factorization of \( p(x) \) is:
\[
p(x) = (x + 6)(x + 1)Q(x)
\]
For the last root, we solve:
\[
Q(x) = x - 3 \implies x = 3, \text{ since the quadratic gives } (x - 3) = 0
\]

Thus, the roots of the polynomial \( p(x) = x^3 + 4x^2 - 15x - 18 \) are:
1. \( x = -6 \)
2. \( x = -1 \)
3. \( x = 3 \)

Therefore, the correct answer is:
**\( x = -6, x = -1, \text{ and } x = 3 \)**.