Question

Find all zeroes of each function.
p(x)=x^3-3x^2+9x-7

Answer 1

To find the zeroes of the polynomial function \( p(x) = x^3 - 3x^2 + 9x - 7 \), we can start by using the Rational Root Theorem to find possible rational roots. The Rational Root Theorem suggests that any rational solution \( p/q \) (where \( p \) divides the constant term \(-7\) and \( q \) divides the leading coefficient \( 1\)) could be \( \pm 1, \pm 7 \).

Now we can test each of these candidates:

1. **Testing \( x = 1 \)**:
\[
p(1) = 1^3 - 3(1^2) + 9(1) - 7 = 1 - 3 + 9 - 7 = 0
\]
So, \( x = 1 \) is a root.

Next, we can factor \( p(x) \) by using synthetic division on the polynomial by \( x - 1 \).

### Synthetic Division
1. Set up synthetic division with root \( 1 \):
\[
\begin{array}{r|rrrr}
1 & 1 & -3 & 9 & -7 \\
& & 1 & -2 & 7 \\
\hline
& 1 & -2 & 7 & 0 \\
\end{array}
\]
The quotient is \( x^2 - 2x + 7 \).

So, we have:
\[
p(x) = (x - 1)(x^2 - 2x + 7)
\]

### Finding Zeroes of the Quadratic
Next, we need to find the zeroes of the quadratic \( x^2 - 2x + 7 \). We use the quadratic formula:
\[
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]
where \( a = 1, b = -2, c = 7 \).

Calculating the discriminant:
\[
b^2 - 4ac = (-2)^2 - 4(1)(7) = 4 - 28 = -24
\]

Since the discriminant is negative, there are two complex roots:
\[
x = \frac{-(-2) \pm \sqrt{-24}}{2(1)} = \frac{2 \pm \sqrt{24}i}{2} = 1 \pm \sqrt{6}i
\]

### Final Root Summary
Thus, the complete set of zeroes for the function \( p(x) = x^3 - 3x^2 + 9x - 7 \) is:
\[
\text{Zeroes: } x = 1, \quad x = 1 + \sqrt{6}i, \quad x = 1 - \sqrt{6}i
\]