Question
Find the zeros and fully factor f(x)=x^3+x^2-8x-6, including factors for irrational zeros. Use radicals, not decimal approximations.
The zeros are
The fully factored form is f(x)=
The zeros are
The fully factored form is f(x)=
Answers
GPT-4o mini
To find the zeros and fully factor the polynomial \( f(x) = x^3 + x^2 - 8x - 6 \), we can start by looking for rational roots using the Rational Root Theorem. The possible rational roots are factors of the constant term (-6) divided by factors of the leading coefficient (1). The possible rational roots are thus:
\[
\pm 1, \pm 2, \pm 3, \pm 6
\]
Next, we can test these possible roots by substituting them into \( f(x) \):
1. Testing \( x = 1 \):
\[
f(1) = 1^3 + 1^2 - 8 \cdot 1 - 6 = 1 + 1 - 8 - 6 = -12 \quad \text{(not a root)}
\]
2. Testing \( x = -1 \):
\[
f(-1) = (-1)^3 + (-1)^2 - 8 \cdot (-1) - 6 = -1 + 1 + 8 - 6 = 2 \quad \text{(not a root)}
\]
3. Testing \( x = 2 \):
\[
f(2) = 2^3 + 2^2 - 8 \cdot 2 - 6 = 8 + 4 - 16 - 6 = -10 \quad \text{(not a root)}
\]
4. Testing \( x = -2 \):
\[
f(-2) = (-2)^3 + (-2)^2 - 8 \cdot (-2) - 6 = -8 + 4 + 16 - 6 = 6 \quad \text{(not a root)}
\]
5. Testing \( x = 3 \):
\[
f(3) = 3^3 + 3^2 - 8 \cdot 3 - 6 = 27 + 9 - 24 - 6 = 6 \quad \text{(not a root)}
\]
6. Testing \( x = -3 \):
\[
f(-3) = (-3)^3 + (-3)^2 - 8 \cdot (-3) - 6 = -27 + 9 + 24 - 6 = 0 \quad \text{(is a root)}
\]
Since \( x = -3 \) is a root, we can use synthetic division to factor \( f(x) \) by \( x + 3 \):
Performing synthetic division:
\[
\begin{array}{r|rrr}
-3 & 1 & 1 & -8 & -6 \\
& & -3 & 6 & 6 \\
\hline
& 1 & -2 & -2 & 0 \\
\end{array}
\]
The result is \( x^2 - 2x - 2 \).
Next, we need to factor \( x^2 - 2x - 2 \). We can find the roots using the quadratic formula:
\[
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{2 \pm \sqrt{(-2)^2 - 4 \cdot 1 \cdot (-2)}}{2 \cdot 1} = \frac{2 \pm \sqrt{4 + 8}}{2} = \frac{2 \pm \sqrt{12}}{2} = \frac{2 \pm 2\sqrt{3}}{2} = 1 \pm \sqrt{3}
\]
Thus, the roots of \( f(x) \) are \( x = -3, 1 + \sqrt{3}, 1 - \sqrt{3} \).
The fully factored form of \( f(x) \) is:
\[
f(x) = (x + 3)(x - (1 + \sqrt{3}))(x - (1 - \sqrt{3}))
\]
This can also be expressed as:
\[
f(x) = (x + 3)(x - 1 - \sqrt{3})(x - 1 + \sqrt{3})
\]
To summarize:
The zeros are:
\[
-3, 1+\sqrt{3}, 1-\sqrt{3}
\]
The fully factored form is:
\[
f(x) = (x + 3)\left(x - (1 + \sqrt{3})\right)\left(x - (1 - \sqrt{3})\right)
\]
\[
\pm 1, \pm 2, \pm 3, \pm 6
\]
Next, we can test these possible roots by substituting them into \( f(x) \):
1. Testing \( x = 1 \):
\[
f(1) = 1^3 + 1^2 - 8 \cdot 1 - 6 = 1 + 1 - 8 - 6 = -12 \quad \text{(not a root)}
\]
2. Testing \( x = -1 \):
\[
f(-1) = (-1)^3 + (-1)^2 - 8 \cdot (-1) - 6 = -1 + 1 + 8 - 6 = 2 \quad \text{(not a root)}
\]
3. Testing \( x = 2 \):
\[
f(2) = 2^3 + 2^2 - 8 \cdot 2 - 6 = 8 + 4 - 16 - 6 = -10 \quad \text{(not a root)}
\]
4. Testing \( x = -2 \):
\[
f(-2) = (-2)^3 + (-2)^2 - 8 \cdot (-2) - 6 = -8 + 4 + 16 - 6 = 6 \quad \text{(not a root)}
\]
5. Testing \( x = 3 \):
\[
f(3) = 3^3 + 3^2 - 8 \cdot 3 - 6 = 27 + 9 - 24 - 6 = 6 \quad \text{(not a root)}
\]
6. Testing \( x = -3 \):
\[
f(-3) = (-3)^3 + (-3)^2 - 8 \cdot (-3) - 6 = -27 + 9 + 24 - 6 = 0 \quad \text{(is a root)}
\]
Since \( x = -3 \) is a root, we can use synthetic division to factor \( f(x) \) by \( x + 3 \):
Performing synthetic division:
\[
\begin{array}{r|rrr}
-3 & 1 & 1 & -8 & -6 \\
& & -3 & 6 & 6 \\
\hline
& 1 & -2 & -2 & 0 \\
\end{array}
\]
The result is \( x^2 - 2x - 2 \).
Next, we need to factor \( x^2 - 2x - 2 \). We can find the roots using the quadratic formula:
\[
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{2 \pm \sqrt{(-2)^2 - 4 \cdot 1 \cdot (-2)}}{2 \cdot 1} = \frac{2 \pm \sqrt{4 + 8}}{2} = \frac{2 \pm \sqrt{12}}{2} = \frac{2 \pm 2\sqrt{3}}{2} = 1 \pm \sqrt{3}
\]
Thus, the roots of \( f(x) \) are \( x = -3, 1 + \sqrt{3}, 1 - \sqrt{3} \).
The fully factored form of \( f(x) \) is:
\[
f(x) = (x + 3)(x - (1 + \sqrt{3}))(x - (1 - \sqrt{3}))
\]
This can also be expressed as:
\[
f(x) = (x + 3)(x - 1 - \sqrt{3})(x - 1 + \sqrt{3})
\]
To summarize:
The zeros are:
\[
-3, 1+\sqrt{3}, 1-\sqrt{3}
\]
The fully factored form is:
\[
f(x) = (x + 3)\left(x - (1 + \sqrt{3})\right)\left(x - (1 - \sqrt{3})\right)
\]