Answer :
To solve the equation [tex]\(x^4 + x^3 - 14x^2 - 9x + 45 = 0\)[/tex], we need to find the values of [tex]\(x\)[/tex] that make the equation true. This is a polynomial equation of degree 4.
### Step-by-step solution:
1. Identify potential rational roots:
We can use the Rational Root Theorem to identify potential rational roots. The theorem states that any rational solution, expressed as a fraction [tex]\(\frac{p}{q}\)[/tex], has [tex]\(p\)[/tex] as a factor of the constant term (45) and [tex]\(q\)[/tex] as a factor of the leading coefficient (1).
Factors of 45: ±1, ±3, ±5, ±9, ±15, ±45
Factors of 1: ±1
Potential rational roots: ±1, ±3, ±5, ±9, ±15, ±45.
2. Test potential rational roots:
We substitute these values into the polynomial to check for roots.
- Testing [tex]\(x = -3\)[/tex], we find that it makes the polynomial equation zero, so [tex]\(x = -3\)[/tex] is a root.
- Testing [tex]\(x = 3\)[/tex], it also makes the polynomial equation zero, thus [tex]\(x = 3\)[/tex] is another root.
3. Perform polynomial division:
Once we find a root, we can factor it out of the polynomial using polynomial division. Given roots [tex]\(x = -3\)[/tex] and [tex]\(x = 3\)[/tex], perform synthetic or long division to divide the original polynomial by [tex]\((x + 3)\)[/tex] and [tex]\((x - 3)\)[/tex] to simplify the polynomial.
4. Solve the remaining quadratic equation:
The remaining equation is obtained after factoring out the terms associated with the roots you found. It's typically a quadratic equation that can be solved using the quadratic formula, [tex]\(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)[/tex].
- After division, the remaining quadratic is [tex]\(2x^2 + x - 21 = 0\)[/tex].
Using the quadratic formula, we have:
[tex]\( a = 2, b = 1, c = -21 \)[/tex]
[tex]\[
x = \frac{-1 \pm \sqrt{1^2 - 4 \times 2 \times (-21)}}{2 \times 2}
\][/tex]
[tex]\[
x = \frac{-1 \pm \sqrt{1 + 168}}{4}
\][/tex]
[tex]\[
x = \frac{-1 \pm \sqrt{169}}{4}
\][/tex]
[tex]\[
x = \frac{-1 \pm 13}{4}
\][/tex]
This gives two solutions:
[tex]\[
x = \frac{12}{4} = 3
\][/tex]
[tex]\[
x = \frac{-14}{4} = -\frac{1 + \sqrt{21}}{2}
\][/tex]
[tex]\[
x = -\frac{1}{2} + \frac{\sqrt{21}}{2}
\][/tex]
[tex]\[
x = -\frac{1}{2} - \frac{\sqrt{21}}{2}
\][/tex]
5. Summary of all solutions:
Therefore, the solutions to the equation are:
[tex]\[
x = -3, \quad x = 3, \quad x = -\frac{1}{2} + \frac{\sqrt{21}}{2}, \quad x = -\frac{1}{2} - \frac{\sqrt{21}}{2}
\][/tex]
These are all the roots of the polynomial equation [tex]\(x^4 + x^3 - 14x^2 - 9x + 45 = 0\)[/tex].
### Step-by-step solution:
1. Identify potential rational roots:
We can use the Rational Root Theorem to identify potential rational roots. The theorem states that any rational solution, expressed as a fraction [tex]\(\frac{p}{q}\)[/tex], has [tex]\(p\)[/tex] as a factor of the constant term (45) and [tex]\(q\)[/tex] as a factor of the leading coefficient (1).
Factors of 45: ±1, ±3, ±5, ±9, ±15, ±45
Factors of 1: ±1
Potential rational roots: ±1, ±3, ±5, ±9, ±15, ±45.
2. Test potential rational roots:
We substitute these values into the polynomial to check for roots.
- Testing [tex]\(x = -3\)[/tex], we find that it makes the polynomial equation zero, so [tex]\(x = -3\)[/tex] is a root.
- Testing [tex]\(x = 3\)[/tex], it also makes the polynomial equation zero, thus [tex]\(x = 3\)[/tex] is another root.
3. Perform polynomial division:
Once we find a root, we can factor it out of the polynomial using polynomial division. Given roots [tex]\(x = -3\)[/tex] and [tex]\(x = 3\)[/tex], perform synthetic or long division to divide the original polynomial by [tex]\((x + 3)\)[/tex] and [tex]\((x - 3)\)[/tex] to simplify the polynomial.
4. Solve the remaining quadratic equation:
The remaining equation is obtained after factoring out the terms associated with the roots you found. It's typically a quadratic equation that can be solved using the quadratic formula, [tex]\(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)[/tex].
- After division, the remaining quadratic is [tex]\(2x^2 + x - 21 = 0\)[/tex].
Using the quadratic formula, we have:
[tex]\( a = 2, b = 1, c = -21 \)[/tex]
[tex]\[
x = \frac{-1 \pm \sqrt{1^2 - 4 \times 2 \times (-21)}}{2 \times 2}
\][/tex]
[tex]\[
x = \frac{-1 \pm \sqrt{1 + 168}}{4}
\][/tex]
[tex]\[
x = \frac{-1 \pm \sqrt{169}}{4}
\][/tex]
[tex]\[
x = \frac{-1 \pm 13}{4}
\][/tex]
This gives two solutions:
[tex]\[
x = \frac{12}{4} = 3
\][/tex]
[tex]\[
x = \frac{-14}{4} = -\frac{1 + \sqrt{21}}{2}
\][/tex]
[tex]\[
x = -\frac{1}{2} + \frac{\sqrt{21}}{2}
\][/tex]
[tex]\[
x = -\frac{1}{2} - \frac{\sqrt{21}}{2}
\][/tex]
5. Summary of all solutions:
Therefore, the solutions to the equation are:
[tex]\[
x = -3, \quad x = 3, \quad x = -\frac{1}{2} + \frac{\sqrt{21}}{2}, \quad x = -\frac{1}{2} - \frac{\sqrt{21}}{2}
\][/tex]
These are all the roots of the polynomial equation [tex]\(x^4 + x^3 - 14x^2 - 9x + 45 = 0\)[/tex].
