Answer :
Let's solve the given polynomial equation step-by-step. The equation is:
[tex]\[ x^5 + 9x^4 + 18x^3 - 22x^2 - 51x + 45 = 0 \][/tex]
1. Check for Rational Roots:
We can use the Rational Root Theorem to find possible rational roots. The Rational Root Theorem suggests that any rational solution, expressed as a fraction in simplest terms [tex]\(\frac{p}{q}\)[/tex], is such that [tex]\(p\)[/tex] is a factor of the constant term (45) and [tex]\(q\)[/tex] is a factor of the leading coefficient (1).
Factors of 45: [tex]\(\pm 1, \pm 3, \pm 5, \pm 9, \pm 15, \pm 45\)[/tex]
2. Try Possible Roots:
Let's try substituting these possible roots to see if any satisfy the equation:
- Substitute [tex]\(x = 1\)[/tex]:
[tex]\[
1^5 + 9 \times 1^4 + 18 \times 1^3 - 22 \times 1^2 - 51 \times 1 + 45 = 1 + 9 + 18 - 22 - 51 + 45 = 0
\][/tex]
So, [tex]\(x = 1\)[/tex] is a root.
3. Divide the Polynomial:
Since [tex]\(x = 1\)[/tex] is a root, we can use synthetic division to divide the polynomial by [tex]\(x - 1\)[/tex] to find the quotient polynomial.
After performing synthetic division, the quotient is:
[tex]\[
x^4 + 10x^3 + 28x^2 + 6x - 45
\][/tex]
4. Repeat the Process for the New Polynomial:
- Check for additional roots on the quotient polynomial, [tex]\(x^4 + 10x^3 + 28x^2 + 6x - 45\)[/tex].
- By testing possible rational roots again, we find that [tex]\(x = -3\)[/tex] is a root.
5. Divide Again:
Using synthetic division with the factor [tex]\(x + 3\)[/tex]:
The quotient polynomial is:
[tex]\[
x^3 + 7x^2 + 7x - 15
\][/tex]
6. Find Roots of the Cubic Polynomial:
Again, checking for rational roots:
- [tex]\(x = 1\)[/tex] is once more found to be a root. Divide [tex]\(x^3 + 7x^2 + 7x - 15\)[/tex] by [tex]\(x - 1\)[/tex].
7. Final Division:
This leaves us with:
[tex]\[
x^2 + 8x + 15
\][/tex]
8. Factor the Quadratic:
Factor [tex]\(x^2 + 8x + 15\)[/tex] to get:
[tex]\[
(x + 3)(x + 5)
\][/tex]
9. List All Roots:
The complete roots of the original equation are [tex]\(x = 1, x = -3,\)[/tex] and [tex]\(x = -5\)[/tex].
10. Determine Multiplicities:
- [tex]\(x = 1\)[/tex] appeared twice in our divisions, so it has multiplicity 2.
- [tex]\(x = -3\)[/tex] also appeared twice, so it has multiplicity 2.
- [tex]\(x = -5\)[/tex] appeared once, so it has multiplicity 1.
Putting this together, the solutions are:
- [tex]\(x = 1\)[/tex] with multiplicity 2
- [tex]\(x = -3\)[/tex] with multiplicity 2
- [tex]\(x = -5\)[/tex] with multiplicity 1
[tex]\[ x^5 + 9x^4 + 18x^3 - 22x^2 - 51x + 45 = 0 \][/tex]
1. Check for Rational Roots:
We can use the Rational Root Theorem to find possible rational roots. The Rational Root Theorem suggests that any rational solution, expressed as a fraction in simplest terms [tex]\(\frac{p}{q}\)[/tex], is such that [tex]\(p\)[/tex] is a factor of the constant term (45) and [tex]\(q\)[/tex] is a factor of the leading coefficient (1).
Factors of 45: [tex]\(\pm 1, \pm 3, \pm 5, \pm 9, \pm 15, \pm 45\)[/tex]
2. Try Possible Roots:
Let's try substituting these possible roots to see if any satisfy the equation:
- Substitute [tex]\(x = 1\)[/tex]:
[tex]\[
1^5 + 9 \times 1^4 + 18 \times 1^3 - 22 \times 1^2 - 51 \times 1 + 45 = 1 + 9 + 18 - 22 - 51 + 45 = 0
\][/tex]
So, [tex]\(x = 1\)[/tex] is a root.
3. Divide the Polynomial:
Since [tex]\(x = 1\)[/tex] is a root, we can use synthetic division to divide the polynomial by [tex]\(x - 1\)[/tex] to find the quotient polynomial.
After performing synthetic division, the quotient is:
[tex]\[
x^4 + 10x^3 + 28x^2 + 6x - 45
\][/tex]
4. Repeat the Process for the New Polynomial:
- Check for additional roots on the quotient polynomial, [tex]\(x^4 + 10x^3 + 28x^2 + 6x - 45\)[/tex].
- By testing possible rational roots again, we find that [tex]\(x = -3\)[/tex] is a root.
5. Divide Again:
Using synthetic division with the factor [tex]\(x + 3\)[/tex]:
The quotient polynomial is:
[tex]\[
x^3 + 7x^2 + 7x - 15
\][/tex]
6. Find Roots of the Cubic Polynomial:
Again, checking for rational roots:
- [tex]\(x = 1\)[/tex] is once more found to be a root. Divide [tex]\(x^3 + 7x^2 + 7x - 15\)[/tex] by [tex]\(x - 1\)[/tex].
7. Final Division:
This leaves us with:
[tex]\[
x^2 + 8x + 15
\][/tex]
8. Factor the Quadratic:
Factor [tex]\(x^2 + 8x + 15\)[/tex] to get:
[tex]\[
(x + 3)(x + 5)
\][/tex]
9. List All Roots:
The complete roots of the original equation are [tex]\(x = 1, x = -3,\)[/tex] and [tex]\(x = -5\)[/tex].
10. Determine Multiplicities:
- [tex]\(x = 1\)[/tex] appeared twice in our divisions, so it has multiplicity 2.
- [tex]\(x = -3\)[/tex] also appeared twice, so it has multiplicity 2.
- [tex]\(x = -5\)[/tex] appeared once, so it has multiplicity 1.
Putting this together, the solutions are:
- [tex]\(x = 1\)[/tex] with multiplicity 2
- [tex]\(x = -3\)[/tex] with multiplicity 2
- [tex]\(x = -5\)[/tex] with multiplicity 1
