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To solve the polynomial equation [tex]\( x^4 - 3x^3 + 3x^2 - 3x + 2 = 0 \)[/tex], we need to find all the roots of the polynomial. Let’s go through the steps to find these roots analytically:
1. Write down the polynomial:
[tex]\[ f(x) = x^4 - 3x^3 + 3x^2 - 3x + 2 \][/tex]
2. Try to factor the polynomial:
One way to factorize a polynomial is to use the Rational Root Theorem, which suggests that potential rational roots are factors of the constant term (in this case, [tex]\(2\)[/tex]) divided by factors of the leading coefficient (which is [tex]\(1\)[/tex]).
The potential rational roots are [tex]\( \pm 1, \pm 2 \)[/tex].
3. Test potential rational roots:
Let's test [tex]\(x = 1\)[/tex]:
[tex]\[ f(1) = 1^4 - 3(1)^3 + 3(1)^2 - 3(1) + 2 = 1 - 3 + 3 - 3 + 2 = 0 \][/tex]
This indicates that [tex]\(x = 1\)[/tex] is a root.
Next, we use synthetic division or polynomial division to divide [tex]\(f(x)\)[/tex] by [tex]\(x - 1\)[/tex].
4. Perform polynomial division by [tex]\(x - 1\)[/tex]:
[tex]\[ x^4 - 3x^3 + 3x^2 - 3x + 2 \div (x - 1) \][/tex]
Using synthetic division:
[tex]\[ \begin{array}{r|rrrrr} 1 & 1 & -3 & 3 & -3 & 2 \\ & & 1 & -2 & 1 & -2\\ \hline & 1 & -2 & 1 & -2 & 0\\ \end{array} \][/tex]
The quotient is [tex]\(x^3 - 2x^2 + x - 2\)[/tex].
5. Repeat the process for [tex]\(x^3 - 2x^2 + x - 2\)[/tex]:
Test [tex]\(x = 1\)[/tex] for [tex]\(x^3 - 2x^2 + x - 2\)[/tex]:
[tex]\[ f(1) = 1^3 - 2(1)^2 + 1 - 2 = 1 - 2 + 1 - 2 = -2 \neq 0 \][/tex]
Test [tex]\(x = -1\)[/tex]:
[tex]\[ f(-1) = (-1)^3 - 2(-1)^2 + (-1) - 2 = -1 - 2 - 1 - 2 = -6 \neq 0 \][/tex]
Test [tex]\( x = 2\)[/tex]:
[tex]\[ f(2) = 2^3 - 2(2)^2 + 2 - 2 = 8 - 8 + 2 - 2 = 0 \][/tex]
This indicates [tex]\(x = 2\)[/tex] is a root.
6. Perform polynomial division by [tex]\(x - 2\)[/tex]:
[tex]\[ x^3 - 2x^2 + x - 2 \div (x - 2) \][/tex]
Using synthetic division:
[tex]\[ \begin{array}{r|rrrr} 2 & 1 & -2 & 1 & -2 \\ & & 2 & 0 & 2\\ \hline & 1 & 0 & 1 & 0\\ \end{array} \][/tex]
The quotient is [tex]\(x^2 + 1\)[/tex].
7. Solve the quadratic equation [tex]\(x^2 + 1 = 0\)[/tex]:
[tex]\[ x^2 + 1 = 0 \Rightarrow x^2 = -1 \Rightarrow x = \pm i \][/tex]
8. Combine all the roots found:
Therefore, the solutions to the polynomial equation are:
[tex]\[ \boxed{x = 1, x = 2, x = -i, x = i} \][/tex]
The correct choice is:
[tex]\[ x = \pm i, x = 1, x = 2 \][/tex]
1. Write down the polynomial:
[tex]\[ f(x) = x^4 - 3x^3 + 3x^2 - 3x + 2 \][/tex]
2. Try to factor the polynomial:
One way to factorize a polynomial is to use the Rational Root Theorem, which suggests that potential rational roots are factors of the constant term (in this case, [tex]\(2\)[/tex]) divided by factors of the leading coefficient (which is [tex]\(1\)[/tex]).
The potential rational roots are [tex]\( \pm 1, \pm 2 \)[/tex].
3. Test potential rational roots:
Let's test [tex]\(x = 1\)[/tex]:
[tex]\[ f(1) = 1^4 - 3(1)^3 + 3(1)^2 - 3(1) + 2 = 1 - 3 + 3 - 3 + 2 = 0 \][/tex]
This indicates that [tex]\(x = 1\)[/tex] is a root.
Next, we use synthetic division or polynomial division to divide [tex]\(f(x)\)[/tex] by [tex]\(x - 1\)[/tex].
4. Perform polynomial division by [tex]\(x - 1\)[/tex]:
[tex]\[ x^4 - 3x^3 + 3x^2 - 3x + 2 \div (x - 1) \][/tex]
Using synthetic division:
[tex]\[ \begin{array}{r|rrrrr} 1 & 1 & -3 & 3 & -3 & 2 \\ & & 1 & -2 & 1 & -2\\ \hline & 1 & -2 & 1 & -2 & 0\\ \end{array} \][/tex]
The quotient is [tex]\(x^3 - 2x^2 + x - 2\)[/tex].
5. Repeat the process for [tex]\(x^3 - 2x^2 + x - 2\)[/tex]:
Test [tex]\(x = 1\)[/tex] for [tex]\(x^3 - 2x^2 + x - 2\)[/tex]:
[tex]\[ f(1) = 1^3 - 2(1)^2 + 1 - 2 = 1 - 2 + 1 - 2 = -2 \neq 0 \][/tex]
Test [tex]\(x = -1\)[/tex]:
[tex]\[ f(-1) = (-1)^3 - 2(-1)^2 + (-1) - 2 = -1 - 2 - 1 - 2 = -6 \neq 0 \][/tex]
Test [tex]\( x = 2\)[/tex]:
[tex]\[ f(2) = 2^3 - 2(2)^2 + 2 - 2 = 8 - 8 + 2 - 2 = 0 \][/tex]
This indicates [tex]\(x = 2\)[/tex] is a root.
6. Perform polynomial division by [tex]\(x - 2\)[/tex]:
[tex]\[ x^3 - 2x^2 + x - 2 \div (x - 2) \][/tex]
Using synthetic division:
[tex]\[ \begin{array}{r|rrrr} 2 & 1 & -2 & 1 & -2 \\ & & 2 & 0 & 2\\ \hline & 1 & 0 & 1 & 0\\ \end{array} \][/tex]
The quotient is [tex]\(x^2 + 1\)[/tex].
7. Solve the quadratic equation [tex]\(x^2 + 1 = 0\)[/tex]:
[tex]\[ x^2 + 1 = 0 \Rightarrow x^2 = -1 \Rightarrow x = \pm i \][/tex]
8. Combine all the roots found:
Therefore, the solutions to the polynomial equation are:
[tex]\[ \boxed{x = 1, x = 2, x = -i, x = i} \][/tex]
The correct choice is:
[tex]\[ x = \pm i, x = 1, x = 2 \][/tex]
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