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To factorize the polynomial [tex]\( P(x) = x^3 + 2x^2 + x + 2 \)[/tex], we will use algebraic techniques to find its factors. Here’s the step-by-step solution:
1. Identify possible rational roots using the Rational Root Theorem:
The Rational Root Theorem states that any potential rational root of a polynomial equation [tex]\( P(x) \)[/tex] with integer coefficients is a factor of the constant term divided by a factor of the leading coefficient.
For [tex]\( P(x) = x^3 + 2x^2 + x + 2 \)[/tex]:
- The constant term is [tex]\( 2 \)[/tex].
- The leading coefficient is [tex]\( 1 \)[/tex].
Possible rational roots are the factors of [tex]\( \frac{\text{constant term}}{\text{leading coefficient}} = \frac{2}{1} \)[/tex], i.e., [tex]\( \pm 1, \pm 2 \)[/tex].
2. Test the possible rational roots:
- Substitute [tex]\( x = 1 \)[/tex] into [tex]\( P(x) \)[/tex]:
[tex]\[ P(1) = 1^3 + 2(1^2) + 1 + 2 = 1 + 2 + 1 + 2 = 6 \neq 0 \][/tex]
So, [tex]\( x = 1 \)[/tex] is not a root.
- Substitute [tex]\( x = -1 \)[/tex] into [tex]\( P(x) \)[/tex]:
[tex]\[ P(-1) = (-1)^3 + 2(-1)^2 + (-1) + 2 = -1 + 2 - 1 + 2 = 2 \neq 0 \][/tex]
So, [tex]\( x = -1 \)[/tex] is not a root.
- Substitute [tex]\( x = 2 \)[/tex] into [tex]\( P(x) \)[/tex]:
[tex]\[ P(2) = 2^3 + 2(2^2) + 2 + 2 = 8 + 8 + 2 + 2 = 20 \neq 0 \][/tex]
So, [tex]\( x = 2 \)[/tex] is not a root.
- Substitute [tex]\( x = -2 \)[/tex] into [tex]\( P(x) \)[/tex]:
[tex]\[ P(-2) = (-2)^3 + 2(-2)^2 + (-2) + 2 = -8 + 8 - 2 + 2 = 0 \][/tex]
So, [tex]\( x = -2 \)[/tex] is a root.
3. Factorize [tex]\( P(x) \)[/tex] using [tex]\( x + 2 \)[/tex] as a factor:
Given that [tex]\( x = -2 \)[/tex] is a root, [tex]\( x + 2 \)[/tex] is a factor of [tex]\( P(x) \)[/tex]. Use polynomial division or synthetic division to divide [tex]\( P(x) \)[/tex] by [tex]\( x + 2 \)[/tex].
4. Polynomial division of [tex]\( P(x) \)[/tex] by [tex]\( x + 2 \)[/tex]:
[tex]\[ \begin{array}{r|rrrr} & 1 & 2 & 1 & 2 \\ \hline -2 & & -2 & 0 & -2 & 0 \\ \hline & 1 & 0 & 1 & 0 \\ \end{array} \][/tex]
So, [tex]\( P(x) = (x + 2)(x^2 + 1) \)[/tex].
5. Write the final factorized form:
Therefore, the factored form of [tex]\( P(x) = x^3 + 2x^2 + x + 2 \)[/tex] is:
[tex]\[ P(x) = (x + 2)(x^2 + 1) \][/tex]
Thus, [tex]\( P(x) = (x + 2)(x^2 + 1) \)[/tex].
1. Identify possible rational roots using the Rational Root Theorem:
The Rational Root Theorem states that any potential rational root of a polynomial equation [tex]\( P(x) \)[/tex] with integer coefficients is a factor of the constant term divided by a factor of the leading coefficient.
For [tex]\( P(x) = x^3 + 2x^2 + x + 2 \)[/tex]:
- The constant term is [tex]\( 2 \)[/tex].
- The leading coefficient is [tex]\( 1 \)[/tex].
Possible rational roots are the factors of [tex]\( \frac{\text{constant term}}{\text{leading coefficient}} = \frac{2}{1} \)[/tex], i.e., [tex]\( \pm 1, \pm 2 \)[/tex].
2. Test the possible rational roots:
- Substitute [tex]\( x = 1 \)[/tex] into [tex]\( P(x) \)[/tex]:
[tex]\[ P(1) = 1^3 + 2(1^2) + 1 + 2 = 1 + 2 + 1 + 2 = 6 \neq 0 \][/tex]
So, [tex]\( x = 1 \)[/tex] is not a root.
- Substitute [tex]\( x = -1 \)[/tex] into [tex]\( P(x) \)[/tex]:
[tex]\[ P(-1) = (-1)^3 + 2(-1)^2 + (-1) + 2 = -1 + 2 - 1 + 2 = 2 \neq 0 \][/tex]
So, [tex]\( x = -1 \)[/tex] is not a root.
- Substitute [tex]\( x = 2 \)[/tex] into [tex]\( P(x) \)[/tex]:
[tex]\[ P(2) = 2^3 + 2(2^2) + 2 + 2 = 8 + 8 + 2 + 2 = 20 \neq 0 \][/tex]
So, [tex]\( x = 2 \)[/tex] is not a root.
- Substitute [tex]\( x = -2 \)[/tex] into [tex]\( P(x) \)[/tex]:
[tex]\[ P(-2) = (-2)^3 + 2(-2)^2 + (-2) + 2 = -8 + 8 - 2 + 2 = 0 \][/tex]
So, [tex]\( x = -2 \)[/tex] is a root.
3. Factorize [tex]\( P(x) \)[/tex] using [tex]\( x + 2 \)[/tex] as a factor:
Given that [tex]\( x = -2 \)[/tex] is a root, [tex]\( x + 2 \)[/tex] is a factor of [tex]\( P(x) \)[/tex]. Use polynomial division or synthetic division to divide [tex]\( P(x) \)[/tex] by [tex]\( x + 2 \)[/tex].
4. Polynomial division of [tex]\( P(x) \)[/tex] by [tex]\( x + 2 \)[/tex]:
[tex]\[ \begin{array}{r|rrrr} & 1 & 2 & 1 & 2 \\ \hline -2 & & -2 & 0 & -2 & 0 \\ \hline & 1 & 0 & 1 & 0 \\ \end{array} \][/tex]
So, [tex]\( P(x) = (x + 2)(x^2 + 1) \)[/tex].
5. Write the final factorized form:
Therefore, the factored form of [tex]\( P(x) = x^3 + 2x^2 + x + 2 \)[/tex] is:
[tex]\[ P(x) = (x + 2)(x^2 + 1) \][/tex]
Thus, [tex]\( P(x) = (x + 2)(x^2 + 1) \)[/tex].
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