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To find a polynomial function of the lowest degree with rational coefficients that has the given numbers as some of its zeros, follow these steps:
1. Identify the given zeros: The zeros provided are [tex]\( 1+i \)[/tex] and [tex]\( 4 \)[/tex].
2. Determine the conjugate zero for complex roots: For polynomials with rational coefficients, complex roots must come in conjugate pairs. Since [tex]\( 1+i \)[/tex] is a zero, its conjugate [tex]\( 1-i \)[/tex] must also be a zero.
Thus, the zeros are [tex]\( 1+i \)[/tex], [tex]\( 1-i \)[/tex], and [tex]\( 4 \)[/tex].
3. Form the factors: Write the polynomial in factored form using its zeros:
[tex]\[ (x - (1 + i)), \quad (x - (1 - i)), \quad \text{and} \quad (x - 4). \][/tex]
4. Expand the polynomial: Calculate the product of these factors step-by-step.
First, multiply the factors involving the complex conjugates:
[tex]\[ (x - (1 + i))(x - (1 - i)) \][/tex]
Use the difference of squares formula [tex]\( (a - b)(a + b) = a^2 - b^2 \)[/tex], where [tex]\( a = x - 1 \)[/tex] and [tex]\( b = i \)[/tex]:
[tex]\[ (x - 1)^2 - i^2 \][/tex]
Note that [tex]\( i^2 = -1 \)[/tex], so:
[tex]\[ (x - 1)^2 - (-1) = (x - 1)^2 + 1 \][/tex]
Expand [tex]\( (x-1)^2 \)[/tex]:
[tex]\[ (x - 1)^2 = x^2 - 2x + 1 \][/tex]
Therefore:
[tex]\[ (x - 1)^2 + 1 = x^2 - 2x + 1 + 1 = x^2 - 2x + 2 \][/tex]
Now we have the intermediate polynomial:
[tex]\[ (x^2 - 2x + 2) \][/tex]
5. Include the remaining root: The factor corresponding to the root [tex]\( 4 \)[/tex] is [tex]\( (x - 4) \)[/tex].
Multiply the intermediate polynomial by this factor:
[tex]\[ (x^2 - 2x + 2)(x - 4) \][/tex]
6. Expand the final expression: Apply the distributive property to expand:
[tex]\[ (x^2 - 2x + 2)(x - 4) = x(x^2 - 2x + 2) - 4(x^2 - 2x + 2) \][/tex]
Perform the multiplications separately:
[tex]\[ x(x^2 - 2x + 2) = x^3 - 2x^2 + 2x \][/tex]
[tex]\[ -4(x^2 - 2x + 2) = -4x^2 + 8x - 8 \][/tex]
Combine the terms:
[tex]\[ x^3 - 2x^2 + 2x - 4x^2 + 8x - 8 = x^3 - 6x^2 + 10x - 8 \][/tex]
Therefore, the polynomial of the lowest degree with rational coefficients that has the given zeros [tex]\( 1+i \)[/tex] and [tex]\( 4 \)[/tex] is:
[tex]\[ x^3 - 6x^2 + 10x - 8 \][/tex]
1. Identify the given zeros: The zeros provided are [tex]\( 1+i \)[/tex] and [tex]\( 4 \)[/tex].
2. Determine the conjugate zero for complex roots: For polynomials with rational coefficients, complex roots must come in conjugate pairs. Since [tex]\( 1+i \)[/tex] is a zero, its conjugate [tex]\( 1-i \)[/tex] must also be a zero.
Thus, the zeros are [tex]\( 1+i \)[/tex], [tex]\( 1-i \)[/tex], and [tex]\( 4 \)[/tex].
3. Form the factors: Write the polynomial in factored form using its zeros:
[tex]\[ (x - (1 + i)), \quad (x - (1 - i)), \quad \text{and} \quad (x - 4). \][/tex]
4. Expand the polynomial: Calculate the product of these factors step-by-step.
First, multiply the factors involving the complex conjugates:
[tex]\[ (x - (1 + i))(x - (1 - i)) \][/tex]
Use the difference of squares formula [tex]\( (a - b)(a + b) = a^2 - b^2 \)[/tex], where [tex]\( a = x - 1 \)[/tex] and [tex]\( b = i \)[/tex]:
[tex]\[ (x - 1)^2 - i^2 \][/tex]
Note that [tex]\( i^2 = -1 \)[/tex], so:
[tex]\[ (x - 1)^2 - (-1) = (x - 1)^2 + 1 \][/tex]
Expand [tex]\( (x-1)^2 \)[/tex]:
[tex]\[ (x - 1)^2 = x^2 - 2x + 1 \][/tex]
Therefore:
[tex]\[ (x - 1)^2 + 1 = x^2 - 2x + 1 + 1 = x^2 - 2x + 2 \][/tex]
Now we have the intermediate polynomial:
[tex]\[ (x^2 - 2x + 2) \][/tex]
5. Include the remaining root: The factor corresponding to the root [tex]\( 4 \)[/tex] is [tex]\( (x - 4) \)[/tex].
Multiply the intermediate polynomial by this factor:
[tex]\[ (x^2 - 2x + 2)(x - 4) \][/tex]
6. Expand the final expression: Apply the distributive property to expand:
[tex]\[ (x^2 - 2x + 2)(x - 4) = x(x^2 - 2x + 2) - 4(x^2 - 2x + 2) \][/tex]
Perform the multiplications separately:
[tex]\[ x(x^2 - 2x + 2) = x^3 - 2x^2 + 2x \][/tex]
[tex]\[ -4(x^2 - 2x + 2) = -4x^2 + 8x - 8 \][/tex]
Combine the terms:
[tex]\[ x^3 - 2x^2 + 2x - 4x^2 + 8x - 8 = x^3 - 6x^2 + 10x - 8 \][/tex]
Therefore, the polynomial of the lowest degree with rational coefficients that has the given zeros [tex]\( 1+i \)[/tex] and [tex]\( 4 \)[/tex] is:
[tex]\[ x^3 - 6x^2 + 10x - 8 \][/tex]
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