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Which polynomial function has a leading coefficient of 1 and roots (7 + 1) and (5 - i) with multiplicity 1? 0 f(x) = (x + 7)(x-i)(× + 5)(× + 1) 0 f(x) = (x - 7)(x-1)(x -5)(x+ 1) •fX)=(メー(7ー0)(メー(5+1)(Xー(7+1)(メー(5-1)) • f(x) = (x + (7 - i))(x + (5 + i))(x+ (7 + 1)(×+ (5 - 1))

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Final answer:

The correct polynomial function is f(x) = (x - 8)((x - 5)^2 + 1), constructed using the given roots and satisfying the conditions of having a leading coefficient of 1 and the specified roots with multiplicity 1.

Step-by-step explanation:

The student is asking for a polynomial function with a leading coefficient of 1 and given roots. Given the roots (7 + 1) and (5 - i) with multiplicity 1, we must construct a polynomial using these roots. Since complex roots come in conjugate pairs, the root (5 - i) implies that there is also a root (5 + i).

The polynomial function with a leading coefficient of 1 and roots (7 + 1) and (5 - i) with multiplicity 1 is:

f(x) = (x + 7)(x - i)(x + 5)(x + 1)

Therefore, the polynomial with the stated properties is f(x) = (x - (7 + 1))(x - (5 + i))(x - (5 - i)).

After simplifying, this expands to f(x) = (x - 8)(x - 5 - i)(x - 5 + i). Multiplying the complex factors, we get (x - 5)^2 + 1 (since (a - bi)(a + bi) = a^2 + b^2), and the polynomial becomes f(x) = (x - 8)((x - 5)^2 + 1).

User Andreas Hagen
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