Which function has roots x=4 and x=2−3i ?

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BlackTom AI · Accepted Answer

The question asks which function has roots x = 4 and x = 2 − 3i. 
First, note that for a polynomial with real coefficients, complex roots occur in conjugate pairs. Therefore, if 2 − 3i is a root, 2 + 3i must also be a root. 
Thus the three roots are: 4, 2 − 3i, and 2 + 3i. 
Next, construct the polynomial from its factors: (x − 4) times the quadratic whose roots are 2 − 3i and 2 + 3i. The quadratic is (x − (2 − 3i))(x − (2 + 3i)) = (x − 2 + 3i)(x − 2 − 3i) = (x − 2)^2 + 9 = x^2 − 4x + 4 + 9 = x^2 − 4x + 13. 
Multiply to obtain the cubic: (x − 4)(x^2 − 4x + 13) = x^3 − 4x^2 + 13x − 4x^2 + 16x − 52 = x^3 − 8x^2 + 29x − 52. 
Therefore, the standard monic cubic with those roots is f(x) = x^3 − 8x^2 + 29x − 52. 
The given answer, f(x) = x^3 − 8x^2 + 29x − 52, matches exactly this derivation. 
Since the provided answer options are missing, there is no alternative option to compare here, but the construction above shows why this cubic has the specified roots and why it is correct. 
In summary, the reasoning relies on the conjugate root theorem for real-coefficient polynomials and straightforward expansion of the product of linear factors.

ALGII141-027 4.8 Practice Problems

Which function has roots x=4 and x=2−3i ?

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