QuestionJuly 15, 2025

Find a cubic polynomial in standard form with real coefficients having the given zeros. 0 and 4+5i P(x)= square (Simplify your answer.)

Find a cubic polynomial in standard form with real coefficients having the given zeros. 0 and 4+5i P(x)= square (Simplify your answer.)
Find a cubic polynomial in standard form with real coefficients having the given zeros. 0 and 4+5i P(x)= square (Simplify your answer.)

Solution
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Answer

P(x) = x^3 - 8x^2 + 41x Explanation 1. Identify all zeros Given zeros are 0 and 4+5i. Since coefficients are real, the conjugate 4-5i is also a zero. 2. Write polynomial from zeros The polynomial can be expressed as P(x) = (x - 0)(x - (4+5i))(x - (4-5i)). 3. Simplify complex factors Multiply (x - (4+5i))(x - (4-5i)) using the formula for difference of squares: **(a-b)(a+b) = a^2 - b^2**. Here, a = x - 4 and b = 5i. (x - 4)^2 - (5i)^2 = (x^2 - 8x + 16) - (-25) = x^2 - 8x + 41. 4. Expand the polynomial Now multiply by the remaining factor: x(x^2 - 8x + 41) = x^3 - 8x^2 + 41x.

Explanation

1. Identify all zeros<br /> Given zeros are $0$ and $4+5i$. Since coefficients are real, the conjugate $4-5i$ is also a zero.<br /><br />2. Write polynomial from zeros<br /> The polynomial can be expressed as $P(x) = (x - 0)(x - (4+5i))(x - (4-5i))$.<br /><br />3. Simplify complex factors<br /> Multiply $(x - (4+5i))(x - (4-5i))$ using the formula for difference of squares: **$(a-b)(a+b) = a^2 - b^2$**. Here, $a = x - 4$ and $b = 5i$.<br /> $(x - 4)^2 - (5i)^2 = (x^2 - 8x + 16) - (-25) = x^2 - 8x + 41$.<br /><br />4. Expand the polynomial<br /> Now multiply by the remaining factor: $x(x^2 - 8x + 41) = x^3 - 8x^2 + 41x$.
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