Question

Give a third degree polynomial that has zeros of 13, 5i, and -5i, and has a value of -680 when x=3. Write the polynomial in standard form ax³ + bx² + cx + d.

Answer 1

A third degree polynomial with given zeros 13, 5i, and -5i is constructed using factors and the additional information that the polynomial equals -680 when x=3. The resulting polynomial in standard form is 2x^3 - 26x^2 + 50x - 650.

Step-by-step explanation:

The student is asking for a third degree polynomial with specific zeros and a value at a certain point. To construct such a polynomial, we can use the zeros 13, 5i, and -5i to create factors of the polynomial:

(x - 13)(x - 5i)(x + 5i)

Since complex roots occur in conjugate pairs, the polynomial will have real coefficients. Multiplying the factors involving the imaginary zeros gives:

(x - 13)(x^2 + (5i)^2) = (x - 13)(x^2 - 25)

Expanding this and multiplying by the leading coefficient 'a' gives us:

a(x^3 - 13x^2 + 25x - 325)

To find the value of 'a', we use the fact that the polynomial has a value of -680 when x=3:

a(3^3 - 13(3)^2 + 25(3) - 325) = -680

This simplifies to 27a - 117a + 75a - 325a = -680

-340a = -680, so a = 2.

Therefore, the polynomial in standard form is:

2x^3 - 26x^2 + 50x - 650