Question

Find a polynomial with real coefficients that has the given zeros.

4+ 3i, 4-3i, - 1
One such polynomial P(x) can be defined as P(x) = x - 7x² +___x+25
What does this even mean?

Answer 1

To find the polynomial with given zeros 4+3i, 4-3i, and -1, we use the fact that complex zeros appear in conjugate pairs and multiply the factors (x - (4+3i))(x - (4-3i))(x + 1). Expanding these factors yields the polynomial P(x) = x^3 - 7x^2 + 18x + 25.

Step-by-step explanation:

To find a polynomial with real coefficients that has the given zeros 4+3i, 4-3i, and -1, one can use the fact that complex roots in polynomials with real coefficients come in conjugate pairs. Knowing this, the factors of the polynomial can be written down and then expanded to find the polynomial in standard form.

For the zeros 4+3i and 4-3i, we can write:

• (x - (4+3i))(x - (4-3i))

For the zero -1, we just have:

• (x - (-1)) or (x + 1)

Multiplying these factors:

• ((x - 4) - 3i)((x - 4) + 3i)(x + 1)

Expands to:

• ((x - 4)^2 - (3i)^2)(x + 1)
• ((x^2 - 8x + 16) - (-9))(x + 1)
• ((x^2 - 8x + 25)(x + 1)

When fully expanded, this will give a polynomial with an x term that fills in the blank in the original question P(x) = x - 7x² + ___x + 25. The correct term is found through the expansion process, which results in 18x.

Therefore, the polynomial with the given zeros is:

• P(x) = x^3 - 7x^2 + 18x + 25