Given a degree 5 polynomialg(x) with real coecients that has the leading coecient 2, andhas the zeros5i,1 and 3, and 0.

Answer 1

A degree 5 polynomial means that the highest exponent is 5 (x⁵) and the polynomial has 5 zeros. To solve this type of problem, use the factored form. Since there are 5 zeros, there are 5 factors:

P(x) = a·(x-p)(x-q)(x-r)(x-s)(x-t)

where a is the leading coefficient and p, q, r, s, and t are the roots or zeros. The problem statement tells you that a = 2 and it gives you 4 of the 5 zeros. Since the polynomial has real coefficients, we can use the Conjugate Root Theorem to find the 5th root. The Conjugate Root Theorem is:

For a polynomial with real coefficients, if a+bi is a root, so is a-bi. Conversely, if a-bi is a root, so is a+bi.

So we have 5i as one root, which we can represent as 0 + 5i. That means that 0 - 5i, or just -5i, is also a root. So now we have all five zeros: 5i, -5i, 1, 3, and 0. The polynomial is:

P(x) = 2·(x-5i)(x-(-5i))(x-1)(x-3)(x-0) = 2·(x-5i)(x+5i)(x-1)(x-3)x

You can leave it as it is in factored form or multiply it all out to put it into standard form.