Find a polynomial with real coefficients that has the given zeros. -1 and 5-2i One such polynomial P(x) can be defined as P(x) = x³-9x²+x+29.

Question

asked Jun 6, 2024 203k views

1 Answer

Ramya · Answer 1 · 2024-06-09T16:53:20+0000

Step-by-step explanation

The polynomial we are given is

P(x) = x^3 - 9x^2 + kx + 29

where we'll replace k with an actual number later.

If x = -1 is a root then P(-1) = 0 must be the case.

We'll use this to determine k.

$P(x) = x^3 - 9x^2 + kx + 29\\\\P(-1) = (-1)^3 - 9(-1)^2 + k(-1) + 29\\\\0 = -1 - 9(1) - k + 29\\\\0 = 19 - k\\\\k = 19\\\\$

Therefore,

P(x) = x^3 - 9x^2 + 19x + 29

will have the roots x = -1, x = 5-2i and x = 5+2i.

Recall that complex roots come in conjugate pairs. This happens if and only if each of the polynomial's coefficients are real numbers.