Polynomials

Question

A polynomial P(x) of degree 5 with real coefficients has 7 as a zero of multiplicity 3 and 1+2i as a complex zero. Find all zeros of P(x) and give a possible form of the polynomial.

Answer 1

Hi Naomi,

 

P(x) is of degree 5,therefore it must have 5 roots. Since P(x) has real coefficients, the remaining root must be 1-2i according to the Complex Conjugates Theorem.

Now, we can write P(x) as below: 

P(x) = (x - 7)^3*Q(x)

Q(x) is of degree 2 with the roots 1 + 2i and 1 - 2i because P(x) is of degree 5.

Q(x) = x^2 - (sum of the roots)x + product of the roots

= x^2 - (1-2i+(1+2i))x + ((1-2i)(1+2i))

= x^2 -2x + 5

Then, 

P(x) = (x - 7)³(x² - 2x + 5)