Find all of the zeros: x^5 – 3x^4 – 15x^3 + 45x^2 – 16x + 48 = 0

Question

It will most likely have an imaginary number and two real numbers.

Answer 1

Notice a strange thing about the polynomial? The ratio of the coefficients of the first two terms, the middle two and the last two are all the same: 1:-3. The second term's coefficient is -3 times the first's; the fourth term's coefficient is -3 times the third's; and the last term's coefficient is -3 times the penultimate's. Let's start factorising!

x^4(x-3)-15x^2(x-3)-16(x-3)=0; so (x^4-15x^2-16)(x-3)=0 and we can take it further: (x^2-16)(x^2+1)(x-3)=0, and further still: (x-4)(x+4)(x-i)(x+i)(x-3)=0. So the roots are -4, 3, 4, i, -i. Two imaginary roots, just like you thought!