X^5 3x^4-3x^3-9x^2-10x-30

Question

problem says state the number of complex zeros, the possible number of real and imaginary zeros, and the possible rational zeros for each function.then find all zeros.

Answer 1

x⁵+3x⁴-3x³-9x²-10x-30 looks like it can be factorised:

x⁴(x+3)-3x²(x+3)-10(x+3)=(x⁴-3x²-10)(x+3)=

(x²-5)(x²+2)(x+3)=

(x-√5)(x+√5)(x-i√2)(x+i√2)(x+3), from which all zeroes (real and complex) can be seen.

A degree 5 polynomial must have at least one real zero (because 5 is an odd number and complex roots come in pairs). The possibilities for the remaining zeroes are:

4 real zeroes;

2 real zeroes, and 2 complex zeroes (conjugates);

4 complex zeroes, (2 pairs of conjugate zeroes).

Rational zeroes come from the factors of 30 = (1, 2, 3, 5). However, having factorised the polynomial, the zeroes are: 

√5, -√5, i√2, -i√2, -3. So there are 3 real zeroes and 2 complex zeroes. Amongst the real zeroes there is a pair of irrational zeroes. Only one rational zero (-3) appears in the list of actual zeroes. The other zeroes contain an irrational number (√2 or √5, and 2 and 5 are two of the rational zeroes).