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If the polynomial has an even degree, meaning that the highest power of the variable (for example: x) is an even number (for example: x^4), then all the zeroes could be complex. If odd, there must be at least one real zero.

A complex zero is given by the complex expression: a+ib, where a and b are real, and i=sqrt(-1). If the polynomial has real coefficients (all the numbers in at are, or represent, real numbers), as is usually the case, then the imaginary part of the complex zeroes must combine to make a real number. For example, if one complex zero is identified as a+ib, then there will be another zero a-ib such that (x-a-ib)(x-a+ib)=x^2-2ax+a^2+b^2. (Note that, although a and b are real, they are not necessarily rational numbers.) There are no imaginary components in this expansion.

Example: x^4+x^3-x-1 is degree 4 polynomial. It has real zeroes 1 and -1 so that it factorises to (x-1)(x+1)(x^2+x+1).

The zeroes of x^2+x+1 are found by setting the quadratic to zero and solving using the formula; so x=(-1+sqrt(1-4))/2=(-1+sqrt(-3))/2=(-1+isqrt(3))/2. The two complex roots are -1/2-isqrt(3)/2 and -1/2+isqrt(3)/2. In this case a=-1/2 and b=sqrt(3)/2. Note that a^2+b^2=1.

Example: you're told that a degree 5 polynomial has a real zero 1, and two complex zeroes 1+i and -2-i. Find all zeroes and the polynomial. The two missing zeroes will be 1-i and -2+i.

The polynomial is (x-1)(x-1-i)(x-1+i)(x+2+i)(x+2-i)=(x-1)(x^2-2x+2)(x^2+4x+5)=x^5+x^4-3x^3-x^2+12x-10.

Strictly speaking, the polynomial is a(x^5+x^4-3x^3-x^2+12x-10), where a is a real number, because multiplying a polynomial by a constant doesn't affect its zeroes.

by Top Rated User (711k points)

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