Resolve `Z^5+ 1` into linear & quadratic factors with real coefficients.

If the expression z^(5)-32 can be factorised into linear and quadratic factors over real coefficients as (z^(5)-32)=(z-2)(z^(2)-pz+4)(z^(2)-qz+4) then find the value of (p^(2)+2p)

Find all the quadratic factors of x^(8)+x^(4)+1 with real coefficients.

Find the four roots of the equation z^(4)+16=0 and use them to factorize z^(4)+16 into quadratic factors with real coefficients.

The quadratic equation with real coefficients has one root 2i^(*)s

Introduction and State Quadratic Equation with real coefficients

Statement -1 : There is just one quadratic equation with real coefficient one of whose roots is 1/(sqrt2 +1) and Statement -2 : In a quadratic equation with rational coefficients the irrational roots are in conjugate pairs.

Statement -1 : There is just one quadratic equation with real coefficient one of whose roots is 1/(sqrt2 +1) and Statement -2 : In a quadratic equation with rational coefficients the irrational roots are in conjugate pairs.

For what values of m will the expression 5x^(2)+5y^(2)+4mxy-2x+2y+m be capable of resolution into two linear factors with real coefficients?

Statement -1 : There is just on quadratic equation with real coefficients, one of whose roots is 1/(3+sqrt7) and Statement -2 : In a quadratic equation with rational coefficients the irrational roots occur in pair.

Statement -1 : There is just on quadratic equation with real coefficients, one of whose roots is 1/(3+sqrt7) and Statement -2 : In a quadratic equation with rational coefficients the irrational roots occur in pair.