Question

(x+2)(x-3)(x+4)(x-6)= 6x²

Answer 1

$(x+2)(x-3)(x+4)(x-6)= 6x^2\\ \\(x^2-x-6)(x^2-2x-24)=6x^2\\ \\x^4-2x^3-24x^2-x^3+2x^2+24x-6x^2+12x+144=6x^2\\ \\x^4-3x^3-34x^2+36x+144 =0$

Factors of 144/1 are the possible rational factors.  They are
    +- of 1,2,3,4,6,8,9,12,16,18,24,36,48,72,144

On trying them, the polynomial equation is not satisfied.  So the roots are irrational or imaginary.   We try factorizing as follows:

$x^4-5x^3-12x^2+2x^3-22x^2+36x+144=0\\ \\x^2(x^2-5x-12)+2x^3-10x^2-12x^2-24x+60x+144=0\\ \\x^2(x^2-5x-12)+2x(x^2-5x-12)-12x^2+60x+144=0\\ \\x^2(x^2-5x-12)+2x(x^2-5x-12)-12(x^2-5x-12)=0\\ \\(x^2-5x-12)(x^2+2x-12)=0$

Solving each quadratic equation separately we get:

$x = \frac{5+-\sqrt{25+48}}{2}=\frac{5+-\sqrt{73}}{2}\\ \\.\ \ \ \ \ and\ \ \ \ x=\frac{-2+-\sqrt{4+48}}{2}=-1+-\sqrt{13}$

So given polynomial equation has four irrational roots.