Answers

2014-10-19T04:32:47+05:30

This Is a Certified Answer

×
Certified answers contain reliable, trustworthy information vouched for by a hand-picked team of experts. Brainly has millions of high quality answers, all of them carefully moderated by our most trusted community members, but certified answers are the finest of the finest.
(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.

1 5 1
I hope it is clear enough. This does not seem to be in secondary school syllabus. which class is it?