Factor completely over the set of polynomials with integer coefficients: \(4(x + 5)(x + 6)(x + 10)(x + 12) - 3x^2.\)

Guest May 13, 2019

#1**+2 **

Lets start by multiplying it all out.

\(4(x+5)(x+6)(x+10)(x+12) - 3x^2 = 4(x^4+[5+6+10+12]x^3+[5*6+5*10+5*12+6*10+6*12+10*12]x^{2}+[5*6*10+5*6*12+5*10*12+6*10*12]x+[5*6*10*12]\)(Sorry it's so long, I wanted to be complete.)

That simplifies to the more manageable, \(4(x^4+33x^3+200x^2+1980x+3600)-3x^2=4x^4+132x^3+800x^2-3x^2+7920x+14400=4x^4+132x^3+797x^2+7920x+14400\)

As for factoring it... you need to use a calculator (maybe this one?) Let's see how:

solve(4x^4+132x^3+797x^2+7920x+14400=0) = __{x=-((sqrt(33*sqrt(193)+(901/4))/2))-((sqrt(193)/2))-((33/4)), x=(sqrt(33*sqrt(193)+(901/4))/2)-((sqrt(193)/2))-((33/4)), x=(sqrt(193)/2)-((sqrt((901/4)-(33*sqrt(193)))/2))-((33/4)), x=(sqrt(193)/2)+(sqrt((901/4)-(33*sqrt(193)))/2)-((33/4))}__

That's your answer if you want the roots, but honestly, I doubt whoever gave you this problem was looking for this sort of answer (at least, I hope not )

Hope this helped!

helperid1839321 May 16, 2019