We use cookies to personalise content and advertisements and to analyse access to our website. Furthermore, our partners for online advertising receive pseudonymised information about your use of our website. cookie policy and privacy policy.

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

 May 13, 2019

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 smiley)


Hope this helped!

 May 16, 2019

8 Online Users