# Solve $(4x+3)^2(2x+1)(x+1)=75$?

How do I find in set of complex numbers the solutions of the following equation?

$$(4x+3)^2(2x+1)(x+1)=75$$ I hope you'll give me just a hint.

Thank you very much!

• x=1/2 is one solution. 25X3 – evil999man May 6 '14 at 12:45

$$(4x+3)^2(2x+1)(x+1)=75$$ $$(4x+3)^2(4x+2)(4x+4)=75\times 8$$ $$(16x^2+24x+9)(16x^2+24x+8)=600$$ let $$16x^2+24x+8=t$$ then $$t(t+1)=600$$ then it is easy to solve it
We note that $75=5^2\times 3$
Hence, we try to see whether we can be lucky to have integer solutions. Setiing $4x+3=\pm5$ gives $2$ values for $x$,i.e. $x=1/2,-2$ for which above equation is satisfied.
Now you can divide to get a quadratic and get other $2$ roots.