How to solve the equation $(25{ x }^{ 2 }-1)(10x+1)(2x+1)=11$?

Question

How to solve this equation? $$(25{ x }^{ 2 }-1)(10x+1)(2x+1)=11$$

Answer 1

HINT:

$$(25x^2-1)(10x+1)(2x+1)=11\Longleftrightarrow$$ $$(2x+1)(10x+1)(25x^2-1)=11\Longleftrightarrow$$ $$500x^4+300x^3+5x^2-12x-1=11\Longleftrightarrow$$ $$500x^4+300x^3+5x^2-12x-12=0\Longleftrightarrow$$ $$\left(10x^2+3x-2\right)\left(50x^2+15x+6\right)=0\Longleftrightarrow$$ $$10x^2+3x-2=0\space\space\vee\space\space 50x^2+15x+6=0$$

Use the ABC-formula, and you'll get the answer!

Answer 2

If we expand this we get $-1 -12x + 5x^2 + 300x^3 + 500x^4 = 11$ or $-12 -12x + 5x^2 + 300x^3 + 500x^4 = 0$. We can then factor that into the following $(-2 + 3x + 10x^2) (6 + 15x + 50x^2) = 0$. This we then see that we have two second degree polynomials multiplied together equal to 0. This means that we can get 4 (some possibly imaginary) values using the quadratic formula. We get the solutions $x = -3/20 - \sqrt{89}/20$ or $x = \sqrt{89}/20 - 3/20$, with the other two solutions being imaginary.

Note: I did my expanding and factoring using wolframalpha.com to save time.