<s>[INST] How to solve for the zeros of $3x^4 - 7x^3 - 12x^2 + 12x + 11$? I’ve tried to see if a rational factor would work from $q/p$, but that hasn’t worked.
I tried grouping and then factoring but that does not work either.
I do not know any further methods that I could use since it doesn’t seem like I could use the quadratic formula or completing the square.
 [/INST] Hints: You can convert the equation such that factorization becomes easier:
$3x^4-7x^3-12x^2+12x+11=0$
rewrite as:
$x^4-\frac 73 x^3-\frac 4x^2+4x+\frac {11}3=0$
Let y=kx :
$y^4-\frac 73 \cdot \frac{y^3}{k^3}-4\cdot \frac{y^2}{k^2}+4\cdot\frac yk +\frac{11}3=0$
let $k=3$:
$y^4-7y^3-36y^2+108y+297=0$
Now you can factor and tranform equation to:
$y^2(y^2-7y-18)-18y^2+108y+297=0$
$y^2(y-9)(y+2)-9[2(y-9)(y+3)+21]=0$
$(y-9)[y^2(y+2)-18(y+3)]=1\times 9\times 21$
Now equate factors on both sides; for example:
$y-9=1\Rightarrow y=10\rightarrow x=\frac{10}3=3.3$
$y^2(y+2)-18(y+3)= 9\times 21$
apply similar method to find more solutions.
Wolfram gives $x=-1.345, -0.69, 1.28, 3.08$.
</s>