Finding roots of cubic equation by factorisation

Question

I want to find solutions of a cubic equation which is in $x$. I know a cubic equation will have three roots. Consider this cubic equation

$(a-xb)[(c-xd)(e-fx)-g]=0$

Here $a,b,c,d,e,f,g$ are constants. You can see that above cubic equation have a factor of $(a-xb)$ as a multiple of other term i.e. $[(c-xd)(e-fx)-g]$

Now can I equate both terms separately to zero i.e.

$(a-xb) =0$

$[(c-xd)(e-fx)-g]=0$

Now I can solve for $(a-xb)=0$ for x. I will find one root from this equation and for other two roots i can solve the quadratic equation $[(c-xd)(e-fx)-g]=0$.

Can I do that? Will than the roots that I obtained will be the solution of original cubic equation i.e.$(a-xb)[(c-xd)(e-fx)-g]=0$?

Answer 1

Over the real numbers if $x\cdot y=0$ then either $x$ or $y$ or both must be $0$. The reals are a Field. That means there are two operations, multiplication and addition where both multiplication and division follow the usual requirements for a group (1: the binary operation of two members of the group yields a member of the group, 2: the operation is associative, 3: there's an identity element so that xe=ex=x, 4: every element has an inverse so that $xx^{-1}=e$ ), further they are both commutative, $x+y=y+x, xy=yx$. Finally, it has the property that if the product of two reals is 0, then at least one of the factors is 0, this isn't always the case. Thus, Fields are also called Commutative Division Rings.

So you in your problem you can break it up into $(Ax+B)(Cx^2+Dx+E)=0$

So either $x=-B/A$ or $x=\frac{-D \pm \sqrt{D^2-4CE}}{2C}$

This is expected, a cubic should have 3 solutions up to multiplicity, though they might not all be real.

Answer 2

Logically, it makes sense that if the product of several terms is 0, then at least one of those terms has to be 0. So the solution could be found by simply equating each of the factors to 0, like you have done.