#when a equation has x(...)= 0, can u divide the 0 by x, or will the x always be a solution

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limber merlin
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i'm not sure if the heading makes sense but i have a sample question that might make it clearer

$$x^4+ax^3+2x^2=0$$, if a is a real constant, when will the equation have one unique solution?

can you factorise out the x^2 and then divide 0 with x^2? or would you always use the null factor law when you carry out operations with x?

dense vectorBOT
subtle saddleBOT
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nashipear
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dense vectorBOT
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limber merlin
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when a equation has x(...)= 0, can u divide the 0 by x, or will the x always be a solution

chilly current
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If you have some polynomial $p_N(x)=0$
Where you can factor out $x^k$ to yield you
$x^kp_{N-k}(x)=0$ that yields the constraint that $x^k=0$ or $p_{N-k}(x)=0$

subtle saddleBOT
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Lumberdude #LeaveWolfAsHeIs

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chilly current
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You can only divide by x provided that $x\neq 0$, which should be the case if you factored out all x.