You cannot get rid of it (unless dividing everything by but then it gets tricky). But you can still factorize this equation :
But I agree, it is not always obvious to factorize such equations. Have you learnt the quadratic formula yet ? It always works and even tells you when no factorization is possible instead of letting you struggle with an unsolvable problem in R. Here is a spoiler, if you want to know about it :
Spoiler:
Quadratic formula (I won't put the details, they are found on Wiki) :
Say :
You can define a "discriminant" (to know where it comes from, wikipedia) :
If this discriminant is negative, then the equation has no solution in R and cannot be factorized (you'll see later it has solutions, though).
If this discriminant is zero, then the equation has one root (solution). In fact, it has two, but they are the same. This solution is :
If this discriminant is positive, then the equation has two distinct roots :
you cannot get rid of the 2, unless you divided the whole equation by 2 which would make the quadratic more difficult. simply factorise the equation:
2x2-3x-5=0
(2x-5)(x+1)=0
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Quadratic eq., solve for x when first variable is not 1 
Linear Mode
