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X's possible solutions are -1/2, 0, and 3.
Howd you get that.?
I would start by subtracting 2 from each side, to give \[2x^3-5x^2-3x=0\] Now we start factoring: there's a common x in all 3 terms: \[x(2x^2-5x-3)=0\] so x = 0 is one solution
Now you can factor \[2x^2-5x-3\] or use the quadratic formula to get the other two.
Subtract 2 from both sides, you will be left with- \[2 x^3 - 5 x^2 -3x =0\] Now take x common from it, and factor the remaining...
To factor, we'll have something like (2x+a)(x+b) for appropriate values of a and b, giving: (2x+a)(x+b) = 2x^2 + ax + 2bx + ab = 2x^2 + (a+2b)x + ab To make that the same as 2x^2-5x-3, we need to find a and b such that ab = -3 and (a+2b)=-5. a moment of thought should tell you that a=1, b=-3 so our remaining terms factor to (2x+1)(x-3) and the whole thing is \[x(2x+1)(x-3)=0\] with solutions at x=0, x=-1/2, x=3
(the solutions simply being the values of x that make the product terms = 0)
Soo.. is the answer 0 or all of the above O.o
All 3 of those values will make the polynomial = 0. x=0: \[2(0)^3-5(0)^2-3(0) = 0\] x=3: \[2(3)^3-5(3)^2-3(3) = 54-45-9 = 0\] x=-1/2: check this one yourself :-)
In general, you'll have as many solutions as the power of the highest order term in the polynomial. We have an x^3, so there are 3 solutions. Some of them may be complex numbers, but in this case, they were all real numbers.
I remember now(: Thanks that helped a lot!
It's just like riding a bicycle, right? :-)
A bicycle that tries to grab your pants in the chain send you crashing to the ground! :-)
Except all that you have to remember is to pedal and keep balance.. Algebra you have to remember a lot more and youll never use it haha
Lol yes that is very true!
Well, I suppose it depends on what path you take through life, but I encounter algebra quite often — it just doesn't always say "ALGEBRA" in big red letters on the cover. Now, finding the roots of cubic polynomials might be something that you won't do very often, agreed...but doesn't it feel good to be able to look at something that so many people around you are likely to say "oh, that's hard, I can't do that" and think "I can do that!"?
Here's a secret: a lot of it is simply not believing that you can't do it! The math they teach before calculus really isn't that difficult; becoming proficient at it just requires the right explanation, and practice. Unfortunately, not every teacher can supply the right explanation for every student, and the practice takes time which some aren't willing to devote.
Yes! It is awesome knowing I can do something most older people cant even do! But, I don't think I'll be encountering any of this in the area I'm going into except for maybe the basics My last school the kids didn't even try to learn they just starred at it as if it were a foreign language and it could have been the simplest thing ever