Question

2t^3=5t-11t^2

Answer 1

I brought everything over so... 2t^3-5t+11t^2
Then factored out a t so.. t(2t^2-5+11t)
then i cant factor..so im stuck

Answer 2

@pita0001 .. first of all when we take all the terms to single side..
then we are supposed to change the signs of those terms whose position is being changed..
and since we take all the terms to a single side therefore u need to equate the equation with 0

Answer 3

I did change the signs..it started off as being 2t^3=5t-11t^2
And when I brought it over it became 2t^3-5t+11t^2
And then to set it to zero.. 2t^3-5t+11t^2=0

Did I do something wrong so far?

Answer 4

No... u didn't..
now...divide both the sides with "t"

Answer 5

and you would get a quadratic equation...

Answer 6

2t^2-5+11t = 0
now try making its factors...:D

Answer 7

Then wouldn't I multiply 11*2? And thats 22
And 22 and 5 dont factor.. still lost

Answer 8

@gudden

In Math it is wrong to divide both sides of an equation by a variable.
If you do that, you lose a zero or a solution of that equation.

Answer 9

@pita0001

What you did was right in the beginning.

"Then factored out a t so.. t(2t^2-5+11t) = 0"

Let's continue from there! :D

You should use the Quadratic Equation Formula to solve the quadratic equation you get in the bracket.

And then you'll get it in the factored form.

Got it? :)

Answer 10

Akash...
actually in this equation i said divide it by t...cuz, t was coming out to be common...
so i said it..
and if u would divide any equation by a variable...u still donot loose any of the zeros...

Answer 11

@gudden
Even your method is correct.
It is obvious that t = 0 is one of the solutions of the given equation.
So we can divide the equation with 't' to find other (non-zero) solutions.

Answer 12

@gudden

This polynomial given is cubic, so it must have 3 zeroes.
If you divide everything by 't' you love that one 't' thus giving you a quadratic polynomial which has only 2 zeroes. Which is not needed.
We need 3 zeroes and if you divide by 't' one zero which, as a matter of fact, is ZERO! :D, goes away.

Thus, in mathematics [excluding limits and derivatives] we cannot divide f(x) = 0 by x.

Answer 13

5+- √ 5^2-4(2)(11)/2(2)
which equals 5+- √ 25-4(2)(11)/2(2)
so 5+- √ -63/4

Answer 14

@akashd..
if u see in this question there is clearly no constant term... therefore it would not be wrong to divide it with a linear variable of power 1...
cuz its simply giving the answer to be the value of that variable as 0

Answer 15

I would refrain from doing something like that, and it is needless to say I do not want to argue about this anymore.
I think, I'd still prefer not dividing by a variable [Causing a loss of a zero!]

2t^3=5t-11t^2
2t^3+11t^2-5t = 0
t(2t^2 + 11t - 5) = 0

**[Attached are the roots]**

Finally,
A cubic polynomial will give 3 zeroes, 2 of which we get from the Quadratic polynomial and the other from the lone variable 't'. :D

Now if the question was to factorize which it may not even be [because it is an equation] I suppose you could divide the terms by a 't'. But I'll still not do it, because that does not show complete factorization!

:D

Answer 16

Auh..finally..the white flag..
Can's say I didn't enjoy this thread :P

Answer 17

@LastDayWork ... u had a great time :p