Question

x^3-5x^2-2x+10=0

Answer 1

We know that if this has integer roots then they will be factors of 10, so try 1, 2, 5 and 10 (and the negative values of these). 5 works, so we know (x-5) is a factor. Divide the polynomial by (x-5) and we get:
x^3-5x^2-2x+10=(x-5)(x^2-2)=(x-5)(x+sqrt2)(x-sqrt2)
The solutions: x=5, x=sqrt2, x=-sqrt2

Answer 2

Hi to you friend.
Let's first begin by recognizing the problem in question, I assume you want to factor that, in other words, "find the values of x, that makes the whole thing zero".
Now, that is, indeed, a 3rd degree polynomial or a "3rd degree equation".
why?
Because the variable with the highest exponent, has an exponent of three.
Let's refresh that concept, every variable has a very basic structure:
\[(3)x ^{2}_{b}\]
The three we have on the left side is what we call "coeficent", and we call that to every number that multiplies a variable.
Now, a "variable", in notion, is a "changing number" wich is usually represented with a letter or whatever you prefer, as long as it's not a number.
To the top right, we have what we call "exponent", wich you might know in depth what it is some day, the exponent, in equations is the indicator of the "grade" or "degree" of an equation.
To the right bottom, we have what I call the "index" of the variable, wich is used when we work with multiple variables. Not this case.

Moving on with the problem, having refreshed all that, we can now predict what we are in front of, a 3rd degree equality because the "x" with the highest exponent is the x with an exponent of 3.
\[x^3 -5x^2 - 2x +10=0\]
as we can clearly see.
Since it is equal zero, we will ask ourselves "what vale of x will make the whole thing zero?".
In order to do that, we have to apply any technique known to find the solution of these kinds of equalities.
I'll chose by ruffini's diagram.
In other to use ruffinis I will look at the "constant" of the equation.
A "constant" is a special case of a variable, when all the values are the same, in other words, it's just a number.
In case of the problem, it's the "10".
I will take, and I will choose any pair of values whose product makes 10.
here they are: (2,+5), (-2,-5) (1,10) (-1,-10)
So, I can choose any of those numbers, (2,-2,5,-5,10,-10)
I will take the 5 and do the division:

As a recap, The first value goes down, multiplies and goes to the next one and sum, repeat.
Now, since the "residue" of the division gave me "0", it must mean that 5, is a root or "solution" if you prefer, so now we found one of the values, x=5.
But it doesn't stop there, now, since we have found a root, the grade of the equality goes down by one and it looks like this:
\[x^2 - 2=0\]
And I'll let you take over from here.

Answer 3

Tom, your solution made a little more sense. Thanks to both of you, however.

Answer 4

My pleasure, glad you understand it now. Feel free to give me a shout if you get stuck with any others :)