Solve for $m^3 + m - 10 = 0$ By the theorem of rational roots, I know that the only rational root is $m = 2$. But I want to solve by factoring too, from Wolfram I know that $(m - 2)$ and $(m^2 + 2m + 5)$ are the factors, but I don't know how to find out these factors. Knowing the factors, I think its something to do with difference of cubes, but I was unnable to relate to it, since the equation have three terms instead of the two from the diference of cube formula. Can you explain me how to find these factors? Thanks in advance.
 A: By Vieta's formulas the three roots of the cubic $m^3 + m - 10 = 0$ have sum $0$ and product $10$. Once you found that one root is $2$, that leaves two other roots with sum $0 - 2 = \color{red}{-2}$ and product $10 / 2 = \color{blue}{5}$, so,  again by Vieta's formulas, the quadratic factor is $m^2 \color{red}{+ 2}m \color{blue}{+ 5}$.
A: If you know $m=2$ is a root, that means $(m-2)$ is a factor of $m^3+m-10$. So you have two ways to find the remaining factor.

*

*let $m^3+m-10=(m-2)(m^2+am+b)$ Expand the RHS and compare coefficients to find $a$ and $b$.

*Use long division for polynomials to find it. Here is a link maybe helps.

A: Once you know that $m = 2$ is a root, you can take advantage of such fact to obtain the desired result:
\begin{align*}
m^{3} + m - 10 & = (m^{3} - 4m) + (5m - 10)\\\\
& = m(m^{2} - 4) + 5(m - 2)\\\\
& = m(m + 2)(m - 2) + 5(m - 2)\\\\
& = [m(m + 2) + 5](m - 2)\\\\
& = (m^{2} + 2m + 5)(m - 2)
\end{align*}
Hopefully this helps!
A: You can you the following rule maybe better known outside Italy as Horner's Rule:
$$
\begin{array}{c|ccc|c}
 & 1 &0&1 & -10 \\
2 & & 2 & 4&10 \\
\hline
 & 1 & 2 &5& 0 \\
\end{array}
$$
The numbers $(1,2,5)$ are the coefficients of the second factor: $m^2+2m+5$
A: This is a way to find the other factor; it is certainly not the quickest (many of the methods described in other posts and comments are what people, including myself, would generally do first), but this ties the factorization to other topics your pre-calculus course may cover.  [Please pardon the repetition if these are already familiar to you.]
The fact that this cubic polynomial, $ \ x^3 + x - 10 \ $ has no quadratic term and that the linear coefficient is positive tells us that the function curve has no "turning point".  This is easy to show using calculus, but we can do it using the properties of the polynomial.  If we "shift" the function upward 10 units to make it $ \ x^3 + x \ \ , $ the curve passes through the origin.  If $ \ x \ $ is positive and increasing, both terms are as well, and so the sum is.  This is an odd-symmetry function, so $ \ f(-x) \ = \ -f(x) \ \ ; $ if we take larger and larger negative values of $ \ x \ $ then, this function only becomes more and more negative.  So $ \ x^3 + x \ $ never "doubles back on itself", which means that it only crosses the $ \ x-$axis once; this is the same as saying that  there is only one real number for which this function equals zero $ ( \ x \ = \ 0 \  $ in this case).
That property is not affected by "shifting the curve vertically", so all cubic functions $ \ x^3 + cx + d \ \ $ with $ \ c \ > \ 0 \ $ have only one real zero.  The other two zeroes of the cubic polynomial must then be complex numbers (a cubic polynomial has three zeroes altogether, counting possible duplications [multiplicities > 1] ).  We also learn that if all of the coefficients are real numbers, any complex zeroes must be in "conjugate pairs" $ \ \alpha + \beta·i \ $ and  $ \ \alpha - \beta·i \ \ . $
So the complete factorization of this polynomial is $ \ x^3 + x - 10 $ $ = \ (x - 2) ·( \ x - [\alpha + \beta·i] \ )·( \ x - [\alpha - \beta·i] \ ) \ \ , $ from which we obtain
$$ (x - 2) ·( \ x - [\alpha + \beta·i] \ )·( \ x - [\alpha - \beta·i] \ ) $$ $$ = \ \ (x - 2) ·( \ [x -  \alpha] + \beta·i  \ )·( \ [x -  \alpha] - \beta·i  \ )  $$
$$ = \ \ (x \ - \ 2) · [ \ ( \  x \ - \ \alpha \ )^2 \ -  \ ( \ \beta·i  \ )^2 \ ] \ \ = \ \ (x \ - \ 2) · ( \ x^2 \ - \ 2 \alpha x \ + \ [\alpha^2 + \beta^2] \ )  $$
$$ = \ \  x^3 \ - \ ( \ 2 + 2 \alpha \ )·x^2 \ + \ ( \ [\alpha^2 + \beta^2] + 4 \alpha \ )·x \ - \ 2·[\alpha^2 + \beta^2] \ \ .  $$
Now that all the background theory has been laid out, we can compare the coefficients of the polynomials.  The quadratic coefficient, as we said, is zero, so $ \ 2 + 2 \alpha \ = \ 0 \ \Rightarrow \ \alpha \ = \ -1 \ \ . $  From the linear coefficient, we obtain $ \ [\alpha^2 + \beta^2] + 4 \alpha \ = \ [ \ (-1)^2 + \beta^2] - 4 \ = \ 1 \ \Rightarrow \ \beta^2 \ = \ 4 \ \ . $  This is confirmed by the result for the constant term $ \ -2·[\alpha^2 + \beta^2] \ = \ -2·[ \ (-1)^2 + 4 \ ] \ = \ -10 \ \ . $
We conclude that the "factorization over real numbers" is
$$  x^3  \ +  \ x \ - \  10 \ \ = \ \ (x \ - \ 2) · ( \ x^2 \ - \ 2 · (-1) · x \ + \ [ \ (-1)^2 + 4 \ ] \ ) $$ $$ = \ \ (x \ - \ 2) · ( \ x^2 \ + \ 2x \ + \ 5 \ ) \ \ . \ $$   We have also found along the way that the three zeroes of the polynomial are $ \ 2 \ , \ -1 + 2i \ , \ -1 - 2i \ \ , $ so we could also write the complete factorization over complex numbers.
