algebra-precalculuscubicsfactoringpolynomials
Trying to figure this one out but I see no logical approach to this at all.
$$x^3-3x^2-4x+12$$
I know that it will be 3 parts most likely and that each will start with x but beyond that I will just guess at evrything most likely. How do I factor something weird like this?
With the polynomial $x^4 -7x^3 + 3x^2 + 31x + 20$ where $(x+1)$ is a factor for example.
Do I use the same procedure but start a degree higher, like $(x+1)(\alpha x^3 + \beta x^2 + \gamma x + \delta)$ and move on down to figure out the 2nd degree after that?
There are other techniques, namely the polynomial long division as mentioned in a comment by pedja, the Ruffini's rule and for polynomials with integer coefficients the rational root theorem, but you can apply the method of equating coefficients to
$$\begin{eqnarray*} P(x) &:&=x^{4}-7x^{3}+3x^{2}+31x+20 =\left( x+1\right) \left( \alpha x^{3}+\beta x^{2}+\gamma x+\delta \right). \end{eqnarray*}\;\; \tag{1}$$
Expanding the RHS and comparing with the LHS, where the coefficient of $x^{4}$ is $1$, you would conclude that $\alpha =1$ and are left with a simpler equality
$$\begin{eqnarray*} P(x) &:&=x^{4}-7x^{3}+3x^{2}+31x+20=\left( x+1\right) \left( x^{3}+\beta x^{2}+\gamma x+\delta \right) \\ &=&x^{4}+\left( \beta +1\right) x^{3}+\left( \beta +\gamma \right) x^{2}+\left( \gamma +\delta \right) x+\delta . \end{eqnarray*}$$ Equating coefficients you get the simple system of four linear equations $$ \begin{equation*} \left\{ \begin{array}{c} \delta =20 \\ \gamma +\delta =31 \\ \beta +\gamma =3 \\ \beta +1=-7, \end{array} \right. \end{equation*}$$ whose solution is $\delta =20,\beta =-8,\gamma =11$. So
$$\begin{equation*} x^{4}-7x^{3}+3x^{2}+31x+20=\left( x+1\right) \left( x^{3}-8x^{2}+11x+20\right). \end{equation*}\tag{2}$$ Now by direct inspection (or by the rational root theorem, since $1,2,4$ and $5$ are the positive divisors of $20$) you can find that $x=-1$ is a zero of the cubic polynomial $Q(x):=x^{3}-8x^{2}+11x+20$, ie $Q(-1)=0$. Applying a similar method to $Q(x)$ you would find that
$$\begin{equation*} Q(x)=x^{3}-8x^{2}+11x+20=\left( x+1\right) \left( x-4\right) \left( x-5\right) . \end{equation*}$$
Consequently, $$\begin{equation*} x^{4}-7x^{3}+3x^{2}+31x+20=\left( x+1\right) ^{2}\left( x-4\right) \left( x-5\right). \end{equation*}\tag{3}$$
A shorter method is the Ruffini's rule applied to the polynomial division of $P(x)$ by $(x-r)$. This case (with $P(x)=x^{4}-7x^{3}+3x^{2}+31x+20$ and $r=-1,P(r)=0$, ) is shown bellow, where the remainder is $P(r)=s$.
While the expression cannot be factored (it cannot be represented strictly as the product of factors, we can proceed to manipulate it in a way that involves two factors, with an added (subtracted) constant: $$ \begin{align} 4x^2 + 4x - 9y^2 - 1 &= 4x^2 + 4x + \color{red}{1} - 9y^2 - 1 \color{red}{-1} \\ &= (2x+1)^2 - (9y^2 + 2) \\ &= (2x+1)^2 - (3y)^2 -2 \\ &= (2x + 1 +3y)(2x + 1 - 3y) -2 \\ &= (2x + 3y +1)(2x - 3y + 1) -2 \;? \end{align} $$
I'm "simply" simplifying the expression you gave (which is not an equation). If you meant $$4x^2 + 4x - 9y^2 - 1 = 0$$ you could write:
$$4x^2 + 4x - 9y^2 - 1=0 $$ $$ \iff (2x+1)^2 - (3y)^2 = 2 $$ $$\iff (2x + 1 +3y)(2x + 1 - 3y) = 2 $$ $$\iff (2x + 3y +1)(2x - 3y + 1) = 2$$
Best Answer
HINT $\rm\quad f(x)\ =\ x^3 - a\ x^2 - b\ x + a\:b\: \ =\: \ x^2\ (x- a) - b\ (x - a)\ =\ \cdots$
Alternatively, by the Rational Root Test, the only possible rational roots are integer factors of $\rm\:a\:b\:.\:$ But clearly $\rm\ x = a\ $ is a root since it makes the first and last pair of terms cancel out. Therefore, since $\rm\:f(a) = 0\:$ we deduce that $\rm\:f(x)\:$ has the factor $\rm\:x-a\:$ by the Factor Theorem.
For more efficient polynomial factorization algorithms see my post here and the following survey.
Kaltofen, E. Factorization of Polynomials, pp. 95-113 in:
Computer Algebra, B. Buchberger, R. Loos, G. Collins, editors, Vienna, Austria, 1982.