$$2x^3 + 9x^2 +7x -6$$
This equation doesn't factor by grouping, and other than that I have no idea how to solve this problem. Will someone please help?
[Math] How to factor a four term polynomial without grouping
algebra-precalculusfactoring
Related Solutions
HINT
$x^3 - 3x^2 + 4 = x^3 + 1 -3x^2 + 3 = (x + 1)(...) - 3(x^2 - 1) = ...$
The only really general way of which I am aware is to guess at the form of the factorization. Since it is monic (the highest term has coefficient 1), you know that the factors should also be so. Thus, there are really only 2 possible factorizations you need to think of, at least at start, which may then be further reducible through easier methods. If we denote the polynomial by $P(x)$, we produce the following candidate factorization equations:
- one is factorization to a linear term and cubic term, i.e.
$$P(x) = (x + a)(x^3 + b_2 x^2 + b_1 x + b_0)$$
- the other is factorization to two quadratic terms, i.e.
$$P(x) = (x^2 + a_1 x + a_0)(x^2 + b_1 x + b_0)$$
The "obvious" next case of this would simply result in now getting a third-degree polynomial on the left and first on the right, but that's just case 1 thanks to the commutative property, so this is exhaustive. The second case is what you have here. The first case is most easily tested and solved by a simple application of the rational root theorem which will, if it's possible, give the value for $a$ - followed by a polynomial long division to get the rest.
For the second case, there isn't really a much better method than to just multiply it all out:
$$(x^2 + a_1 x + a_0)(x^2 + b_1 x + b_0) = x^4 + c_3 x^3 + c_2 x^2 + c_1 x + c_0$$
where we've introduced for notational cleanliness (I had the computer multiply this out for me because it's there)
$$c_3 := a_1 + b_1$$ $$c_2 := a_0 + a_1 b_1 + b_0$$ $$c_1 := a_1 b_0 + a_0 b_1$$ $$c_0 := a_0 b_0$$
Then you just set the $c_j$ equal to the appropriate coefficient values read off from the terms of the given polynomial (i.e. $c_0 = 50$ in your given example), and try to find whole number values for $a_j$ and $b_j$ that work. You'd probably want to start with $c_3$ and $c_0$ first.
Best Answer
Use the Rational Root Theorem to find possible candidates for roots. For example, $-2$ is a candidate, and it is indeed a zero of your polynomial (the polynomial is equal to zero when $x = -2$), hence $(x - (-2)) = (x + 2)$ is a factor.
Now use synthetic division on the polynomial, call it $p(x)$, to obtain the quotient when divided by $(x + 2)$.
You may simply be able to simply "eye up" the polynomial-quotient, this time with the resulting quadratic equation to $p(x)/(x+2).\;$ If the factorization of the quadratic is readily apparent, then you'll have three factors in all: $(x+ 2)$ and the factors of the quadratic.
Otherwise, use the coefficients of the quadratic and the Rational Root Theorem to find a second zero/root, $x = b$, and use polynomial division to divide the quadratic by the second factor $(x - b)$, to find the resulting quotient, which will be the "last" factor. If all had gone well, (and it should, in this case), you'll now havethree factors in all!