How to fully factorise $2x^4+7x^3+4x^2-4x$?
I'm struggling to factorise polynomials like this one. I'm not sure how to best approach this problem. I've tried using the remainder and factor theorems but they seem to work only when at least one factor is known. Since I only know the polynomial itself, I'm struggling to efficiently solve these types of questions. Any help would be greatly appreciated
Hint:
First factor out $x$: $$2x^4+7x^3+4x^2-4x=x(2x^3+7x^2+4x-4)$$ Next, you can factor the cubic polynomial using the Rational roots theorem. After you've found a root $\alpha$, divide by $x-\alpha$. The quotient will be a standard quadratic polynomial, of which the roots are easy to determine.
Taking out a factor of $x$, we get $$ x(2x^3+7x^2+4x-4) \, . $$ To factorise this further, we need to find a root of the equation $$ 2x^3+7x^2+4x-4 = 0 \, . $$ The rational root theorem states that if $p/q$ is a root of a polynomial, then $p$ must be an integer factor of the constant term, and $q$ must be an integer factor of the leading coefficient. Here, this means that $p$ must go into $-4$, and $q$ must go into $2$. Once you have found a root, you can factorise the cubic further, and you are left with a quadratic, which can be broken down further into two linear factors.
The common $x$ is a dead give-away so we start with writing it as $x(2x^3 + 7x^2 + 4x -4)$.
Then you basically hope and pray that the remaining third degree term has an integer or easy rational root which you can then pull out as a linear term. Any third degree polynomial over $\Bbb R$ has a real root after all. But applying classic methods like Cardano can be nasty.. So hope for a nice root.
Candidates follow from the rational root theorem: divisors of $-4$ so $-1,1,-2,2$ (maybe halved because of the leading $2$). It turns out that $x=-2$ is a root. So $x+2 = (x - (-2))$ divides your polynomial
and we get $$x(x+2)(2x^2 + 3x -2)$$
Now the final quadratic also has a root $x=-2$, found the same way
and we finally get $$x(x+2)^2(2x-1)$$
as the factorisation.
The AC method works with this polynomial. First factor an $x$ out to get $x(2x^3+7x^2+4x-4)$, and multiply the cubic by $4$ to get $8x^3+28x^2+16x-16 = (2x)^3 + 7(2x)^2 + 8(2x) - 16$. Then you can substitute $x = 2X$ to get $X^3 + 7X^2+8X-16$, where it is apparent that $X = 1$ is a root by the factor theorem.
Once you have one divisor, use a method such as polynomial long division to get the other factor, which must be a quadratic. Using synthetic division:
| $1$ | $1$ | $7$ | $8$ | $-16$ |
|---|---|---|---|---|
| $+$ | $1$ | $8$ | $16$ | |
| $1$ | $8$ | $16$ | $0$ |
thus $X^3 + 7X + 8X - 16 = (X - 1)(X^2 + 8X + 16)$. Find the roots of this polynomial, and finally use $x = 2X$ to get the original roots back. $x = 0$ also from earlier.
As $x$ is the common factor:
$$2x^4+7x^3+4x^2-4x$$ $$x\left(2x^3+7x^2+4x-4\right)$$ $$x\left(2x-1\right)\left(x+2\right)\left(x+2\right)$$ $$\boxed{x\left(2x-1\right)\left(x+2\right)^2}$$
