Solving $2x^4+x^3-11x^2+x+2 = 0$ I am having no idea how I can solve this problem.
I need help!
Here's the problem
$2x^4+x^3-11x^2+x+2 = 0$
I am learning Quadratic Expressions and this is what I need to solve, and I can't understand how :C
 A: $2x^4+x^3-11x^2+x+2=0$
Note that the coefficients: $2,1,-11,1,2$ are symmetrical.
$2(x^4+1)+(x^3+x)-11x^2=0$
$2(x^4+4x^3+6x^2+4x+1)-7(x^3+x)-23x^2=0$
$2(x^4+4x^3+6x^2+4x+1)-7(x^3+2x^2+x)-9x^2=0$
$2(x+1)^4-7(x+1)^2x-9x^2=0$
$2\left(\frac{(x+1)^2}{x}\right)^2-7\left(\frac{(x+1)^2}{x}\right)-9=0$
A: $$2x^4+x^3-11x^2+x+2 = (x-2)(2x^3+5x^2-x-1)=(x-2)(2x-1)(x^2+3x+1)$$
Now you can solve it easily
EDIT: You can use the rational root theorem to get the possible rational roots 
A: By the rational root theorem, there are $6$ possibilities to check: $\pm\frac12,\pm1,$ or $\pm2$.  The easiest values to test are $\pm1$ and neither works.  The next easiest value to test is $2$, which is a solution.  In addition, using Kenny Lau's observation from his comment, the equation can be re-written as
$$2x^2+x-11+\frac1x+\frac2{x^2}=0$$
Since $x=0$ is not a solution, there is no harm from this division.  Note that the substitution $x=\frac1y$ gives the same equation back in $y$.  So if $y$ is a solution, so is $\frac1y$.  If $2$ is a solution, so is $\frac12$.  Factoring out $(x-2)(x-\frac12)$ from the original equation should yield a quadratic which, in the worst case, can be solved by completing the square or quadratic formula.
A: Not entirely sure what context you're encountering the problem. In the college algebra I teach, this question is presented after learning about synthetic division, factor theorem, the rational roots theorem and the quadratic formula.


*

*Synthetic division is a compactified version of long division. It only uses three lines, no matter how big the polynomial is, and it doesn't require writing all those $x$s. Additionally, some students get tripped up by subtracting monomials with negative coefficients in long division (more specifically, they forget that they're subtracting them too), and synthetic division makes this simpler since it uses addition instead of subtraction. On the other hand, having long division use subtraction and synthetic division use addition can be confusing. Also, synthetic division won't work if you ever wanted to divide by a nonlinear polynomial (or one that has leading coefficient other than the implicit $1$, at least directly). If you're not familiar with synthetic division, you can always just use polynomial long division. If you don't know that, then it's a good thing to read up on.

*Factor theorem says that a value $c$ is a root of a polynomial function $f(x)$ if and only if $(x-c)$ appears in a factorization of $f(x)$ as $(x-c)q(x)$. (Letter "q" for quotient.)

*Rational roots theorem says that any rational root $p/q$ of $a_nx^n+\cdots+a_1x+a_0$ has the form $p/q=\pm(\textrm{divisor of }a_0)/(\textrm{divisor of }a_n)$.

*Quadratic formula says the roots of a quadratic $ax^2+bx+c$ are given by $$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a} $$


We can combine these to factorize (not all but) a general class of higher degree polynomials completely by hand calculations. First, employ the rational roots theorem to list out a finite set of potential rational zeros. Then, use synthetic division to divide the polynomial by each potential zero (or long division by each potential linear factor $x-c$) for every value in the list. Once you hit a potential zero that yields a remainder of $0$, you know that is an actual zero, hence $f(x)$ may be factored part-way as $f(x)=(x-c)q(x)$ where $q(x)$ is the quotient. 
Then you can repeat this process on $q(x)$, and so on to keep factoring $f(x)$ further and further, and once the quotient is finally a quadratic $ax^2+bx+c$, you can use the quadratic formula to find the roots $r_1$ and $r_2$ and thus factor the quadratic itself as $a(x-r_1)(x-r_2)$. In the problems we give, the final two zeros found $r_1$ and $r_2$ are not necessarily rational like the other ones, hence the need for the quadratic formula.
A: You can factorise the expression using the factor theorem:
$$P(x) = 2x^4+x^3-11x^2+x+2 $$
$$P(1) \neq 0 $$
$$\therefore (x-1) \ is \ NOT \ a \ root.$$
$$P(2) = 0$$
$$\therefore (x-2) \ IS \ a \ root.$$
$$\therefore P(x) = 2x^4+x^3-11x^2+x+2 = (x−2)(2x^3+5x^2−x−1)$$
Where $(2x^3+5x^2−x−1)$ can be obtained by either of long division or synthetic division (which is just a short cut for long division).
You can then treat $(2x^3+5x^2−x−1)$ as P(x) and, using the factor theorem, continued finding linear factors.
