Solving $2x^4+x^3-11x^2+x+2 = 0$

Question

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

One can use the general theory of palindromic polynomials. Or else we can note that we get lucky, $2$ is a root and therefore so is $1/2$. — André Nicolas, Jun 25 '16 at 06:05
Yes, they go 4,3,2,1 and What I think is that I need to take x^2 common — dada wilson, Jun 25 '16 at 06:06
Does your textbook cover the quadratic formula, rational roots theorem and either synthetic division or long division before this problem? Does the text give an example of this kind of polynomial being factored? — anon, Jun 25 '16 at 06:27
See http://math.stackexchange.com/questions/480102/quadratic-substitution-question-applying-substitution-p-x-frac1x-to-2x4x — lab bhattacharjee, Jun 25 '16 at 07:45
As pointed out, this is reciprocal equation (the coefficients are symmetric). You can have a look at this question and maybe also some questions linked there to learn more about this type of equations. — Martin Sleziak, Jun 25 '16 at 07:45
http://math.stackexchange.com/a/1044258/11206
http://math.stackexchange.com/a/499377/11206 — miracle173, Jun 25 '16 at 10:40

score 5 · Answer 1 · answered Jun 25 '16 at 06:06

5

$2x^4+x^3-11x^2+x+2=0$

Note that the coefficients: $2,1,-11,1,2$ are symmetrical.

$2(x^4+1)+(x^3+x)-11x^2=0$

$2(x^4+4x^3+6x^2+4x+1)-7(x^3+x)-23x^2=0$

$2(x^4+4x^3+6x^2+4x+1)-7(x^3+2x^2+x)-9x^2=0$

$2(x+1)^4-7(x+1)^2x-9x^2=0$

$2\left(\frac{(x+1)^2}{x}\right)^2-7\left(\frac{(x+1)^2}{x}\right)-9=0$

answered Jun 25 '16 at 06:06

Kenny Lau

25,049

My textbook shows the answer as -3(+ or -)rootof(5)/2,1/2, 2 – dada wilson Jun 25 '16 at 06:07
@UdayKiran That is just a transformation into quadratic equation, so that you can solve it by yourself. – Kenny Lau Jun 25 '16 at 06:08
Kinda hard for me :C – dada wilson Jun 25 '16 at 06:10
@UdayKiran Just treat $\frac{(x+1)^2}x$ as one variable. – Kenny Lau Jun 25 '16 at 06:12
2

@UdayKiran i.e. Let $\frac{(x+1)^2}{x} = y$, and then you have simply
$$2y^2 - 7y - 9 = 0$$

which is a lot easier to solve.
– Edward Evans Jun 25 '16 at 06:20
Really nice factorization! Does this only simplify to a quadratic when the coefficients are symmetrical? Also, pretty sure that if you had more terms you could also simplify it to a cubic or higher right? – Ovi Jun 25 '16 at 06:20
@Ovi This only applies when the coefficients are symmetrical. This could be applied to quartic equations, 6-ic equations, 8-ic equations, etc. – Kenny Lau Jun 25 '16 at 06:21

score 4 · Accepted Answer · edited Jun 25 '16 at 07:46

4

$$2x^4+x^3-11x^2+x+2 = (x-2)(2x^3+5x^2-x-1)=(x-2)(2x-1)(x^2+3x+1)$$ Now you can solve it easily

EDIT: You can use the rational root theorem to get the possible rational roots

edited Jun 25 '16 at 07:46

Martin Sleziak

53,687

answered Jun 25 '16 at 06:06

Let me try this – dada wilson Jun 25 '16 at 06:08
I got it!! @Navaro!! Can I know how you simplified that equation? – dada wilson Jun 25 '16 at 06:10
@UdayKiray : Now it is straightforward :) – Jun 25 '16 at 06:10
How do I simplify it so I can do it soo easily? :O – dada wilson Jun 25 '16 at 06:11
Uhhh Rational Root theorem? Like this? alpha = -b(+or-)rootof(b^2 - 4ac)/2a – dada wilson Jun 25 '16 at 06:14
@UdayKiran: using the rational root theorem you can see that the solutions can be : $1, -1, -1/2, 1/2, 2,2$ , you try one by one to figure out the roots – Jun 25 '16 at 06:15

score 3 · Answer 3 · edited Jun 25 '16 at 07:47

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.

score 0 · Answer 4 · answered Jun 25 '16 at 06:21

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.

Just referred my syllabus, I will be learning about these in my next class :D. Thanks for the text, I will bookmark this for future reference — dada wilson, Jun 25 '16 at 06:26

score 0 · Answer 5 · answered Jun 25 '16 at 06:25

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.

Solving $2x^4+x^3-11x^2+x+2 = 0$

5 Answers5