Solve $2x^4 + 7x^3 -34x^2 -21x + 18 = 0$

Solve $2x^4 + 7x^3 -34x^2 -21x + 18 = 0$ over the real numbers.

I know the final answers but I want a logical way to solve it. I must add you can't use derivation or other advanced formulas. You can use ways such multiplying two sides or using quadratic formulas and such ways. Also you can use simple factorization methods or changing the variable but no advanced calculus or differential formulas.

I want to know a clear way to solve it. For example, if you multiply it by a number or variable, say the reason for doing that and how did you find the appropriate number (or polynomial).

Thanks and sorry for my English.

• How about guessing the solution $x+1=0$ and then use polynomial factorization/division.
– ctst
Oct 10, 2016 at 11:49
• The first thing you should check is whether there are any rational roots. Since $18 = 2\cdot 3^2$, by rational root theorem, you only need to check 18 numbers:$\pm 2^e 3^f$ where $-1 \le e \le 1$, $0 \le f \le 2$. It turns out all four roots has this form, so you are done. Oct 10, 2016 at 11:55
• It can be done by guessing but I want a way which find the answers directly and not by checking the numbers. I know a way but the first part of it (which divide the equation by a polynominal ($x^2$)) is done without explaining why $x^2$ is selected. I want a strategy to solve this question (and other similiar questions) without guessing. Thanks. Oct 10, 2016 at 11:59
• There is a quartic formula – like the quadratic formula, only for quartics instead of quadratics. Type "quartic formula" into the internet and see what comes back at you. Oct 10, 2016 at 12:17

For higher degree polynomials, I recommend first checking with the rational roots theorem.

It's very simple, take the first coefficient $(2)$ and the last coefficient $(18)$, and factor them:

$$2=1\times2$$

$$18=1\times2\times3^2$$

Thus, all real rational roots are of the form

$$r=\pm\frac{\{1,2,3,6,9,18\}}{\{1,2\}}$$

So, we run through and test:

$$r\ne+1\\\boxed{r=-1}\\\boxed{r=+\frac12}\\r\ne-\frac12\\r\ne+2\\r\ne-2\\\boxed{r=+3}\\r\ne-3\\r\ne+\frac32\\r\ne-\frac32\\r\ne+6\\\boxed{r=-6}$$

And look at that, we've got all four roots! So we can then put it back into factored form:

$$2x^4+7x^3-34x^2-21x+18=2(r+1)(r-\frac12)(r-3)(r+6)$$

• Hi. Thanks for your good explanation. I will mark your answer as the best answer tomorrow but until then, I want to know does any other logical way exist that doesn't need number checking or not. As I said before, A way is to divide the equation by $x^2$ and after that, It can be done with changing the variable and some quadratic-equation solving but there isn't any good reason for dividing it by $x^2$ so I want to know does any other solution exists or not? Oct 10, 2016 at 15:08
• @titansarus as others have said, there exists a purely algebraic method to factor all quartic polynomials but as you can see, it is quite unwieldy. I highly recommend against attempting algebraic factoring before the rational roots theorem. The rational roots theorem is not simply 'guess and check', but is a methodical use of the fundamental theorem of algebra. If you are interested in methods to factor (if possible) higher degree polynomials, see Galois Theory. Oct 10, 2016 at 15:24
• @titansarus in the scenario of no rational roots, I recommend not the algebraic solutions involving radicals, but the solutions involving trig functions. Oct 10, 2016 at 15:29