1
$\begingroup$

The polynomial equation is: $x^4-5x^3+5x^2+5x-6=0$.
How do i simplify this equation so that i can find its roots.
Please, can anyone teach me how to find roots of equations of degree 4 and degree 3.

$\endgroup$
  • $\begingroup$ By trivial guessing, we immediately notice that $x = 1$ is a factor. See if you can make another guess for the residue cubic to reduce the problem into a quadratic. $\endgroup$ – Yiyuan Lee Nov 19 '16 at 6:43
2
$\begingroup$

.The key is first substituting integral small values of $x$, which are factors of the constant term, which is $6$. In this case, we try to substitute $1$. We see that if $f(x) = x^4-5x^3+5x^2+5x-6$, then: $f(1)=0$ and $f(-1)=0$. Hence, $x-1$ and $x+1$ are factors of $x^4-5x^3+5x^2+5x-6$.

Now perform polynomial division: divide $ x^4-5x^3+5x^2+5x-6$ by the product $(x-1)(x+1) = x^2-1$, and you should get $(x^2-5x+6)$. Once this is done, you can either use the quadratic formula, or the same technique as above to see that $2$ and $3$ are roots of $x^2-5x+6$, which means that $x^2-5x+6 = (x-2)(x-3)$.

Hence, $f(x) = (x+1)(x-1)(x-2)(x-3)$.

If it were a general cubic or quartic equation, then although formulas (like Cardano) are known, they are cumbersome to say the least, and very hard to work out on hand. So if you get a problem like this to solve by hand, it is most likely either an observation or hit-and-trial with the factors of the constant term (if there is no constant term, then $x$ is a factor, so divide by $x$ until you get a constant term) that is likely to give a breakthrough.

$\endgroup$
  • $\begingroup$ First of all ,thank you for answering. So i keep substituting $f(1),f(-1),f(2),f(-2) $by guess work . Suppose , along with$f(1)$ and $f(-1)$, $f(2)$ also turns out to be 0, would i have to perform polynomial division by $(x-1)(x+1)(x-2)$ then? $\endgroup$ – ICE burger Nov 19 '16 at 6:58
  • $\begingroup$ @ICEburger see RobertZ's answer for detailed approach to such kind of problems $\endgroup$ – vidyarthi Nov 19 '16 at 7:12
  • $\begingroup$ @ICEburger Yes, my friend, you would. And after division, you would obtain $x-3$, which is already a linear factor. Hence, you would be done. $\endgroup$ – астон вілла олоф мэллбэрг Nov 19 '16 at 7:22
  • $\begingroup$ For this equation:$x^4-2x^3+3x^2-2x+1$ . I used $f(1),f(-1),f(2),f(-2)$ all the way to -4 but didnt get$ f(x)=0$. What step should i use now. $\endgroup$ – ICE burger Nov 19 '16 at 8:23
  • $\begingroup$ This is a significantly difficult question, because this will not have any linear factors in that case. The question comes down to how you test for quadratic factors. Suppose that $x^2+bx+c$ divides this polynomial, then putting $x=0$ gives you that $c$ must divide the constant, so that $c = \pm 1$. Now a little bit of trial and error with $b$ shows you that $x^2-x+1$ divides this expression (and the answer is $(x^2-x+1)^2$). But then, this is hardly a systematic method, and if you were searching for one I would throw in the towel. $\endgroup$ – астон вілла олоф мэллбэрг Nov 19 '16 at 8:39
2
$\begingroup$

Hint. Look first for rational solutions: for a polynomial of degree $n$, $$a_n x^n+\cdots +a_1x+a_0$$ with integer coefficients, if $p/q\in\mathbb{Q}$ is a solution then $p$ divides $a_0$ and $q$ divides $a_n$.

In your case, a very "lucky" one I would say, try with the divisors of $-6$ (note that $a_n=1$) that is: $\pm 1$,$\pm2$,$\pm3$,$\pm6$.

$\endgroup$
  • $\begingroup$ So ,what you meant was that i should divide the constant term by the coefficient of the term with highest degree , and then try the divisors of that answer? In this case -6/1. And thanks for the answer. $\endgroup$ – ICE burger Nov 19 '16 at 7:22
  • $\begingroup$ No, $a_n=4$ and $a_0=6$ then you try $p/q$ where $p$ is $\pm 1$,$\pm2$,$\pm3$,$\pm6$ and $q$ is $\pm 1$,$\pm2$,$\pm4$. There a lot of fractions $p/q$ but when you get one solution you can divide the polynomial by $(x-p/q)$, and consider the quotient. $\endgroup$ – Robert Z Nov 19 '16 at 7:40
1
$\begingroup$

A quartic polynomial is always of form $$ax^4+bx^3+cx^2+de+e$$. Now we know that in a quartic equation the product of the roots is equal to $e/a$. Here $e$ is equal to $-6$ and the coefficient of $x^4$ is also 1 so the product of the zeroes is equal to $-6$. We also know that the sum of the zeroes of a quartic polynomial is equal to $-b/a$ and since here also $a$ is $1$ the sum of the zeroes is equal to $-b$ I.e $5$. Thus we conclude that we have to find a set of integers whose sum is $5$ and product is $-6$. Now you just need to factorise $-6$ which will be $2×3×1×-1$ .Now you are very lucky that you got exactly 4 factors which will stand for the $4$ zeroes of the polynomial. Now you need to arrange these numbers in a way that you get $5$ which we get through $2+3+1-1$ thus we get the zeroes of the polynomial as $2,3,1,-1$. If the coefficient of $x^4$ is not 1 then this method won't be good as you might get your zeroes in fractions which will be very difficult to operate upon. But this is the fastest method where $a=1$ . You can carry out this method for all polynomials where $a=1$ and easily get the roots.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.