# Solving the cubic polynomial equation $x^3+3x^2-5x-4=0$

How can I solve the cubic polynomial equation $$x^3+3x^2-5x-4=0$$

I simplified it to: $$x(x^2+3x-5)=4$$ But I don't know where to go from here.

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expand the x^2+3x-5 then bring 4 to the left side and then solve –  gekkostate Aug 25 '12 at 18:24

The manipulation that you performed does not help. Use the rational root test: since the leading coefficient is $1$ and the constant term is $4$, the only possible rational roots of the cubic are fractions of the form $a/b$, where $a$ is a divisor of $4$ and $b$ is a divisor of $1$. In other words, if the cubic has any rational root at all, it must be one of the numbers $\pm1,\pm2$, and $\pm4$. Actually calculation shows that

$$(-4)^3+3(-4)^2-5(-4)-4=-64+48+20-4=0\;,$$

so $-4$ is a root of the cubic $x^3+3x^2-5x-4$. Now use the fact that $r$ is a root of a polynomial if and only if $x-r$ is a factor of that polynomial to conclude that $x^3+3x^2-5x-4$ is divisible by $x-(-4)=x+4$. When you perform this division $-$ either by polynomial long division or by synthetic division $-$ you’ll get a quadratic as the quotient. Let’s say that you get $x^2+bx+c$ as the quotient; then you know that

$$x^3+3x^2-5x-4=(x+4)(x^2+bx+c)\;.$$

This tells you that $x=-4$ is one solution to your original equation, and the others are the solutions to the quadratic equation $x^2+bx+c=0$, which you can find using the quadratic formula.

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Thank you, I understand it now. –  LanguagesNamedAfterCofee Aug 25 '12 at 18:38
By the Rational Root Theorem, such a solution must be a divisor of $4$. Try all of them. We get lucky, there is an integer solution. (Don't forget that the divisors of $4$ include some negative numbers.)
After you have found a solution $x=a$, divide the polynomial by $x-a$. You will end up with a quadratic, which you solve in the usual way to find the remaining roots.