# 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 – Jeel Shah 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

Hint: One might hope that there is a rational solution. There needn't be, but if this is a school exercise there probably is.

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.

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All of the answers are fine, but I feel that it should be expanded upon why your intended approach does not work. It is a mistake that I have seen many, many times, and it is probably worth addressing.

Why do we factor expressions? If we are trying to solve $$f(x) = 0$$ and we can factor $f(x) = h(x)g(x)$, then this is the same as $$h(x)g(x) = 0$$

The key fact now is that if the product of two numbers/functions is identically zero, then one of the two numbers/functions must be zero themselves. That is, we must have either $h(x) = 0$ or $g(x) = 0$. This is why we factor in this setting.

Now, by contrast, if we are trying to solve, say, $$f(x) = 4$$ and we factor $f(x) = h(x)g(x)$, then this yields $$h(x)g(x) = 4$$ which tells us nothing whatsoever! Maybe $h(x) = 1/2$ and $g(x) = 8$. Maybe $h(x) = 2/x$ and $g(x) = 2x$. We can say absolutely nothing about this case, and so the factorization of $f(x)$ in this case is not helpful. We could, however, look at $$f(x) - 4 = 0$$ and then try to factor $f(x) - 4 = r(x)s(x)$, which would then be reduced to solving $r(x) = 0$ and $s(x) = 0$.

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You are given $x^3+3x^2-5x-4=0$

First, find one of the roots of the polynomial. By way of the rational root theorem, we find that -4 is one of the roots, therefore x + 4 is a factor. Next, split the polynomial in accordance with x + 4 as follows:

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

Next factor each pair of terms from left to right as follows: $x^2(x + 4) -x(x + 4) - 1(x + 4) = 0$

Since x + 4 is common to each factorization as expected, factor it out:

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

By zero product property:

$x + 4 = 0$

$x^2 - x - 1 = 0$

Obviously $x = -4$ is one of the solutions. The other solution can be found by completing the square:

$x^2 - x = 1$

$x^2 - x + \left(\dfrac{1}{2}\right)^2 = 1 + \left(\dfrac{1}{2}\right)^2$

$\left(x - \dfrac{1}{2}\right)^2 = 1 + \dfrac{1}{4}$

$\left(x - \dfrac{1}{2}\right)^2 = \dfrac{5}{4}$

$x = \dfrac{\pm\sqrt{5} + 1}{2}$

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