Find roots of $x^4 + 4x^3 - 25x^2 - 16x + 84$

Question

I'm not quite sure how to approach this, how are you supposed to quickly factor a 4th degree polynomial? For $x^4 + 4x^3 - 25x^2 - 16x + 84$ it's factored into $(x-3)(x-2)(x+2)(x+7)$ which like I said, I have no idea how to approach. Would really appreciate some help, thanks in advance!

Answer 1

You can use synthetic division in $$p(x)=x^{4}+4x^{3}-25x^{2}-16x+84.$$

To do this, notice that the divisors of:

• $a=1$: $\{\pm1\}$.
• $b=84$: $\{\pm 1, \pm 2, \pm 3, \pm 4, \pm 6, \pm 7, \pm 12, \pm 14, \pm 28, \pm 42, \pm 84\}$.

Then the synthetic division gives us that $$p(x)=x^{4}+4x^{3}-25x^{2}-16x+84=(x-3)(x-2)(x+2)(x+7).$$

Answer 2

First, apply Descartes' Rule of signs. Now we know this polynomial has either two or zero positive real roots and either two or zero negative roots. (How does this help? If we happen to have a long list of candidate roots and, while checking through this list, we find two positive roots, we can immediately discard all remaining positive candidates.)

Then we apply the rational root theorem. This tells us that the only rational candidate roots are a divisor of $84$ over a divisor of $1$: $$ \{-84, -42, -26, -21, -14, -12, -7, -6, -4, -3, -2, -1, 1, 2, 3, 4, 6, 7, 12, 14, 21, 26, 42, 84\} \text{.} $$ We see the coefficients $4$, $-25$, and $-16$, which are sums of products of roots. These are "small", so we should probably start in the middle of the list and work our way out.

\begin{align*} x& &&|& &x^4 + 4x^3 - 25x^2 - 16x + 84 \\ \hline 1& &&|& &1+4-25-16+84 > 0 \\ 2& &&|& &16+32-100-32+84 = 0 &&\checkmark \\ 3& &&|& &81 + 108 - 225 - 144 + 84 = 0 &&\checkmark \end{align*} Having found two real roots, any further real roots are negative. Continuing the table, starting from $-1$, you will find $-2$ and then $-7$.

But we can go a little faster. By synthetic division (or long division), once we have the root $x = 2$, we can reduce the polynomial. $$ x^4 + 4x^3 - 25x^2 - 16x + 84 = (x-2)(x^3 + 6x^2 - 13x - 42) $$ So any root of $x^3 + 6x^2 - 13x - 42$ is also a root of the original polynomial. By the rule of signs, this has one positive real root and two or zero negative real roots. Any rational root is a divisor of $42$ divided by a divisor of $1$. This reduces the previous list of candidate roots to (excluding $1$, the one candidate we definitely verified is not a root) $$ \{ -42, -21, -14, -7, -6, -3, -2, -1, 2, 3, 6, 7, 14, 21, 42\} \text{.} $$ Testing $2$ in this reduced polynomial, we find $2$ is not a repeated root. And then we find $3$ is a root: $$ 3^3 + 6\cdot 3^2 - 13\cdot 3 - 42 = 0 \text{.} $$ That's the last possible positive root and we are left with $$ \frac{x^3 + 6x^2 - 13x - 42}{x-3} = x^2 + 9x+14 \text{.} $$

Of course, this is quadratic, so there is no reason to avoid directly applying the quadratic formula... Alternatively, we compute the discriminant : $9^2 - 4 \cdot 14 = 25$, so the two roots differ by $\sqrt{25} = 5$. (Since the discriminant is a square, we know the remaining two roots are rational, hence are on our list.) Looking at the negatives on our list, the only pairs that differ by $5$ are $\{-1, -6\}$ and $\{-2, -7\}$. Checking $-1$, we find that $-1$ is not a root, so $-2$ and $-7$ are the final two roots.