1. ## Solving trinomial

Help!

Solve algebraically: 10x^3+x^2-69x+54 = 0

How do i solve this?
How do i find one root so i can find the other 2?

Thanks so much for any help,

mike

2. Hello, Mike!

Solve algebraically: .$\displaystyle 10x^3+x^2-69x+54 \:= \:0$
We have the Rational Roots Theorem.

If a polynomial $\displaystyle P(x)$ has a rational root, it is of the form $\displaystyle \frac{n}{d}$
. . where $\displaystyle n$ is a factor of the constant term
. . and $\displaystyle d$ is a factor of the leading coefficient.

The constant term is $\displaystyle 54$ with factors: .$\displaystyle \pm1,\:\pm2,\:\pm3,\:\pm6,\:\pm9,\:\pm18,\:\pm27,\ :\pm54$

The leading coefficient is $\displaystyle 10$ with factors: .$\displaystyle \pm1,\:\pm2,\:\pm5,\:\pm10$

That's a lot of fractions to test, but we may get lucky . . .

Try $\displaystyle P(1) \:=\:10\!\cdot\!1^3 + 1^2 = 60\!\cdot\!1 + 54 \:=\:-4$ . . . no

Try $\displaystyle P(2) \;=\;10\!\cdot\!2^3 + 2^2 - 69\!\cdot\!2 + 54 \:=\:0$ . . . YES!

Since $\displaystyle x = 2$ is a zero of $\displaystyle P(x)$, then $\displaystyle (x-2)$ is a factor.

Using long division, we have: .$\displaystyle 10x^3 + x^2 - 69x + 54 \;=\;(x-2)(10x^2 + 21x - 27)$

. . And we find that the quadratic also factors.

Therefore: .$\displaystyle (x-2)(x+3)(10x-9) \;=\;0\quad\Rightarrow\quad x \;=\;2,\,-3,\,\frac{9}{10}$

3. We can also try a little factoring.

Rewrite as

$\displaystyle 10x^{3}-9x^{2}+10x^{2}-9x-60x+54$

Now group:

$\displaystyle (10x^{3}+10x^{2}-60x)-(9x^{2}+9x-54)$

Now, factor:

$\displaystyle 10x(x^{2}+x-6)-9(x^{2}+x-6)$

$\displaystyle (10x-9)(x^{2}+x-6)$

$\displaystyle (10x-9)(x-2)(x+3)$