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
Hello, Mike!
We have the Rational Roots Theorem.Solve algebraically: .$\displaystyle 10x^3+x^2-69x+54 \:= \:0$
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: .$\displaystyle (x-2)(x+3)(10x-9) \;=\;0\quad\Rightarrow\quad x \;=\;2,\,-3,\,\frac{9}{10}$
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:
$\displaystyle 10x(x^{2}+x-6)-9(x^{2}+x-6)$
$\displaystyle (10x-9)(x^{2}+x-6)$
But the quadratic factor further:
$\displaystyle (10x-9)(x-2)(x+3)$
And you can easily see the roots now.