Let f(x)=3x$^3$ - 40x$^2$ + 97x + 10
a. Find a rational number r such that f(r) = 0. (Hint: Use the rational roots theorem to narrow down possibilities for r.)
So, I figured this part out. write r as r=a/b where a|10 and b|3. And r = 10/3. I did a little more showing my work, but I feel good about this answer. It is part b. I am struggling with.
b. Find two other numbers s and t such that f(s) = 0 and f(t) = 0. (Hint: The cubic polynomial f(x) can be expressed as (x-r)g(x) where r is as in part (a) and g(x) is a quadratic polynomial.)
So far, I can tell that these numbers are irrational, because the only answer for part (a) is r = 10/3. I don't understand, however, how x and r can be in the same equation because the way I understood it in part (a) is that f(r) just meant plug in a rational number r for x. Am I missing something here? How can I get started on finding s and t?
Thank you so much to anyone who can help! I am quite stuck on this question.
The Hint is telling you that you should use the root you found in part $\textbf{a}$ in order to divide out a monomial from your cubic polynomial.
So since $r=\frac{10}{3}$, you have that $x-\frac{10}{3}=0\Rightarrow 3x-10=0$.
So therefore $f(x)=3x^3-40x^2+97x+10=(3x-10)g(x)$ where
$$g(x)=\frac{f(x)}{3x-10}=\frac{3x^3-40x^2+97x+10}{3x-10}=x^2-10x-1$$
You can verify this via long division of polynomials.
Now we need only find the roots of $g(x)$ to find the desired numbers $s$ and $t$. By the quadratic formula:
$$x=\frac{-(-10)\pm\sqrt{(-10)^2-4\cdot1\cdot(-1)}}{2\cdot1}=\frac{10\pm\sqrt{104}}{2}=\frac{10\pm2\sqrt{26}}{2}=5\pm\sqrt{26}$$
Then let $s=5+\sqrt{26}$ and $t=5-\sqrt{26}$.