# How do you find the x and y intercepts for f(x) = x^3 - 2.91x^2 - 7.668x - 3.8151?

Sep 20, 2015

Scale the polynomial to make one with integer coefficients then use the rational root theorem to help find the zeros.

Intercepts are $\left(0 , - 3.8151\right)$, $\left(- 0.9 , 0\right)$ (twice) and $\left(4.71 , 0\right)$

#### Explanation:

$f \left(x\right) = {x}^{3} - 2.91 {x}^{2} - 7.668 x - 3.8151$

The intercept with the $y$ axis is $\left(0 , f \left(x\right)\right) = \left(0 , - 3.8151\right)$

Let $t = \frac{100}{3} x$

Let $g \left(t\right) = \frac{1000000}{27} f \left(x\right) = {t}^{3} - 97 {t}^{2} - 8520 t - 141300$

By the rational root theorem, any rational zeros of this polynomial are factors of $141300 = {2}^{2} {3}^{2} {5}^{2} 157$

$g \left(157\right) = 3869893 - 2390953 - 1337640 - 141300 = 0$

$g \frac{t}{t - 157} = {t}^{2} + 60 t + 900 = {\left(t + 30\right)}^{2}$

So the zeros of $g \left(t\right)$ are $t = 157$ and $t = - 30$ (twice)

The corresponding values of $x$ are $\frac{3}{100} t$:

$x = 4.71$ or $x = - 0.9$ (twice)

So the intercepts with the $x$ axis are $\left(4.71 , 0\right)$ and $\left(- 0.9 , 0\right)$ (twice)