Question

5-33.
write an equation for a fifth - degree polynomial function with roots at ( x = - 2 ), ( x = 5 ), and triple root ( x = 2 ). the ( y ) - intercept is ( (0,8) ).

Explanation:

Step1: Write polynomial template

If a polynomial has roots $x=r_1, x=r_2, ..., x=r_n$, it can be written as $f(x)=a(x-r_1)(x-r_2)...(x-r_n)$, where $a$ is the leading coefficient. For roots $x=-2$, $x=5$, and triple root $x=2$:
$f(x)=a(x+2)(x-5)(x-2)^3$

Step2: Solve for $a$ using y-intercept

The y-intercept is $(0,8)$, so substitute $x=0$, $f(0)=8$:
$8=a(0+2)(0-5)(0-2)^3$
Calculate the right-hand side:
$(2)(-5)(-8)=2\times(-5)\times(-8)=80$
So $8=80a$, solve for $a$:
$a=\frac{8}{80}=\frac{1}{10}$

Step3: Substitute $a$ into the template

$f(x)=\frac{1}{10}(x+2)(x-5)(x-2)^3$
(Optional: Expand if needed, but factored form is acceptable for a polynomial equation.)

Answer:

$f(x)=\frac{1}{10}(x+2)(x-5)(x-2)^3$
Or expanded form: $f(x)=\frac{1}{10}x^5 - \frac{3}{10}x^4 - \frac{14}{10}x^3 + \frac{52}{10}x^2 - \frac{16}{10}x - 8$ (simplified to $f(x)=\frac{1}{10}x^5 - \frac{3}{10}x^4 - \frac{7}{5}x^3 + \frac{26}{5}x^2 - \frac{8}{5}x - 8$)