1. find polynomial f(x)

I have a test up tomorrow and I really need to know how to solve this; somehow I didn't write explanations in my notes so although I kind of have a vague idea of how to solve it, I don't get it.

Find a polynomial f(x) of degree 4, given that f(1)=f(2)=f(-2)=0 and f(3)=f(-3)=40.

thank you so much in advance.

2. Originally Posted by colloquial
I have a test up tomorrow and I really need to know how to solve this; somehow I didn't write explanations in my notes so although I kind of have a vague idea of how to solve it, I don't get it.

Find a polynomial f(x) of degree 4, given that f(1)=f(2)=f(-2)=0 and f(3)=f(-3)=40.

thank you so much in advance.
Well, we know that 1, 2, and -2 are zeros. So the 4th zero is real as well. (If it were complex then there would be a fifth zero: the complex conjugate of the fourth zero.) So the polynomial is of the form:
$\displaystyle f(x) = (x - 1)(x - 2)(x + 2)(x - a)$
where a is the fourth zero, currently unknown.

$\displaystyle f(x) = x^4 - (a + 1)x^3 +(a - 4)x^2 + (4a +4)x - 4a$

We also know that
$\displaystyle f(3)= f(-3) = 40$

From our polynomial we know that
$\displaystyle f(3) = -10a + 30 = 40$
so $\displaystyle a = -1$.

(The condition $\displaystyle f(-3) = 40$ gives us the same value.)

Thus
$\displaystyle f(x) = x^4 - 5x^2 + 4$

-Dan

3. Originally Posted by colloquial
I have a test up tomorrow and I really need to know how to solve this; somehow I didn't write explanations in my notes so although I kind of have a vague idea of how to solve it, I don't get it.

Find a polynomial f(x) of degree 4, given that f(1)=f(2)=f(-2)=0 and f(3)=f(-3)=40.

thank you so much in advance.
Here's another approach, a much longer one. Set up a simultaneous equation system of five equations and five unknowns.

Let the polynomial be of the form: $\displaystyle ax^4 + bx^3 + cx^2 + dx + e$

Now,

$\displaystyle f(1) = a + b + c + d + e = 0$ ..........................(1)

$\displaystyle f(2) = 16a + 8b + 4c + 2d + e = 0$ ..................(2)

$\displaystyle f(-2) = 16a - 8b + 4c - 2d + e = 0$ ................(3)

$\displaystyle f(3) = 81a + 27b + 9c + 3d + e = 40$ ...............(4)

$\displaystyle f(-3) = 81a - 27b + 9c -3d + e = 40$ ..............(5)

Now solve the above system to get your coefficients