Find a possible formula for the polynomial with these properties: f is the third degree with f(-3)=0, f(1)=0, f(4)=0, f(0)=4.
f(x)=?
*I don't even know where to start. Please help, step by step if possible.
f is the third degree with f(-3)=0, f(1)=0, f(4)=0, f(0)=4.
f is the third degree, in general case we have
f(x)=ax^3+bx^2+cx+d
f(-3)=0
f(-3)=a(-3)^3+b(-3)^2+c(-3)+d
f(-3)=-27a+9b-3c+d, then
-27a+9b-3c+d=0 (I)
f(1)=0
f(1)=a*1^3+b*1^2+c*1+d
f(1)=a+b+c+d, then
a+b+c+d=0 (II)
f(4)=0
f(4)=a*4^3+b*4^2+c*4+d
f(4)=64a+16b+4c+d, then
64a+16b+4c+d=0 (III)
f(0)=4
f(0)=a*0^3+b*0^2+c*0+d
f(0)=d, then
d=4 (IV)
Form (I), (II), (III), (IV) we need to solve the system
-27a+9b-3c+d=0
a+b+c+d=0
64a+16b+4c+d=0
d=4
and find a, b, c
The function f, which is actually called f(x) has 3 roots
so when it says f(-3) = 0
that means at x = -3, there is no y value, so that is one point
the same goes for 1 and 4, those are also "zeroes" which are places where the function, f, touches the x-axis
and as for f(0) = 4, another point is at (0,4)
Now, from these zeroes, -3, 1, and 4, we can get the equation which is
(x+3)(x-1)(x-4)
And so when you multiply it out you get an x^3, which means it is a third degree polynomial, always remember that the highest exponent tells you how many roots/"zeroes" there are and what degree the equation is
I can't explain why it is, but if anyone here can, please feel free to do so...
Math_helper's method is entirely correct, but realintegerz is very very close to a much easier way to do this.
An alternative form of the cubic is A(x+B)(x+C)(X+D). It isn't quite as general as the one math_helper posted but is perfectly sufficient for this question.
one of the factors is 0, using the null factor theorem. Without loss of generality we can say that this factor is -3-B = 0, giving B = 3.
Repeating this for the remaining zeroes gives C=-1 and D = -4. We then solve for A in the usual manner.
When you are using this to solve a problem simply write down the factors as realintegerz did, but don't forget the constant A
f(x) = A(x--3)(x-1)(x-4)
=A(x+3)(x-1)(x-4)
Then use the
remaining peice of information f(0) = 4 to find A
f(0) =4
A3(-1)(-4) = 4
A = 1/3
So the solution, which is equivalent to that found by math_helper is
f(x) = (1/3)(x+3)(x-1)(x-4)