Create a polynomial of degree 3 that has x-int at -1, 1, 3
(x+1)(x-1)(x-3)=0
multiply it out.
To create a polynomial of degree 3 that has x-intercepts at -1, 1, and 3, we need to set up an equation and solve for the polynomial coefficients.
A polynomial of degree 3 is generally written in the form:
f(x) = ax^3 + bx^2 + cx + d
Since we know that the polynomial has x-intercepts at -1, 1, and 3, we can use these values to set up equations.
When x = -1, the polynomial should equal zero:
0 = a(-1)^3 + b(-1)^2 + c(-1) + d
0 = -a + b - c + d
When x = 1, the polynomial should equal zero:
0 = a(1)^3 + b(1)^2 + c(1) + d
0 = a + b + c + d
When x = 3, the polynomial should equal zero:
0 = a(3)^3 + b(3)^2 + c(3) + d
0 = 27a + 9b + 3c + d
Now, we have a system of equations that we can solve to find the values of a, b, c, and d.
Using these equations, we can set up a system of linear equations:
-a + b - c + d = 0
a + b + c + d = 0
27a + 9b + 3c + d = 0
We can solve this system of equations using various methods, such as substitution or elimination, to find the values of a, b, c, and d. Once we have the values, we can substitute them back into the general form of the polynomial to get the final equation.