What is a quartic polynomial function in standard form with zeros 1, –2, –2, and –3?
I would guess at
y = (x-1)(x + 2)(x + 2)(x + 3)
or
y = 17(x-1)(x+3)(x+2)^2
It asked for one such function
expand it if you feel you must.
( placing a constant in front does not change the x-intercepts )
To find a quartic polynomial function in standard form with given zeros, we need to identify the factors corresponding to each zero. In this case, the given zeros are 1, -2, -2, and -3.
Given zero: 1
Factor: (x - 1)
Given zero: -2
Factor: (x + 2)
Given zero: -2 (again)
Factor: (x + 2)
Given zero: -3
Factor: (x + 3)
Next, we multiply these factors together to obtain the quartic polynomial function:
f(x) = (x - 1)(x + 2)(x + 2)(x + 3)
To simplify further, since (-2)(-2) = 4, we can combine the double root (-2)(-2) into (x + 2)^2:
f(x) = (x - 1)(x + 2)^2(x + 3)
Finally, if you want to expand the expression fully, you can perform the multiplication.