Write a polynomial function in standard form with zeros at 6, -4, and 1
what's the problem? You know
f(x) = a(x+4)(x-1)(x-6)
Just expand that out:
f(x) = x^3 - 3x^2 - 22x + 24
You can multiply by any value for "a" if you wish.
To write a polynomial function in standard form with zeros at 6, -4, and 1, we can use the fact that when a polynomial function is in standard form, it is expressed as a product of linear factors.
A polynomial function with zeros at a, b, and c has factors of (x - a), (x - b), and (x - c). In this case, we have zeros at 6, -4, and 1, so the factors will be (x - 6), (x + 4), and (x - 1).
To find the polynomial function, we multiply these factors:
(x - 6)(x + 4)(x - 1)
To simplify this expression, we can use the distributive property:
(x - 6)(x + 4)(x - 1) = (x - 6)(x^2 - x + 4x - 4)
= (x - 6)(x^2 + 3x - 4)
Expanding further:
(x - 6)(x^2 + 3x - 4) = x(x^2 + 3x - 4) - 6(x^2 + 3x - 4)
= x^3 + 3x^2 - 4x - 6x^2 - 18x + 24
= x^3 - 3x^2 - 22x + 24
So, the polynomial function in standard form with zeros at 6, -4, and 1 is:
f(x) = x^3 - 3x^2 - 22x + 24