Answer :
Zeros are the x values which make the function equal to zero. Set it up as you would for a binomial with a constant multiplier "k" to account for the y-intercept (0, -5) given.
f(x) = k(x-2)(x-3)(x-5)
Use the y-intercept (0,-5) to solve for k.
-5 = k(0-2)(0-3)(0-5)
-5 = -30k
-5/-30 = k
1/6 = k
The cubic polynomial function is then ..
f(x) = (1/6)(x-2)(x-3)(x-5)
Linear factors are the linear (line) expressions you can factor out of the polynomial. They are (x-2), (x-3) and (x-5).
f(x) = k(x-2)(x-3)(x-5)
Use the y-intercept (0,-5) to solve for k.
-5 = k(0-2)(0-3)(0-5)
-5 = -30k
-5/-30 = k
1/6 = k
The cubic polynomial function is then ..
f(x) = (1/6)(x-2)(x-3)(x-5)
Linear factors are the linear (line) expressions you can factor out of the polynomial. They are (x-2), (x-3) and (x-5).
The cubic polynomial equation which passes through the coordinate (0,-5) and whose zeroes are (2 , 0), (3 , 0), and (5 , 0) is: [tex]y = \dfrac{1}{6}(x-2)(x-3)(x-5)[/tex] and it can be determine by arithmetic operations.
Given :
- A cubic polynomial whose zeroes are (2 , 0), (3 , 0), and (5 , 0).
- Cubic polynomial equation passes through (0 , -5).
Arithmetic operations can be used to determine the cubic polynomial equation. The generalised cubic polynomial equation is given by:
[tex]y = K(x-a)(x-b)(x-c)[/tex]
Where, K is constant and a, b, and c are the zeroes of the cubic polynomial equation. Therefore, the value of a, b, and c are 2, 3, and 5 respectively.
[tex]y = K(x-2)(x-3)(x-5)[/tex] ----- (1)
It is given that cubic polynomial equation passes through (0 , -5). So, put x = 0 and y = -5 in equation (1) to evaluate the value of K.
[tex]-5=K(0-2)(0-3)(0-5)[/tex]
[tex]5=30K[/tex]
[tex]K = \dfrac{1}{6}[/tex]
Therefore the cubic polynomial equation is given by:
[tex]y = \dfrac{1}{6}(x-2)(x-3)(x-5)[/tex]
For more information, refer the link given below:
https://brainly.com/question/12254880