Find a polynomial function of degree 3 such that f (2) = 24 and x– 1, x and x+ 2 are factors of the polynomial.

The polynomial function is .

Given that,

Now based on the above information, the calculation is as follows:

f(x) =ax(x - 1)(x + 2)

Here a is a  multiplier.

Now

f(2) = 24

f(2) = 2a(1)(4) = 24

8a = 24

a = 3

It can be write as

.

Therefore we can conclude that the polynomial function is .

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f(x) = 3x³ + 3x² - 6x

given the factors of the polynomial , then

f(x) =ax(x - 1)(x + 2) ( a is a multiplier )

to find a use f(2) = 24

f(2) = 2a(1)(4) = 24 ⇒8a = 24 ⇒ a = 3, thus

f(x) = 3x(x² + x - 2) = 3x³ + 3x² - 6x

options a,b,c are the right answer.

step-by-step explanation:

a linear parent function is always represented as y = x

if we write this function in the form of y=mx+c

⇒

or  

now if we plot a graph we will find these characteristics:

1) y intercept of the graph is zero so its a straight line passing through origin.

2) as we can see in both the forms of function written, values of x and y may be either positive or negative for the positive slope means points on the graph are either in first quadrant or in third one.

3) it's slope is one.