How do you find a polynomial function that has zeros 0, -2, -3?

1 Answer
Jan 16, 2017

Answer:

#f(x) = x^3+5x^2+6x#

Explanation:

Since we are given the zeroes of the polynomial function, we can write the solution in terms of factors.

In general, given 3 zeroes of a polynomial function, a, b, and c, we can write the function as the multiplication of the factors #(x-a), (x-b), and (x-c)#

Simply:

#f(x) = (x-a)(x-b)(x-c)#

In this case, we can show that each of a, b, and c are zeroes of the function:

#f(a) = (a-a)(a-b)(a-c) = (0)(a-b)(a-c) = 0#

#f(b) = (b-a)(b-b)(b-c) = (b-a)(0)(b-c) = 0#

#f(a) = (c-a)(c-b)(c-c) = (c-a)(c-b)(0) = 0#

Since the value of the function at x=a, b and c is equal to 0, then the function #f(x) = (x-a)(x-b)(x-c)# has zeroes at a, b, and c.

With the generalized form, we can substitute for the given zeroes, #x=0, -2, and -3#, where #a=0, b=-2, and c=-3#.

#f(x) = (x-0)(x-(-2))(x-(-3))#

Simplifying gives:

#f(x) = x(x+2)(x+3)#

From here, we can put it in standard polynomial form by foiling the right side:

#f(x) = x(x^2+5x+6)#

And distributing the x yields a final answer of:

#f(x) = x^3+5x^2+6x#

To double check the answer, just plug in the given zeroes, and ensure the value of the function at those points is equal to 0.

#f(0) = (0)^3+5(0)^2+6(0) = 0#

#f(-2) = (-2)^3+5(-2)^2+6(-2) = -8+20-12=0#

#f(-3) = (-3)^3+5(-3)^2+6(-3) = -27 + 45 - 18= 0#

Thus, the function has zeroes as given by x=0, -2, and -3.