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

1 Answer
Jan 16, 2017

f(x) = x^3+5x^2+6xf(x)=x3+5x2+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)(xa),(xb),and(xc)

Simply:

f(x) = (x-a)(x-b)(x-c)f(x)=(xa)(xb)(xc)

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) = 0f(a)=(aa)(ab)(ac)=(0)(ab)(ac)=0

f(b) = (b-a)(b-b)(b-c) = (b-a)(0)(b-c) = 0f(b)=(ba)(bb)(bc)=(ba)(0)(bc)=0

f(a) = (c-a)(c-b)(c-c) = (c-a)(c-b)(0) = 0f(a)=(ca)(cb)(cc)=(ca)(cb)(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)f(x)=(xa)(xb)(xc) has zeroes at a, b, and c.

With the generalized form, we can substitute for the given zeroes, x=0, -2, and -3x=0,2,and3, where a=0, b=-2, and c=-3a=0,b=2,andc=3.

f(x) = (x-0)(x-(-2))(x-(-3))f(x)=(x0)(x(2))(x(3))

Simplifying gives:

f(x) = x(x+2)(x+3)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)f(x)=x(x2+5x+6)

And distributing the x yields a final answer of:

f(x) = x^3+5x^2+6xf(x)=x3+5x2+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) = 0f(0)=(0)3+5(0)2+6(0)=0

f(-2) = (-2)^3+5(-2)^2+6(-2) = -8+20-12=0f(2)=(2)3+5(2)2+6(2)=8+2012=0

f(-3) = (-3)^3+5(-3)^2+6(-3) = -27 + 45 - 18= 0f(3)=(3)3+5(3)2+6(3)=27+4518=0

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