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

1 Answer
Jan 16, 2017

f(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 (xa),(xb),and(xc)

Simply:

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)=(aa)(ab)(ac)=(0)(ab)(ac)=0

f(b)=(ba)(bb)(bc)=(ba)(0)(bc)=0

f(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)=(xa)(xb)(xc) has zeroes at a, b, and c.

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

f(x)=(x0)(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(x2+5x+6)

And distributing the x yields a final answer of:

f(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)=0

f(2)=(2)3+5(2)2+6(2)=8+2012=0

f(3)=(3)3+5(3)2+6(3)=27+4518=0

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