How do you write a polynomial in standard form given zeros -1 and 3 + 2i?

1 Answer
Jan 18, 2017

f(x)=x35x2+7x+13

Explanation:

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

Whenever a complex number exists as one of the zeros, there is at least one more, which is the complex conjugate of the first. A complex conjugate is a number where the real parts are identical and the imaginary parts are of equal magnitude but opposite sign. Thus, the problem stated should have 3 zeros:

x1=1
x2=3+2i
x3=32i

In general, given 3 zeros 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, x1=1
x2=3+2i
x3=32i

Where
a=x1=1
b=x2=3+2i
c=x3=32i.

f(x)=(x(1))(x(3+2i))(x(32i))

From here, we can put it in standard polynomial form by multiplying all the terms:

f(x)=(x+1)(x32i)(x3+2i)

=(x+1)(x32i)(x3+2i)

=(x+1)(x32i)(x3+2i)

=(x+1)[x(x3+2i)3(x3+2i)2i(x3+2i)]

=(x+1)[x23x+2ix3x+96i2ix+6i4i2]

Collecting terms, and substituting i=1

f(x)=(x+1)(x26x+13)

Multiplying terms again:

f(x)=x(x26x+13)+1(x26x+13)

=x36x2+13x+x26x+13

Which yields a final answer:

f(x)=x35x2+7x+13