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

Jan 18, 2017

$f \left(x\right) = {x}^{3} - 5 {x}^{2} + 7 x + 13$

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:

${x}_{1} = - 1$
${x}_{2} = 3 + 2 i$
${x}_{3} = 3 - 2 i$

In general, given 3 zeros 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, ${x}_{1} = - 1$
${x}_{2} = 3 + 2 i$
${x}_{3} = 3 - 2 i$

Where
$a = {x}_{1} = - 1$
$b = {x}_{2} = 3 + 2 i$
$c = {x}_{3} = 3 - 2 i$.

f(x)=(x-(-1))(x−(3+2i))(x-(3-2i))

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

$f \left(x\right) = \left(x + 1\right) \left(x - 3 - 2 i\right) \left(x - 3 + 2 i\right)$

$= \left(x + 1\right) \left(x - 3 - 2 i\right) \left(x - 3 + 2 i\right)$

$= \left(x + 1\right) \left(\textcolor{red}{x} \textcolor{b l u e}{- 3} \textcolor{g r e e n}{- 2 i}\right) \left(x - 3 + 2 i\right)$

$= \left(x + 1\right) \left[\textcolor{red}{x} \left(x - 3 + 2 i\right) \textcolor{b l u e}{- 3} \left(x - 3 + 2 i\right) \textcolor{g r e e n}{- 2 i} \left(x - 3 + 2 i\right)\right]$

$= \left(x + 1\right) \left[{x}^{2} - 3 x \textcolor{red}{+ 2 i x} - 3 x + 9 \textcolor{b l u e}{- 6 i} \textcolor{red}{- 2 i x} \textcolor{b l u e}{+ 6 i} - 4 {i}^{2}\right]$

Collecting terms, and substituting $i = \sqrt{- 1}$

$f \left(x\right) = \left(\textcolor{red}{x} \textcolor{b l u e}{+ 1}\right) \left({x}^{2} - 6 x + 13\right)$

Multiplying terms again:

$f \left(x\right) = \textcolor{red}{x} \left({x}^{2} - 6 x + 13\right) \textcolor{b l u e}{+ 1} \left({x}^{2} - 6 x + 13\right)$

$= \textcolor{red}{{x}^{3} - 6 {x}^{2} + 13 x} + \textcolor{b l u e}{{x}^{2} - 6 x + 13}$

Which yields a final answer:

$f \left(x\right) = {x}^{3} - 5 {x}^{2} + 7 x + 13$