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

1 Answer
Jan 18, 2017

Answer:

#f(x)=x^3-5x^2+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:

#x_1=-1#
#x_2=3+2i#
#x_3=3-2i#

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+2i#
#x_3=3-2i#

Where
#a=x_1=-1#
#b=x_2=3+2i#
#c=x_3=3-2i#.

#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(x)=(x+1)(x-3-2i)(x-3+2i)#

#=(x+1)(x-3-2i)(x-3+2i)#

#=(x+1)(color(red)(x)color(blue)(-3)color(green)(-2i))(x-3+2i)#

#=(x+1)[color(red)(x)(x-3+2i)color(blue)(-3)(x-3+2i)color(green)(-2i)(x-3+2i)]#

#=(x+1)[x^2-3x color(red)(+2ix)-3x+9color(blue)(-6i)color(red)(-2ix)color(blue)(+6i)-4i^2]#

Collecting terms, and substituting #i =sqrt(-1)#

#f(x)=(color(red)(x)color(blue)(+1))(x^2-6x+13)#

Multiplying terms again:

#f(x)=color(red)(x)(x^2-6x+13)color(blue)(+1)(x^2-6x+13)#

#=color(red)(x^3-6x^2+13x)+color(blue)(x^2-6x+13)#

Which yields a final answer:

#f(x)=x^3-5x^2+7x+13#