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=-1x1=−1
x_2=3+2ix2=3+2i
x_3=3-2ix3=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)(x−a),(x−b),and(x−c)
Simply:
f(x)=(x−a)(x−b)(x−c)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)=0f(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)=0f(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)=0f(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)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=-1x1=−1
x_2=3+2ix2=3+2i
x_3=3-2ix3=3−2i
Where
a=x_1=-1a=x1=−1
b=x_2=3+2ib=x2=3+2i
c=x_3=3-2ic=x3=3−2i.
f(x)=(x-(-1))(x−(3+2i))(x-(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)f(x)=(x+1)(x−3−2i)(x−3+2i)
=(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)(x−3−2i)(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(x−3+2i)−3(x−3+2i)−2i(x−3+2i)]
=(x+1)[x^2-3x color(red)(+2ix)-3x+9color(blue)(-6i)color(red)(-2ix)color(blue)(+6i)-4i^2]=(x+1)[x2−3x+2ix−3x+9−6i−2ix+6i−4i2]
Collecting terms, and substituting i =sqrt(-1)i=√−1
f(x)=(color(red)(x)color(blue)(+1))(x^2-6x+13)f(x)=(x+1)(x2−6x+13)
Multiplying terms again:
f(x)=color(red)(x)(x^2-6x+13)color(blue)(+1)(x^2-6x+13)f(x)=x(x2−6x+13)+1(x2−6x+13)
=color(red)(x^3-6x^2+13x)+color(blue)(x^2-6x+13)=x3−6x2+13x+x2−6x+13
Which yields a final answer:
f(x)=x^3-5x^2+7x+13f(x)=x3−5x2+7x+13
