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

Question

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

Sadie Alexander · Accepted Answer

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

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#

Madelyn Anderson · Answer

undefined To write a polynomial in standard form given zeros -1 and 3 + 2i, we use the fact that complex zeros come in conjugate pairs. Thus, if 3 + 2i is a zero, then its conjugate, 3 - 2i, is also a zero.

First, we create the factors of the polynomial using the given zeros:

1. For the real zero -1, the factor is (x + 1).
2. For the complex zeros 3 + 2i and 3 - 2i, the factors are (x - (3 + 2i)) and (x - (3 - 2i)) respectively, which simplify to (x - 3 - 2i) and (x - 3 + 2i).

Next, we multiply these factors to obtain the polynomial:

$(x + 1)(x - 3 - 2i)(x - 3 + 2i)$

We can multiply the complex conjugate factors first to simplify:

$(x + 1)((x - 3)^2 - (2i)^2)$

Expanding and simplifying:

$(x + 1)((x^2 - 6x + 9) - (-4))$

$(x + 1)(x^2 - 6x + 9 + 4)$

$(x + 1)(x^2 - 6x + 13)$

Finally, we distribute (x + 1) to each term in the second factor:

$x(x^2 - 6x + 13) + 1(x^2 - 6x + 13)$

$x^3 - 6x^2 + 13x + x^2 - 6x + 13$

$x^3 - 5x^2 + 7x + 13$

So, the polynomial in standard form is $x^3 - 5x^2 + 7x + 13$.