Question

1. Write a polynomial function of minimum degree with real coefficients whose zeros include those listed. Write the polynomial in standard form.
4, -8, and 2 + 5i


f(x) = x4 - 6x3 - 20x2 + 122x - 928

f(x) = x4 - 19x2 + 244x - 928

f(x) = x4 - 6x3 + 20x2 - 122x + 928

f(x) = x4 - 61x2 + 244x - 928

Answer 1

Polynomial with real coefficients always has even number of complex roots. We know that one of them is 2 + 5i so the second one will be  2 - 5i and:

$f(x)=\big(x-4\big)\big(x-(-8)\big)\big(x-(2+5i)\big)\big(x-(2-5i)\big)=\\\\=(x-4)(x+8)(x-2-5i)(x-2+5i)=\\\\=(x^2-4x+8x-32)(x^2-2x+5ix-2x+4-10i-5ix+10i-25i^2)\\\\=\big(x^2+4x-32\big)\big(x^2-4x+4-25\cdot(-1)\big)=\\\\=(x^2+4x-32)(x^2-4x+29)=\\\\=x^4-4x^3+29x^2+4x^3-16x^2+116x-32x^2+128x-928=\\\\=\boxed{x^4-19x^2+244x-928}$

Answer B.

Answer 2

Answer:

Option 2 - $f(x)=x^4-19x^2+244x-928$          

Step-by-step explanation:

Given : Zeros of the polynomial 4, -8, and 2 + 5i.

To find : Write a polynomial function of minimum degree with real coefficients whose zeros include those listed ?

Solution :

The zeros of the polynomial are 4, -8, and 2 + 5i.

There is one conjugate exist i.e. 2-5i.

So, The polynomial form of zeros are      

$f(x)=(x-4)(x+8)(x-(2+5i))(x-(2-5i))$

$f(x)=(x-4)(x+8)(x-2-5i))(x-2+5i))$

Multiply the factors i.e. first two terms and other two terms multiply each other,

$f(x)=(x^2+8x-4x-32)(x^2+5ix-2x-5ix-25i^2+10i-2x-10i+4)$

$f(x)=(x^2+4x-32)(x^2-4x+25+4)$

$f(x)=(x^2+4x-32)(x^2-4x+29)$        

$f(x)=x^4-4x^3+29x^2+4x^3-4x^2+116x-32x^2+128x-928$      

$f(x)=x^4-19x^2+244x-928$          

Therefore, Option 2 is correct.