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.

8 0

3 years ago

Read 2 more answers

Please help. I got an answer but I do not think it is correct.

AVprozaik [17]

Answer: 81

Step-by-step explanation:

(3^2)^2=81

Hope that helped

8 0

3 years ago

Sum of 2x 2 + 3x - 4, 8 - 3x, and -5x 2 + 2.

Salsk061 [2.6K]

(2x²<span> + 3x - 4) + (8 - 3x) + (-5x</span>²<span> + 2)

= </span> 2x² + 3x - 4 + 8 - 3x - 5x² + 2

= -3x² + 6

= 3(-x² + 2)

5 0

3 years ago

Determine if the following relations are functions. if they are functions, also state the domain and range.

Feliz [49]

Answer:

no, they are not functions

Step-by-step explanation:

this is because a single domain has multiple range.

i.e. domain 2 has ranges 2 and 4

4 0

3 years ago