How would I solve this? (x-4+i)(x-4-i)(x-3)It's an algebra 2 problem

To solve :(x-4+i)(x-4-i)(x-3).

Solution:

If you want to solve this , then it should be an equation like:

(x-4+i)(x-4-i)(x-3) = 0. Then you are solving for x , which is unknown and not i, which is known and is equal to (-1)^(1/2). Therefore, by zero product rule, any of the factors of the product (x-4+i)(x-4-i)(x-3) could be zero. So,

x-4i+i = 0 , or x-4-i = 0 or x=3 or

x=4-i or

=x=4+i or

x=3.

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If you want to find the product, then:

{(x-4  +i)(x-4  -i)}(x-3) = {(x-4)^2 - i^2}(x-3}............(1)

But as (a+b)(a-b) = a^2 - b^2 and used  inside the flower bracket. Now, the expression at (1) becomes:

{(x^2-8x+16)-(-1)}(x-3), as x^2 = [(-1)(1/2)]^2 = 1. So, the expression is,

{x^2-8x+17}(x-3)

={x^2-8x+17}x  +  {x^2-8x+17}(-3)

=x^3-8x^2+17x -3x^2+24x-51

=x^3-(8+3)x^2+(17+24)x-51

=x^3-11x^2+41x-51.