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

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### 3 Answers

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.

****......................****.......................****...............****

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.

I believe the term 'i' in this question refers to the imaginary number (-1)^1/2.

To solve this question we will first simplify (x - 4 + i)(x - 4 - i).

We know: (a + b)(a - b) = a^2 - b^6

Therefore: (x - 4 + i)(x - 4 - i) = (x - 4)^2 - i^2

= x^2 - 8x + 16 - 1

= x^2 - 8x + 15

= x^2 - 5x - 3x + 15

= x(x - 5) - 3(x - 5)

= (x - 3)(x - 5)

Substituting this value of (x - 4 + i)(x - 4 - i) in the original expression we get:

(x - 4 + i)(x - 4 - i)(x - 3) = (x - 3)(x - 5)(x - 3)

= (x - 3)^2*(x - 5)

Therefore there are two values of x. These are 3 and 5.

`(x-4+i)(x-4-i)(x-3)`

Tips to remember would be i x i = -1

The first thing i would do are the first parentheses:

`(x-4+i)(x-4-i)`

and then foil:

`x^2 - 4x - x i - 4x + 16 + 4i + x i - 4i -(-1)`

move like terms next to each other to make it easier to combine:

`x^2 - 4x - 4x - x i + x i + 4i - 4i + 16 + 1`

simplify:

`x^2 - 8x + 17`

now multiply this by x - 3

`(x^2 - 8x + 17) (x - 3)`

Foil:

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

combine like terms:

`x^3 - 11x^2 + 41x - 51`