Help me with some Algebra HW :)

[JohnCU]

I need some help with my Algebra III homework. We've never really covered "i" before so it's rather confusing.

(x-5)(x-i)

x^2 - ix - 5x + 5i but what do I do with the i's? I know i^2 is -1, right? Any advice would be valuable. Thanks.

 

[ShizNitz]

Just treat i as a constant. so in the case of (x-5)(x-i)=0, you can conclude that

x-5=0 OR x-i=0 =======> x=5, x=i

 

[BigJohnKC]

<< I need some help with my Algebra III homework. We've never really covered "i" before so it's rather confusing.

(x-5)(x-i)

x^2 - ix - 5x + 5i but what do I do with the i's? I know i^2 is -1, right? Any advice would be valuable. Thanks. >>



(x-5)(x-i)

= x^2 - xi - 5x + 5i

That's your answer.... the imaginary number i must be stated as separate like that. You could reorganize, say

x^2 - 5x - xi + 5i

but it is the same. Treat the i like a variable (like x) but remember its unique properties (i^2 = -1, i^3 = - i, i^4 = i)

 

[NelsonMuntz]

<< Treat the i like a variable (like x) but remember its unique properties (i^2 = -1, i^3 = - i, i^4 = i) >>


Actually, i^4=1, i^5=i and so on.

 

[KEV1N]

Isn't i = sqrt(-1)?
i^2 = 1
i^3 = -1
.
.
.
i^2n = 1
i^2n+1 = -1

 

[NelsonMuntz]

If you square the square root of a number you get that number back, so i^2 is -1 since i is defined as the square root of negative one. If i^2 were to equal 1, then i would equal 1 or -1 not the square root of -1.

 

[BrintonAA]

<< Isn't i = sqrt(-1)?
i^2 = 1
i^3 = -1
.
.
.
i^2n = 1
i^2n+1 = -1 >>


No,
i = root -1
i^2 = -1
i^3 = -1 * i = -i
i^4 = 1
then
i^5 = i = root -1
and then repeat....

i^3 does not equal -1 because root -1 * root -1 * root -1 = -i NOT -1

 

[NelsonMuntz]

Well said BrintonAA, but to the original question, is that an equation that you only gave us half of, or are you just supposed to expand the polynomial (which you did successfully)?

 

[JohnCU]

I was just supposed to write a polynomial function with the given zeros (5, i, -i) with a leading coefficient of 1. The i just threw me off.