Polynomial Functions (plus variables and imaginary numbers)

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Find the values of the function. Remember: i= sqrt(-1). Simplify.

15)
h(x)=2x^2-5x+6

15c) Put in h(1+i) for x.
Click to expand...

I have an answer, but it's wrong. I got -5i+1. The right answer is 1-i. How am I to go and properly solve this problem?

I'm really terrible with functions, especially with imaginary numbers.
 
This is for my Pre-Calculus class, but it's still review of Advanced Algebra.
I know I shouldn't be blaming this on my teacher, but she doesn't teach us anything. Even my friends who are really smart don't understand this.

But anyway. Here is my work.


15c)

h(1+i)=2(1+i)^2-5(1+i)+6
h(1+i)=2+2i^2-5-5i+6 2-5+6=3
h(1+i)=2i^2-5i+3 1^2=-1; 2(-1)=-2; -2+3=1
h(1+i)=-5i+1
Click to expand...

So my answer is -5+i, right? Following my work, I mean.

But the answer in the back of the book (and the one I took from the teachers manuel) are 1-i.

Did I not finish the equation???

Thanks.[/code]
 
In order for us to check your work and help you find your error, you will need to post that work. Please reply showing us what you did. Thank you.

Eliz.

P.S. The provided answer, "1 - i", is correct.

P.P.S. Remember to show all of your steps. For instance, when squaring out that first term, show:

. . .(1 + i)<sup>2</sup> = (1 + i)(1 + i)

. . . . .= 1 + 1i + 1i + i<sup>2</sup>

. . . . .= 1 + 2i - 1

...and so forth. Thank you.
 
I realized that after I posted. So I did. I think I may have found some of my errors by looking at your work....
 
1. h(1+i)=2(1+i)^2-5(1+i)+6 Simplify (1+i)^2 to (1+i)(1+i)
2. h(1+i)=2(1+i)(1+i)-5(1+i)+6 "FOIL"/distribute (1+i)(1+i)
3. h(1+i)=2(1+2i+i^2)-5(1+i)+6 "FOIL"/ distribute 2(1+2i+i^2)
4. h(1+i)=2+4i+2i^2-5(1+i)+6 Combine 2+6
5. h(1+i)=2i^2+4i-5(1+i)+8 simplify i^2 to -1
6. h(1+i)=-2+4i-5(1+i)+8 Combine -2+8
7. h(1+i)=4i-5(1+i)+6 "FOIL"/distribute -5(1+i)
8. h(1+i)=4i-5-5i+6 Combine Like terms
9. h(1+i)=-1i+1 ANSWER!
 
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