# Let p(x)=2x^3-8x^2-23x+63. Use synthetic division to determine: p(-2i)

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Given `p(x)=2x^3-8x^2-23x+63` find `p(-2i)` using synthetic division.

p(-2i) is the remainder found after using synthetic division. (For this reason, synthetic division is often called synthetic substitution.)

-2i | 2 -8 -23 63

-4i -8+16i 32+62i

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2 -8-41 -31+16i 95+62i

**Thus p(-2i)=95+62i.**

You can check using substitution:

`p(-2i)=2(-2i)^3-8(-2i)^2-23(-2i)+63`

`=2(-8i^3)-8(4i^2)-23(-2i)+63`

`=-16i^3-32i^2+46i+63`

`=16i+32+46i+63`

`=95+62i`

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Thanks embizze,

Explaining the synthetic division part helped me understand how to do the problem. Only thing is I'm having problems getting my signs right. Such as when you multiply (-2i)(-4i) I keep getting +8 instead of -8.

I also have one more problem which is the same equation but to determine p(2+i)

I want to be able to learn how to do this myself maybe thru a youtube video to help me understand a little better. I've searched youtube but I couldn't really find anything with p(2+i) or anything similar. But I'm thinking I'm not searching correctly. Do you know where I could maybe find a video that would help explain.

Thanks

The polynomial p(x)=2x^3-8x^2-23x+63

p(-2i) = 2*(-2i)^3-8(-2i)^2-23*(-2i)+63

= 2*-8*i^3-8*4*i^2+ 46i+63

= 2*8*i+8*4+ 46i+63

= 16i + 32 + 46i + 63

= 62i + 95

**The polynomial p(-2i) = 62i + 95**

would this be considered as using synthetic division by replacing x with -2i?

sorry that -65 should have been a -2 as the remainder I typed it in wrong.

I'm confused on this one... thanks justaguide but it says to use synthetic division to determine. I first thought that you would replace x with "-2i" and I did go about solving this problem the way you said above. But all the rest of my synthetic divsion in this chapter dealt with using it like the following

Sorry trying to type a problem on a forum is hard for me. 5 being in the box and the 2 brought straight down.

**5** 2 - 8 - 23 + 63

+10 +10 - 65

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2 + 2 - 13 - 65

Answer being: 2x^2+2x-13+ 2/x-5