How would I write this equation in standard form?

vgamelover123

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Jun 6, 2020
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I tried 5 times and no luck it gives a different format.
 

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You know by now that we do not solve problems for students.
Show us your work and we will inform you where you made any mistakes.
 
Start by multiplying out your factors. Since you know that x = 1 [MATH]\pm[/MATH] i and x = 2 are roots, then what were those factors?
 
You know by now that we do not solve problems for students.
Show us your work and we will inform you where you made any mistakes.
I did show my work in the other screencap but it gives me a different format
 
I did show my work in the other screencap but it gives me a different format
You showed your answer, not your work; and the difference is not in the format, but the specific numbers.

Please show how you got your answer, so we can help you find your error.
 
I did show my work in the other screencap but it gives me a different format
Do you really think that I will consider looking at every possible error that one can make until I figure out your error just because you chose not to show your work? I don't think so. Good luck.
 
I did show my [answer] in the [second screenshot] …
Hello vgamelover. Your answer for the second exercise suggests that you used 4±i instead of -4±i. Please try the second exercise again.

In the first exercise, we start by multiplying (x-2)(x-1-i)(x-1+i). Is that what you did?

If you'd like more help with the first exercise, then please post what you tried. Thanks!

?
 
A polynomial that has zeros at \(\displaystyle x= 1\pm i\) and \(\displaystyle x= 2\) must be a multiple of \(\displaystyle (x- 1- i)(x- 1+ i)(x- 2)\). To multiply \(\displaystyle (x- 1- i)(x- 1+ i)\) write it as \(\displaystyle ((x-1)- i)((x-1)+ i)\) and treat it as \(\displaystyle (a- b)(a+ b)= a^2- b^2\): \(\displaystyle (x- 1)^2- (i)^2= x^2- 2x+ 1- (-1)= x^2- 2x+ 2\). Multiply that by x-2. Finally, multiply by a constant to make f(0)= 4.
 
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