Ok part 2 hey hey it jassie can you answer all my parts and other people to ok here it is
2 answers:
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Answer:
Step-by-step explanation:
You need to use synthetic division to do all of these. The thing to remember with these is that when you start off with a certain degree polyomial, what you get on the bottom line after the division is called the depressed polynomial (NOT because it has to math all summer!) because it is a degree lesser than what you started.
a. 3I 1 3 -34 48
I'm going to do this one in its entirety so you get the idea of how to do it, then you'll be able to do it on your own.
First step is to bring down the first number after the bold line, 1.
3I 1 3 -34 48
_____________
1
then multiply it by the 3 and put it up under the 3. Add those together:
3I 1 3 -34 48
3
----------------------------
1 6
Now I'm going to multiply the 6 by the 3 after the bold line and add:
3I 1 3 -34 48
3 18
_________________
1 6 -16
Same process, I'm going to multiply the -16 by the 3 after the bold line and add:
3I 1 3 -34 48
3 18 -48
___________________
1 6 -16 0
That last zero tells me that x-3 is a factor of that polynomial, AND that the depressed polynomial is one degree lesser and those numbers there under that line represent the leading coefficients of the depressed polynomial:
Factoring that depressed polynomial will give you the remaining zeros. Because this was originally a third degree polynomial, there are 3 zeros as solutions. Factoring that depressed polynomial gives you the remaining zeros of x = -8 and x = 2
I am assuming that since you are doing synthetic division that you have already learned the quadratic formula. You could use that or just "regular" factoring would do the trick on all of them.
Do the remaining problems like that one; all of them come out to a 0 as the last "number" under the line.
You got this!