Please help and show work if possible.
1 answer:
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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!
Answer:
Step-by-step explanation:
I can't believe I'm doing this for 5 points, but ok!
For the first 3, we are going to multiply to find the value of that 3 x 3 matrix by picking up the first 2 columns and plopping them down at the end and then multiplying through using the rules for multiplying matrices:
and from there find the sum of the products of the main axes minus the sum of the products of the minor axes, as follows (I'm not going to state the process in the next 2 problems, so make sure you follow it here. This is called the determinate. The determinate is what you get when you evaluate or find the value of a matrix. Just so you know):
which gives us:
392 + 36 - 192 - [48 + 504 - 112] which simplifies to
236 - 440 which is -204
On to the second one:
and multiplying gives us
which gives us:
-504 + 96 - 9 - [-56 + 216 - 36] which simplifies to
-417 - 124 which is -541, choice c.
Now for the third one:
and multiplying gives us
which gives us:
which simplifies to
-168 - (-262) which is 94, choice c again.
Now for the last one. I'll show you the set up for the matrix equation; I solved it using the inverse matrix. So I'll also show you the inverse and how I found it.
=
and I found the inverse of the 2 x 2 matrix on the left.
Find the inverse by:
* finding the determinate
* putting the determinate under a 1
* multiply that by the "mixed up matrix (you'll see...)
First things first, the determinate:
|A| = (-4*-8) - (-6*-5) which simplifies to
|A| = 32 - 30 so
|A| = 2; now put that under a 1 and multiply it by the mixed up matrix. The mixed up matrix is shown in the next step:
(to get the mixed up matrix, swap the positions of the numbers on the main axis and then change the signs of the numbers on the minor axis). Now we multiply in the 1/2 to get the inverse:
Multiply that inverse by both sides of the equation. This inverse "undoes" the matrix that's already there (like dividing the matrix that's already there by itself) which leaves us with just the matrix of x and y. Multiply the inverse matrix by the solution matrix:
and that right side multiplies out to
x = 20 - 5 which is
x = 15 and
y = -15 + 4 which is
y = -11
(It works, I checked it)