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First attempt to factorise each polynomial.
x^2 - x - 12 = (x - 4)(x + 3)
x^2 - 8x + 16 = (x - 4)(x - 4)
note: These two have a common factor (x - 4)
For the lowest common multiple we only need that factor once and both of the others.
Hence (x )(x )(x ) I'll leave you to complete the brackets.
Bob
The LCM is (x - 4)(x - 4)( x + 3) or (x - 4)^2 (x + 3).
I will not share my name in an open forum. I was born in the Dominican Republic in April 1965. I was raised in NYC. Of course, the city of New York has been completely abandoned and given to illegal aliens by the democratic party.
I live in NYC but I am not a fan of what this place has become. I joined this site to review and revisit mathematics learned long ago. Why not? What else can I do at 58? It's not like I'm popular with females, huh? By the way, I am also former Navy. I am a professional solo guitarist.
This is a good site. It needs to upgrade picture uploads. Make it easy for people to post an image. The steps to post an image here are insane. I tried for months to upload geometric photos using my cell phone but to no avail. There's gotta be a better way.
Science is not part of math. Math is part of Science. The two walk hand in hand. The same applies to physics.
Security work is a thankless job. I work for Mulligan Security at the Falchi Building. The site has a huge turn over rate because the "Chief Engineer" has been given full control of the guard staff. This man loves to verbally abuse the guards. I just lost two of my closest co-workers for the same reason.
They just couldn't take the verbal abuse anymore. What has the guard company done to stop the abuse? They have done nothing at all. Contract security guard companies care more about the contract and not about what the guards actually going through daily.
We are very short-staffed, overworked, verbally abused and weekly having check problems. I am 58 years old. It's hard for people over 50 to land a job. I am also former Navy.
I was a substitute teacher for 8 years in the toughest NYC public schools. I have two college degrees earned back in the 1990s but to no avail. What can I say? My life drags big time.
An electrical circuit contains three resistors connected in parallel. If the resistance of each is R_1, R_2, and R_3 ohms, respectively, their combine resistance R is given by the formula
(1/R) = (1/R_1) + (1/R_2) + (1/R_3)
Express R as a rational expression. Evaluate R for R_1 = 5 ohms, R_2 = 4 ohms, and R_3 = 10 ohms.
Find the LCM of the given polynomial.
x^2 + 4x + 4, x^3 + 2x^2, (x + 2)^3
Find the LCM of the given polynomial.
x^2 - x - 12, x^2 - 8x + 16
Use synthetic division to determine if x - c is a Factor of the given polynomial.
3x^4 + x^3 - 3x + 1; x + (1/3)
agree
Drop down 1.
(-2)(1) = -2
(-2) + (-2) = -4
(-2)(-4) = 8
8 + 3 = 11
(-2)(11) = -22
-22 + 5 = -17
From this I get x^2 -4x + 11, remainder -17.
a = 1, b = -4, c = 11 and d = -17
Sum = a + b + c + d
Sum = 1 - 4 + 11 - 17
Sum = -9
You say?
harpazo1965 wrote:Hi there. I am currently in the review section of my college algebra textbook in my self-study journey. Section R.6 covers Synthetic Division.
Divide 0.1x^3 + 0.2x by x + 1.1
add a term for the missing power:
0.1x^3 + 0x^2 + 0.2x + 0
do the synethtic division setup
synthetic division: -1.1 | 0.1 0.0 0.2 0.0 | -----------------
then do the steps
synthetic division: -1.1 | 0.1 0.00 0.200 0.0000 | -0.11 0.121 -0.3531 -------------------------- 0.1 -0.11 0.321 -0.3531
0.1(-1.1) = -0.11
-0.11 + 0.00 = -0.11
-0.11(-1.1) = 0.121
0.121 + 0.200 = 0.321
0.321 + -1.1 = -0.3531
Answer: 0.1x^2 - 0.11x + 0.321, remainder -0.3531
harpazo1965 wrote:Thank you for your time and effort.
That's fine...and I give up too.
I'm just disappointed I couldn't find the solution to your problem.
It's ok. You tried. It's a Android problem that I just don't know how to fix.
agree
Very good. I will show my work later.
After this question I will be moving on to Section R.7 in the textbook.
Chapter 1 begins real college mathematics. Math is cool. It takes my mind away from the problems of life that try to do me in.
harpazo1965 wrote:Hi there. I am currently in the review section of my college algebra textbook in my self-study journey. Section R.6 covers Synthetic Division.
Divide 0.1x^3 + 0.2x by x + 1.1
add a term for the missing power:
0.1x^3 + 0x^2 + 0.2x + 0
do the synethtic division setup
synthetic division: -1.1 | 0.1 0.0 0.2 0.0 | -----------------
then do the steps
synthetic division: -1.1 | 0.1 0.00 0.200 0.0000 | -0.11 0.121 -0.3531 -------------------------- 0.1 -0.11 0.321 -0.3531
Perfect. Thank you. I will post my work later.
I think the first step is to divide by (x + 2).
This will give me the value of a, b, c, and d.
Yes?
Find the sum of a, b, c, and d if
(x^3 - 2x^2 + 3x + 5)/(x + 2) = ax^2 + bx + c + (d/(x + 2)]
You say?
Hi there. I am currently in the review section of my college algebra textbook in my self-study journey. Section R.6 covers Synthetic Division.
Divide 0.1x^3 + 0.2x by x + 1.1
harpazo1965 wrote:Here is my url:
https://imgur.com/gallery/4Gh1VYy
You told to place between the following:
[ img ] https://imgur.com/gallery/4Gh1VYy [ / img ]
When I do that, there is no picture.
What am I still doing wrong?
Two things:
1. Image tags and the url within the image tags can't contain spaces, but yours do, and must be removed. I've marked the offending spaces with a red x:
(a) before and after the url: xhttps://imgur.com/gallery/4Gh1VYyx
(b) in the opening tag: [ximgx]
(c) in the closing tag: [x/ximgx]
Note: Image tags are a BBCode formatting option that must obey a specific syntax for them to work. See example here: BBCode (near the bottom of the 'Links and images' box).2. The url is to a webpage, and although that page contains the image (+ other stuff), the url must be to the image itself.
See the following...
phrontister (post #10) wrote:...the image extension can be any of '.gif', '.jpeg', '.jpg', '.png' (and probably more)...
The important thing when copying an image url is that the link you're copying must refer to an 'image address' (ie, with an 'image extension' as above), not something else like a 'link address'.
Btw, the direct Imgur url to an actual image begins https://i.imgur.com... ('i.' signifying it's an image link, I suppose).
phrontister (post #14) wrote:This is the URL of the image: https://i.imgur.com/t8F77rc.png
Note that it starts with https://i.imgur.com and ends with an image extension (png, in this case).
And from the image in my post #14:
"Tip: The URL starts with https://i.imgur.com, and ends with an ‘image extension’ (eg, ‘gif’, ‘jpeg’, ‘jpg’ or ‘png’)."Your https://imgur.com/gallery/4Gh1VYy doesn't start with https://i.imgur.com, and doesn't end with an image extension...
This is getting to be too much. The image link or code is not working on my phone. So, I will use this site mainly for help with applications. If a certain problem has an image, I will describe the geometric shape to be best of my ability. This is a good site. I will use it mainly for applications and equations.
Thank you for your time and effort.
harpazo1965 wrote:https://i.imgur.com/t8F77rc.png
I read into your post that you've worked out how to obtain an image link, and to paste it, with your Android! Correct?
If so, all that's left now is for you to enclose the url in img tags to display the image...like so:
[img]image url[/img]
Here is my url:
https://imgur.com/gallery/4Gh1VYy
You told to place between the following:
[ img ] https://imgur.com/gallery/4Gh1VYy [ / img ]
When I do that, there is no picture.
What am I still doing wrong?
I have edited your last post to show this, Bob
Bob,
Here is my url:
https://imgur.com/gallery/4Gh1VYy
I was told to place between the following:
[ img ] https://imgur.com/gallery/4Gh1VYy [ / img ]
When I do that, there is no picture.
I have edited your last post to show this, Bob
Cool.
harpazo1965 wrote:https://i.imgur.com/t8F77rc.png
I read into your post that you've worked out how to obtain an image link, and to paste it, with your Android! Correct?
If so, all that's left now is for you to enclose the url in img tags to display the image...like so:
[img]image url[/img]
You said:
If so, all that's left now is for you to enclose the url in img tags to display the image...like so: [img]image url[/img]
Can you show me what you mean by enclose the url in img tags? Which img tag specifically?
Here is my url: https://i.imgur.com/t8F77rc.png
Like this:
[img]https://i.imgur.com/t8F77rc.png[/img]
and here is your pic , Bob
Enclose that to show me what you mean.
Thanks.
https://i.imgur.com/t8F77rc.png
I found the following instructions on Google Support (here), and adapted them for Imgur + Android:
https://i.imgur.com/t8F77rc.png
This is the URL of the image: https://i.imgur.com/t8F77rc.pngNote that it starts with https://i.imgur.com and ends with an image extension (png, in this case).
I will search YouTube for a video lesson. Let me see if I can solve this problem once and for all.