You are not logged in.
Thanks, Bob.
“It seems to me that you still have to experiment with factors to make the middle term.
I've got to find two suitable factors of 8. Could be 1 and 8; could be 2 and 4. I have to notice that 3x4 + 2x1 = 14. That's no different from what you usually have to do.”
We’re no longer looking for factors of 8, at least not initially. Once we multiply the a term (3) by the c term (8) we get 24.
Then we list the factors of 24. From those we can immediately see that 2 and 12 add to give the b term, the middle term (14).
“I think this is just a slightly different twist on trying to 'fiddle' the middle term by experimenting with factors for the constant term. What I'd usually do is try 3 and 1 with 1 and 8; and try 3 and 1 with 2 and 4. I'd get to 3x4 + 2x1 just the same.”
With this method you don’t have to fiddle in this manner. You don’t have to try 3 times this, and 1 times that, with different factor pairs, in different permutations. All you have to do is choose the pair which add to give the middle term then plug them in to your brackets.
The only extra thing you then have to do is cancel where you can; in this case cancel (3x+12) to make it (x+4)
P.S. I've edited the mistake in the first example in my original post, so's not to confuse any newcomers to the thread!
@amnkb
I don't find the method confusing (see my reply to Bob), as in, how to get the correct answer. I just don't know WHY it works. You asked where I saw it. The teaching assistant in my maths class showed me it. It does save on the trial and error of the first method I showed.
Multiply the a term by the c term (3 x 8 = 24). Then take the pair of factors of 24 that give the b term (14), which would be 2 and 12. Then put the 2 in one bracket, and the 12 in the other. Then cancel the 3x+12 to make it x + 4. And you'll have your factors; (3x+2)(x+4), which equal 3x^2 + 14x + 8.
I'm not seeing the method here. Suppose I have
So I start with (3x )(3x ) but then what?
Bob
Sorry, I made a mistake on the first example. The brackets should've contained 2x, and x, to begin with, not "2^2 and x
paulb203 wrote:Example: 2x^2 + 9x + 4
My way of doing this would be to write out a pair of brackets, 2x^2 in one, x in the other. Then list the factors of 4. Then choose the pair that will give 9 when one of the pair is multiplied by 2, and the other one is multiplied by 1. In this case the pair would be 4, and 1. So:
(2x^2 +1) (x+4)
but this doesnt multiply back to what you started with
you'll get 2x^3+8x^2+x+4paulb203 wrote:But I've been shown a new method which is puzzling me.
Step 1. Multiply the a term by the c term (the 2 by the 4) to give 8. Then write out a pair fo brackets but his time put 2x in both of them, like this:
(2x ) (2x )
where did you see this?
youre right its confusing!
Example:
2x^2 + 9x + 4 (ax^2 + bx + c)
My way of doing this would be to write out a pair of brackets, 2x in one, x in the other. Then list the factors of 4. Then choose the pair that will give 9 when one of the pair is multiplied by 2, and the other one is multiplied by 1. In this case the pair would be 4, and 1. So:
(2x +1) (x+4).
*
But I've been shown a new method which is puzzling me.
Step 1. Multiply the a term by the c term (the 2 by the 4) to give 8. Then write out a pair of brackets but this time put 2x in both of them, like this:
(2x ) (2x )
Then list the factors of 8. Then choose the pair which add to give 9, in this case 1, and 8. Then put the pair into the brackets:
(2x + 1) (2x +8 )
Then see where you can cancel:
(2x +1) remains the same, but (2x + 8) can be divided by 2 to become (x+4)
So our answer is:
(2x+1)(x+4)
Here is where I'm puzzled. I can use the second method to get the correct answer, and it's quicker, I think, than the first method. But I understand the first method. I know why I am taking each step. With the second method I'm left wondering why does it work.
Thanks, Bob.
I found out what it actually is. It's a polygonal chain. So they should be called Frequency Polygonal chains
Is a frequency polygon actually a polygon?
It looks to me like a series of connected lines, a shape, but not a closed one.
Do polygons need to be closed?
If it's not actually a polygon why do they call it one?
Thanks, Bob.
“An acceleration of zero is still an acceleration.”
I Googled this, Bob, and got;
-‘Acceleration of an object can be zero when it is moving with a constant velocity. Since velocity is constant, there will be no change in velocity and so there will be no acceleration.’
But the apple, relative to the hand at least, is not in motion. I get that the two are moving with the Earth, and the solar system, etc, but the apple is at rest relative to the hand, no? So I’m thinking that if the apple has no motion in this respect how can it have velocity. And if it has no velocity, in this respect, how can it have acceleration.
“The situation is more complicated than this however as the person's hand is actually rotating about the polar axis as the Earth rotates. It's not easy to appreciate this if you're just holding an apple, but it's a major factor in causing the world's weather patterns.”
Thanks. It’s always helpful for me to be reminded of this, that nothing is at rest, except relative to something else.
“Replace 'acceleration due to gravity' by 'weight of apple' and it's easy to see why you need a force balancing equation.”
Thanks. I think I get the idea of weight being a force. My mass is 90kg. W=mg means there is a force of gravitational attraction between me and the Earth of 900N, approx, yes? But then I get confused again when I analyse w=mg, which is weight = mass x g, and g is acceleration due to gravity! Back to acceleration. I’m standing on my kitchen floor and, relative to the Earth, I’m at rest. Yet I’m accelerating? At 9.8m/s/s?
One of the first examples of gravity given on this site explains that a force of 0.98 N is required to hold up an apple with a mass of 0.1kg.
F=ma is shown, which translates to F=mg, as a in this context is 9.8m/s/s due to gravity
So, 0.98N = 0.1kg x 9.8m/s/s
I think this is starting to make sense to me but at the same time I'm puzzled regards the use of acceleration in this example.
Is the apple accelerating? It seems to be at rest, in one's hand, no?
Thanks, Bob. I wish the website where I got v=at had put a subscript after v indicating that this equation only works for instantaneous velocity
The equation v=delta x/t usually has a bar line over the v to indicate average v, yes?
On this website, and many others, the velocity of a freefalling object in a vacuum is shown as follows.
After 1,2,3 and 4 seconds respectively;
9.8m/s; 19.6m/s; 29.4m/s; 39.2m/s
I worked out the distance travelled by a freefalling object in a vacuum using d = at^2/2, or, d = 0.5gt^2 and got, for after 1,2,3 and 4 seconds;
4.9m; 19.6m; 44.1m; 78.4m
But when I tried to double check the above using the formula for velocity, v= ΔX/t, I got, for after 1,2,3 and 4 seconds;
4.9m/s; 9.8m/s; 14.7m/s; 19.6m/s
Where have I gone wrong?
I’m always trying to bear in mind average velocity versus instantaneous velocity and I’m aware that there are several formulas for velocity but I thought those were mostly rearrangements of each other.
I see though that v=at works for the velocities at the top of this page.
How do v= Δx/t and v=at relate to each other?
Thanks, Bob.
Here's the comments on the video;
The normal force and its relation to weight, has nothing to do with Newton's third law. It is a coincidence that it is equal and opposite, but it is not the third law counterpart to weight. Its existence has a lot more to do with Newton's second law, as a constraint force to stop the object from accelerating through the surface beneath it. The third law counterpart to your weight, is the weight of Earth in your own gravitational field. The fact that gravity is a two way street, where both the small and large object pull on each other with equal and opposite forces.
*
Careful with invoking 3rd law and saying the normal force is the reaction force to gravity. The 3rd law reaction force to the earth pulling on a mass is the mass pulling on the earth. The normal force is often equal to gravity when a=0 because of Newton's 2nd law
*
Normal force is fun fact a type of elastic force. Normal force is the opposite of pressure force(the F in pressure equation, p=F/A), they are both caused by objects don’t elastic deformation, which our eyes can’t see. So pressure force is caused by your contacting area doing elastic deformation ever so slightly which caused an elastic force and normal force is caused by the other contacting object like the wall doing elastic deformation ever so slightly which also caused an elastic force. If you are wondering, elastic deformation is the cause of elastic force, it is basically a change in an object’s shape caused by a constant force, but will turn back into how it was with the force is not present, example of that is a spring or a rubber band. And elastic force is just a force caused by that. However it also can be explained in energy, it is caused by elastic potential energy turning into kinetic energy.
Thanks, Bob
“...places on Earth where the gravitational pull is more or less; eg up a mountain.”
Because we’re further away from Earth’s centre?
“I don't see what elasticity has got to do with it. Do you have a link to the vid?”
Here you go, Bob. https://www.youtube.com/watch?v=57gKnA94Doc The video doesn’t mention elasticity, it’s someone in the comments. In the first handful of comments 2 people say the Normal force is not an example of Newton’s 3rd Law, and 1 person says it’s a type of elastic force.
“I think they are both N 3rd L.”
Thanks. I watched another video and found out that the Normal force doesn’t only apply to objects on the ground, or on horizontal surfaces, it can apply to, for example, me standing upright pushing on a door.
“They're good enough to land a man on the Moon...”
Did they use stuff like F=ma, and W=mg, etc?
“...Pluto isn't big enough to 'earn' the title planet so it's been downgraded to a minor planet.”
I like telling people it’s a dwarf planet, because they discovered dwarves on it.
“Arithmetic doesn't work in some cases though. 1 + 1 = 1 when you're talking about piles of sand!”
1 pile of sand plus another pile of sand = 2 piles of sand, no? Or do you mean 1 pile piled onto another?
“When Einstein came up with his theory it showed that Newton is only an approximation but relativity isn't consistent with quantum mechanics so the search is on for a theory that fits both.”
So relativity (general relativity) isn’t quite ‘right’ either, similar to Newton’s laws? I’ve heard often that quantum mechanics is spot on, the most accurate theory ever, in terms of predictions, something like that?
“Binary arithmetic was a curiosity for many years. But it is essential in the operation of computers; suddenly an abstract theory blossoms into life as being really useful.”
Base 2? Ones and zeroes?
Thanks, Bob.
“If you put a bag of sugar on some kitchen scales you'll get a reading which may be expressed in kilograms or pounds weight, but is actually a measure of the force the bag is exerting on the scales. It's a force because if you could repeat this test on the Moon you'd get a lower value even though the mass of sugar is unchanged.”
Is that why a person weighs less on the Moon than they do on Earth?
“As the bag isn't moving downwards there must be equilibrium at the point of contact. That means there must be two, not one, force acting here; the force of gravity downwards and what is called the reaction of the scales acting upwards on the bag.”
So for the force of gravity downwards it would be;
F=ma. F=1kgx10m/s^2. F=10N downwards?
Would the reaction force of the scales acting upwards on the bag match/mirror this?
F=ma. F=1kgx?. What would the acceleration upwards be? If you removed the 1kg bag of sugar would the top part of the scale spring up at 10m/s^2?
Nb; I’m using 10m/s^2 as an approximation for g.
“Is the door any different after you have pushed it? I tried this and asked the door but it refused to answer.”
“If you pushed really hard you might break it. I expect every time you open and shut a door the hinge screws cut a little more into the wood until eventually you have to put in some larger ones.”
I saw a Youtube video about the normal force. There was a diagram of a box on a floor. It was given as an example of Newton’s 3rd Law but a few people in the comments said it wasn’t, that it was more about elasticity than about Newton’s 3rd Law. Is that correct? Was I incorrect in thinking that my example of me pushing on a door was an example of Newton’s 3rd Law? Is me pushing on a door different from a box resting on a floor?
“But it's just a model. At an atomic level the door is mostly open space with neutrons etc hanging about inside. Newton's laws are not consistent with Einstein's theories nor with quantum mechanics so my advice is use them when appropriate but don't get too hung up on why they work.”
By model do you mean an approximation? They are roughly correct, and work (in engineering etc), but when we look more closely they’re actually incorrect?
If I push a door the door pushes me, yes?
This seems strange to me. When I push a door I am making an effort to do something, it is a form of exertion. I am trying. I will burn a calorie or two in the process. I am actively doing something.
The door, it seems to me, is not making an effort, not trying, not exerting itself, not burning any calories, not actively doing something, it’s just standing there (I think?)
The same goes for me pushing a car (I think?).
All the inanimate objects/bodies in the textbook physics examples are, it seems to me, just sitting there, or standing there, while the person in the example is actively doing something.
Q. Is the kind of push ‘done’ by the door/car/etc, a different kind of push to the one a person does?
Q. And if a force is an interaction between objects involving a push or a pull, often with one or more of the objects being inanimate, might it be more helpful to think of the push and pull in terms of repulsion/attraction, given the implications of human pushing and pulling?
Q. After I push the door I am changed, for example, out of breath, an ounce or two lighter, etc; is the door any different than it was, having pushed me?
At the end of the Maths is Fun page on, “Algebra – Basic Definitions” there is a brief explanation of like terms.
One of the terms given in the example is (1/3)xy^2, the other two are, -2xy^2, and, 6xy^2.
What is the significance of the brackets around 1/3?
In this context would there be a difference between (1/3)xy^2, and just, 1/3xy^2?
Also, what would (1/3)xy^2 – 2xy^2 + 6xy^2 equal?
I’m confident that it would at least equal; (1/3)xy^2 + 4xy^2?
But could we go further and get; 4 1/3 xy^2? Or do the brackets prevent us from adding the 1/3 and the 4?
There are some problems I can’t seem to come up with real life examples for, e.g;-5x-2=10
If I think of the -5 as a debt of £5 I can’t make sense of multiplying this by -2 as there is no such thing as -2 people (2 people with a debt of £5 each makes sense, but not negative two people etc).
And the inverse process doesn’t help me either (10÷-2=-5). I can strive to remember the rules, and use the inverse process to check my answers, but to really understand this I think I need to be able to come up with real life examples.
Thanks, Bob.
I do find looking at the inverse process helpful when checking my answer.
“But all these definitions amount to the same thing so it is up to you which you prefer.”
One of the first Google results I got for my question was;
“Division in maths is the process of breaking a number up into equal parts, and finding out how many equal parts can be made. For example, dividing 15 by 3 means splitting 15 into 3 equal groups of 5.”
Is this answer wrong, or at least incomplete? Should it read, ‘One way of understanding division is... etc’? It’s not said that another way of looking at 15÷3 is to ask, ‘How many 3s are there in 15?’
Either way, I do find it interesting. With some questions I find it helpful to think of a physical object (often a cake) being shared amongst children. 1÷2, 1÷4, etc. With others a different approach cab be more helpful.
What about negative numbers, when it comes to division?
If I’m faced with -10÷2 I think of a debt of £10 being split into 2 debts of £5 (-10÷2=-5) with maybe 2 people involved, each taking on a debt of £5
When faced with -5x2 I think of someone having 2 separate debts of £5 and therefore being £10 in debt (-5x2=-10)
But when it comes to, for example, -10÷-5, it seems less straightforward. I know the rule (a negative divided by a negative = a positive)so I know it goes as follows; -10÷-5=2; but I can’t think of a ‘debt’ example like above (because there is no such as, ‘-5 people’. Although could I think of it as, ‘How many -5s are there in -10, in the same way as I might think of 10÷5 as, ‘How many 5s are there in 10’? So a debt of £10 is the same as 2 debts of £5
Thanks, Ganesh.
Does, for example, 10÷2 mean 10 split into 2 equal parts?
Or does it mean how many 2s are in 10?
If we split 10 into 2 equal parts we get 2 parts of 5.
If we ask how many 2s are in 10 we get 5 2s.
Almost the same answer for both questions. Well, the same answer if we simply say, ‘5’.
This question never occurred to me when I was solving simple problems like 10÷2, or 1÷4, etc. But when faced with, for example, 10÷1/2, I had to think about it more.
Does 10÷1/2 mean 10 split into... No, that doesn’t seem to work.
Does 10÷1/2 mean how many 1/2s are in 10?
10÷1/2=20. There are 20 1/2s in 10. That seems to work.
Is that my answer; the best way to understand division is by asking how many x’s are in y?
Thanks, Ganesh
Ha! My emoji was supposed to be a smiley face
Thanks a lot, Bob. That was very helpful. I had fun working out the 10+5n example before looking at the answer.
I did;
100-10=90
90/5=18
But I couldn't come up with the expression 10+5n is greater than 100 )
I had noted the warning about changing the sign from the examples on the main website but your examples were good revision. Thanks.