Time to go back to straight line geometry for a bit.
With knowledge of how to calculate the gradient, the perpendicular gradient, the mid-point and the length of the line, you've also done quite a lot of co-ordinate geometry of the circle. This is an interesting concept. A circle on a set of cartesian co-ordinates.
You start from the premise of a circle with centre (0,0). The equation of this circle is x2 + y2 = r2. If you study a diagram of it, or even draw one yourself, the reason for this becomes clear. Like a lot of co-ordinate geometry, it's based on Pythagoras. In this case, drawing a triangle inside the circle to a point somewhere on the circumference. Say that this circle with centre (0,0) passes through the point (3,4).
The line from the centre to the point functions as the hypoteneuse of the right angled triangle, while line x measures the distance travelled horizontally to the point, and the line y measures the vertical distance to the point.
The equation of this particular circle then tells us the radius.
32 + 42 = r2
or 9 + 16 = r2
= 25
r = 5.
The radius of the circle is 5.
Things get a little more complicated if the circle's centre is not (0,0). Say we have a circle with centre (2,3). We still devise the equation of the circle based on its distance from (0,0).
This circle's centre has moved 2 units x-wards and 3 y-wards (across and up) from the origin. Any calculation of the radius needs to take into account the fact that we are already some way from the origin.
So if the circle passes through the point (5,8), the distance from the centre to the point will be 5-2 and 8-3 (otherwise we're talking about including space outside the circle).
So the equation will be: (x-2)2 + (x-3)2 = r2.
In this case: 9 + 25 = 34
So r = √34
It is perfectly acceptable to leave the answer as a surd at C1, indeed often you have to.
Usually you are given the equation, say (x-4)2 + (y-7)2 = 4
or something.
The centre of the circle is ALWAYS THE OPPOSITE OF THE NUMBERS IN BRACKETS, in this case (4,7). Worth remembering. If it is (x-a)2 + (y-b)2 = r2 the centre is always (a,b).
Sunday 14 September 2008
Saturday 13 September 2008
A Note on Revision
Since I've finished C1 and S1 I am sort of both doing C2 and revising C1.
If I were to give one piece of advice, it would be to treat revision as a dynamic process. Do it as soon as you finish a topic or a module. Do it randomly, answer questions out of the blue, keep coming back to it.
Revision is only boring and hard when you have piles of stuff you barely remember to do in a short space of time. If you do it regularly, treating it like another aspect of your learning, then it becomes easier and things go in better. Writing this blog has been a valuable source of revision for me as well - I hope - as some of my readers.
Keep on keeping on revising. That's the sum total of my advice.
If I were to give one piece of advice, it would be to treat revision as a dynamic process. Do it as soon as you finish a topic or a module. Do it randomly, answer questions out of the blue, keep coming back to it.
Revision is only boring and hard when you have piles of stuff you barely remember to do in a short space of time. If you do it regularly, treating it like another aspect of your learning, then it becomes easier and things go in better. Writing this blog has been a valuable source of revision for me as well - I hope - as some of my readers.
Keep on keeping on revising. That's the sum total of my advice.
Blog Purpose Update
Well I know that I don't have all that many readers here but I don't mind too much. To judge from recent Sitemeter reports some people are actually reading what I have written, even if they are not so interested in commenting.
A quick reminder. I am no expert, in anything. Just a vaguely mature student of A Level maths who is doing it in his spare time when he isn't down the pub or surfing the blogosphere with increasing frustration and disappointment. So this blog was intended to be a wholly positive space in which I could devote time and effort to something of thought, however superficial. I do have a couple of problems though, especially with graphics and symbols. I still haven't worked out how to show these in blogposts properly. I have heard of something called TeX or LaTeX but i don't know how to use it.
If you have found anything on here useful, then I'm glad. It will be the first geniune contribution a blog of mine has made to the world.
A quick reminder. I am no expert, in anything. Just a vaguely mature student of A Level maths who is doing it in his spare time when he isn't down the pub or surfing the blogosphere with increasing frustration and disappointment. So this blog was intended to be a wholly positive space in which I could devote time and effort to something of thought, however superficial. I do have a couple of problems though, especially with graphics and symbols. I still haven't worked out how to show these in blogposts properly. I have heard of something called TeX or LaTeX but i don't know how to use it.
If you have found anything on here useful, then I'm glad. It will be the first geniune contribution a blog of mine has made to the world.
Thursday 11 September 2008
Dividing Polynomials (ii)
Alright then, let's have a crack at this properly.
We are armed with some fraction revision and a little bit of knowledge of polynomials, including factorising using the system of comparing coefficients.
We only need to divide a polynomial by a linear factor for C1.
Division has this nasty habit of not being exact, but of leaving remainders. Unfortunately, 5 does not go easily into 24 but leaves a remainder. Polynomials are also subject to this problem. A linear expression might not be a factor of P(x) as such, but might go into something else, leaving factors and a remainder when asked to be slotted into P(x).
Remember that QUADRATIC EQUATIONS sometimes factorise into TWO LINEAR expressions,
while CUBICS factorise into LINEAR + QUADRATIC (and sometimes three linear).
So if you divide a quadratic by a linear, this might happen:
x2 + 2x + 4 divided by x-2.
If we are to use the concept of comparing coefficients, which is very efficient, we first rewrite the equation into the form it would take expressed as factors and remainders:
x2 + 2x + 4 = (x+p)(x+q) + r.
Got it?
We need to divide it by x-2 though, so we already have one desired factor (THIS DOES NOT MEAN THAT x-2 WILL BE A FACTOR - just that we are trying to express the quadratic in terms of how much x-2 goes into it.
So:
x2 + 2x + 4 = (x-2)(x+q) + r.
Then we collect the different terms in this form. We are not completely, expanding the brackets, we are re-arranging them in terms of their own factors in order to look for coefficients which will tell us the other factors and remainders.
x2 + 2x + 4 = x2 + (-2 + q)x + (-2q+r)
F(x) is in this case equal to x2 and (-2+q) lots of x, and then the two constants, with no x variable, (-2q and r, the possible remainder).
Therefore 2= (-2+q) so q= 4.
4 = (-2x4 + r)
4=8+r
r = -4
x2 + 2x + 4 = (x-2)(x+4) -4.
This only expresses the quadratic as factors and a remainder. The division isn't over yet.
Now let's actually divide the quadratic by (x-2).
x2 + 2x + 4
-------------
x-2
=
(x+4) - 4
----
(x-2)
I'll explain it later. Proper work beckons.
We are armed with some fraction revision and a little bit of knowledge of polynomials, including factorising using the system of comparing coefficients.
We only need to divide a polynomial by a linear factor for C1.
Division has this nasty habit of not being exact, but of leaving remainders. Unfortunately, 5 does not go easily into 24 but leaves a remainder. Polynomials are also subject to this problem. A linear expression might not be a factor of P(x) as such, but might go into something else, leaving factors and a remainder when asked to be slotted into P(x).
Remember that QUADRATIC EQUATIONS sometimes factorise into TWO LINEAR expressions,
while CUBICS factorise into LINEAR + QUADRATIC (and sometimes three linear).
So if you divide a quadratic by a linear, this might happen:
x2 + 2x + 4 divided by x-2.
If we are to use the concept of comparing coefficients, which is very efficient, we first rewrite the equation into the form it would take expressed as factors and remainders:
x2 + 2x + 4 = (x+p)(x+q) + r.
Got it?
We need to divide it by x-2 though, so we already have one desired factor (THIS DOES NOT MEAN THAT x-2 WILL BE A FACTOR - just that we are trying to express the quadratic in terms of how much x-2 goes into it.
So:
x2 + 2x + 4 = (x-2)(x+q) + r.
Then we collect the different terms in this form. We are not completely, expanding the brackets, we are re-arranging them in terms of their own factors in order to look for coefficients which will tell us the other factors and remainders.
x2 + 2x + 4 = x2 + (-2 + q)x + (-2q+r)
F(x) is in this case equal to x2 and (-2+q) lots of x, and then the two constants, with no x variable, (-2q and r, the possible remainder).
Therefore 2= (-2+q) so q= 4.
4 = (-2x4 + r)
4=8+r
r = -4
x2 + 2x + 4 = (x-2)(x+4) -4.
This only expresses the quadratic as factors and a remainder. The division isn't over yet.
Now let's actually divide the quadratic by (x-2).
x2 + 2x + 4
-------------
x-2
=
(x+4) - 4
----
(x-2)
I'll explain it later. Proper work beckons.
Dividing Polynomials
As promised....
One of the key elements of this is the ability to split off fractions effectively.
Think of a normal, but improper fraction, such as 4/3. This could be expressed as 3/3 + 2/3 (or any other combination but this is clean and tidy and useful for us now). 3/3 is of course 1. We are left with 1 + 1/3.
The same thing can be done algebraically, so as to solve basic division questions.
Take (x + 6)/x. As above, we can think of this as x/x + 6/x. x/x = 1, so the answer is 1 + 6/x.
We use this property, making part of the numerator match the denominator, in dividing polynomials.
One of the key elements of this is the ability to split off fractions effectively.
Think of a normal, but improper fraction, such as 4/3. This could be expressed as 3/3 + 2/3 (or any other combination but this is clean and tidy and useful for us now). 3/3 is of course 1. We are left with 1 + 1/3.
The same thing can be done algebraically, so as to solve basic division questions.
Take (x + 6)/x. As above, we can think of this as x/x + 6/x. x/x = 1, so the answer is 1 + 6/x.
We use this property, making part of the numerator match the denominator, in dividing polynomials.
Sunday 7 September 2008
Rounding Up Factorisation
...so after all that you now have the technology (all right, the skills) to factorise a polynomial completely. You first do trial and error to find the first linear factor, then you multiply it by ax2 + bx + c and compare coefficients to find the quadratic. Then you check if the quadratic factorises further.
Onto dividing polynomials proper....
Onto dividing polynomials proper....
Factor Theorem (ii)
Well that's all very neat and tidy, you might be thinking, but what good is this theorem?
It gives you a lovely and easy way of trying to find linear factors by substituting values into the polynomial you are given. Sometimes this does not work easily but it is often worth a try for a minute or so.
It also enables you to do slightly trickier things with polynomials.
You often get questions like this (we'll use the cubic from the previous post for this). A polynomial, P(x) = x3 + ax2 + bx + 6, has (x+1) and (x+2) as factors. Find the values of a and b.
This looks tricky but it is where the factor theorem comes in.
P(-1) = 0. You know this from the theorem.
that means that (-1) + a(1) + b(-1) + 6 = 0
so -1 + a - b + 6 = 0
therefore a - b = -5
P(-2) = 0.
So -8 + 4a -2b + 6 = 0
therefore 4a - 2b = 2
You now have two equations to solve SIMULTANEOUSLY!!
a - b = -5
4a - 2b = 2
Times the first equation by 2:
2a - 2b = -10
4a - 2b = 2
SUBTRACT THE TWO TO ELIMINATE b
-2a = -12
a = 6
Plug this back into one of the equations:
2(6) - 2b = -10
-2b = -22
b = 11
a = 6, b = 11
There is nothing that hard here and yet in just writing this post I have made LOADS of mistakes, which is why it's taken me an hour or so to write.
Here are the errors I made:
1) Using the wrong constant value - ie reading the cubic wrong;
2) Forgetting to cube or square the values in the cubic equation;
3) subtracting with minus signs incorrectly;
4) multiplying incorrectly.
All because I was surfing the blogs while writing this post!
It gives you a lovely and easy way of trying to find linear factors by substituting values into the polynomial you are given. Sometimes this does not work easily but it is often worth a try for a minute or so.
It also enables you to do slightly trickier things with polynomials.
You often get questions like this (we'll use the cubic from the previous post for this). A polynomial, P(x) = x3 + ax2 + bx + 6, has (x+1) and (x+2) as factors. Find the values of a and b.
This looks tricky but it is where the factor theorem comes in.
P(-1) = 0. You know this from the theorem.
that means that (-1) + a(1) + b(-1) + 6 = 0
so -1 + a - b + 6 = 0
therefore a - b = -5
P(-2) = 0.
So -8 + 4a -2b + 6 = 0
therefore 4a - 2b = 2
You now have two equations to solve SIMULTANEOUSLY!!
a - b = -5
4a - 2b = 2
Times the first equation by 2:
2a - 2b = -10
4a - 2b = 2
SUBTRACT THE TWO TO ELIMINATE b
-2a = -12
a = 6
Plug this back into one of the equations:
2(6) - 2b = -10
-2b = -22
b = 11
a = 6, b = 11
There is nothing that hard here and yet in just writing this post I have made LOADS of mistakes, which is why it's taken me an hour or so to write.
Here are the errors I made:
1) Using the wrong constant value - ie reading the cubic wrong;
2) Forgetting to cube or square the values in the cubic equation;
3) subtracting with minus signs incorrectly;
4) multiplying incorrectly.
All because I was surfing the blogs while writing this post!
Saturday 6 September 2008
The Factor Theorem
Well that stuff in the previous post leads us nicely on to the factor theorem. Although this sounds like a low budget 70s sci-fi drama, it is in fact a key part of A Level maths.
The factor theorem is dead simple. If you have a polynomial x3 + 6x2 + 11x + 6, it has factors, namely a linear and a quadratic.
Only this time, the quadratic itself factorises, so that the cubic has 3 linear factors, namely (x+1)(x+2)(x+3).
If I now did something a bit sneaky, I could prove that these are factors of this cubic.
Watch. P(-1) = -13 + 6(-12) + 11(-1) + 6
= -1 + 6 - 11 + 6
= 0
When x=-1 the cubic equation = 0.
This proves that (x+1) is a factor of x3 + 6x2 + 11x + 6
In fact, if (x-a) is a factor of P(x) then P(a) = 0. Always and everywhere.
Why?
Well, look at (x+1)(x+2)(x+3).
These are all factors of x3 + 6x2 + 11x + 6.
So if you put x= -1 into the first factor, it will equal 0 (-1+1) and therefore the whole polynomial will equal 0. The same goes for all the other factors.
If a value, a, put into P(x) does not equal 0, then (x-a) IS NOT a factor.
The factor theorem is dead simple. If you have a polynomial x3 + 6x2 + 11x + 6, it has factors, namely a linear and a quadratic.
Only this time, the quadratic itself factorises, so that the cubic has 3 linear factors, namely (x+1)(x+2)(x+3).
If I now did something a bit sneaky, I could prove that these are factors of this cubic.
Watch. P(-1) = -13 + 6(-12) + 11(-1) + 6
= -1 + 6 - 11 + 6
= 0
When x=-1 the cubic equation = 0.
This proves that (x+1) is a factor of x3 + 6x2 + 11x + 6
In fact, if (x-a) is a factor of P(x) then P(a) = 0. Always and everywhere.
Why?
Well, look at (x+1)(x+2)(x+3).
These are all factors of x3 + 6x2 + 11x + 6.
So if you put x= -1 into the first factor, it will equal 0 (-1+1) and therefore the whole polynomial will equal 0. The same goes for all the other factors.
If a value, a, put into P(x) does not equal 0, then (x-a) IS NOT a factor.
Dividing Polynomials (1)
This can actually be a bit tricky but it's not impossible.
Any polynomial of order 3 - which is what you deal with at C1 - is the product of either linear factors, or a linear and a quadratic factor.
Finding a factor like this, you don't need to worry about remainders but these crop up soon enough, and we'll come onto them in due course.
A basic example of what I mean is: x(x2 + 2x - 4). Multiplying these two factors gives the cubic equation x3 + 2x2 - 4x.
But what do you do if you know one factor and the polynomial?
Say you know the linear factor to be (x-2) and the polynomial to be x3 + x2 - 11x + 10
The other factor must be a quadratic.
(x-2)(ax2+ bx + c) = x3 + x2 - 11x + 10
expand the two brackets as you would normally:
ax3 + bx2 + cx - 2ax2 -2bx - 2c
and then find common factors to tidy this expression up and make it usable:
= ax3 + (b-2a)x2 + (c-2b)x -2c
That bit is worth checking. You're simply collecting together the different values of x2 or whatever, as you would normally in an equation. You're only putting brackets round them because you don't know their values and you need to understand them as together making x2.
This expression is now completely identical to our original polynomial:
= ax3 + (b-2a)x2 + (c-2b)x -2c
x3 + x2 - 11x + 10
We now go through the fairly simple but again easy to cock up process of comparing the co-efficients. Co-efficients are of course the numbers in front of variables (3 is the coefficient of x in the expression 3x).
Comparing the co-efficients here means we look at our original polynomial and compare it with the unknown result - which we can see is a cubic - of our algebraic multiplication. The co-efficients in the known polynomial will tell us the missing co-efficients of our unknown quadratic.
In this case: a = 1
(b-2a) = 1
b - 2 = 1
b= 3
(c-2b) = -11
c-6 = -11
c = -5
We can check this last one by seeing -2c at the end of our algebraic expression. If c is right it will match -2 x -5 = 10, so yes it's right.
This process has given us the three co-efficients for our missing quadratic: 1, 3 and -5
so the two factors of our polynomial are (x-2) and (x2 + 3x -5).
Just be careful at this point: can the quadratic factor be factorised itself, to give three linear factors? Give it a quick b2 - 4ac test: 9 - (4x1x-5) = 9 - (-20) = 29. NO. It has to stay as a quadratic.
Often it will. You'll find that you're often - but not always - asked to find the three factors of p(x) or to factorise p(x) completely when this is the case.
Any polynomial of order 3 - which is what you deal with at C1 - is the product of either linear factors, or a linear and a quadratic factor.
Finding a factor like this, you don't need to worry about remainders but these crop up soon enough, and we'll come onto them in due course.
A basic example of what I mean is: x(x2 + 2x - 4). Multiplying these two factors gives the cubic equation x3 + 2x2 - 4x.
But what do you do if you know one factor and the polynomial?
Say you know the linear factor to be (x-2) and the polynomial to be x3 + x2 - 11x + 10
The other factor must be a quadratic.
(x-2)(ax2+ bx + c) = x3 + x2 - 11x + 10
expand the two brackets as you would normally:
ax3 + bx2 + cx - 2ax2 -2bx - 2c
and then find common factors to tidy this expression up and make it usable:
= ax3 + (b-2a)x2 + (c-2b)x -2c
That bit is worth checking. You're simply collecting together the different values of x2 or whatever, as you would normally in an equation. You're only putting brackets round them because you don't know their values and you need to understand them as together making x2.
This expression is now completely identical to our original polynomial:
= ax3 + (b-2a)x2 + (c-2b)x -2c
x3 + x2 - 11x + 10
We now go through the fairly simple but again easy to cock up process of comparing the co-efficients. Co-efficients are of course the numbers in front of variables (3 is the coefficient of x in the expression 3x).
Comparing the co-efficients here means we look at our original polynomial and compare it with the unknown result - which we can see is a cubic - of our algebraic multiplication. The co-efficients in the known polynomial will tell us the missing co-efficients of our unknown quadratic.
In this case: a = 1
(b-2a) = 1
b - 2 = 1
b= 3
(c-2b) = -11
c-6 = -11
c = -5
We can check this last one by seeing -2c at the end of our algebraic expression. If c is right it will match -2 x -5 = 10, so yes it's right.
This process has given us the three co-efficients for our missing quadratic: 1, 3 and -5
so the two factors of our polynomial are (x-2) and (x2 + 3x -5).
Just be careful at this point: can the quadratic factor be factorised itself, to give three linear factors? Give it a quick b2 - 4ac test: 9 - (4x1x-5) = 9 - (-20) = 29. NO. It has to stay as a quadratic.
Often it will. You'll find that you're often - but not always - asked to find the three factors of p(x) or to factorise p(x) completely when this is the case.
Thursday 4 September 2008
Multiplying polynomials
This is fairly easy too: it is, at its basic level, expanding brackets, though you can make use of the distributive law of algebra to help you do it more systematically.
For example.
(x+2)(x-5) can either be expanded bit by bit (ac+ad+bc+bd) or you can write it like this: x(x-5) + 2(x-5). Instead of adding four separate products individually, you are adding two sets of them. It's basically the same thing, just a little more organised. You could also write it x(x+2) -5(x+2).
You can use this principle to help you multiply polynomials.
(x+9)(3x3 -4x2 + 3)
x(3x3 -4x2 + 3)
+ 9(3x3 -4x2 + 3)
= (3x4 - 4x3 + 3x)
+ (27x3 - 36x2 + 27)
using the knowledge of collecting polynomials from the previous post:
= 3x4 + 23x3 - 36x2 + 3x + 27.
It gets fiddlier than this but usually the problem is signs. Pay really close attention to them because they are an easy ways of losing lots of marks quickly.
For example.
(x+2)(x-5) can either be expanded bit by bit (ac+ad+bc+bd) or you can write it like this: x(x-5) + 2(x-5). Instead of adding four separate products individually, you are adding two sets of them. It's basically the same thing, just a little more organised. You could also write it x(x+2) -5(x+2).
You can use this principle to help you multiply polynomials.
(x+9)(3x3 -4x2 + 3)
x(3x3 -4x2 + 3)
+ 9(3x3 -4x2 + 3)
= (3x4 - 4x3 + 3x)
+ (27x3 - 36x2 + 27)
using the knowledge of collecting polynomials from the previous post:
= 3x4 + 23x3 - 36x2 + 3x + 27.
It gets fiddlier than this but usually the problem is signs. Pay really close attention to them because they are an easy ways of losing lots of marks quickly.
Collecting Polynomials
This is not difficult in principle, but it can become messy if you are not careful with your signs.
You can only add or subtract values with the same powers of x.
Be careful: this means that you CANNOT add 3x4 and 3x3 because they have different powers of x and are therefore completely different values.
You also CANNOT add 3x2 and 3r2 because they are different variables and so different values.
Here is a fairly basic example: add 3x3 + 5x to 2x3 -4x
It helps to collect the two expressions into brackets first:
(3x3 + 5x) + (2x3 -4x)
Now you can sort out the signs more easily:
3x3 + 5x + 2x3 -4x
As it happens there is no sort out needed here!
Now collect the like terms:
(3x3 + 2x3) + (5x - 4x)
= 5x3 + x
It does get a lot more fiddly than this sometimes:
(4x4 + 5x3 - 3x2) - (2x4 - 6x3 + x2 + 9)
But use the same principles. Here the two expressions are already in brackets so we can SORT THE SIGNS OUT:
4x4 + 5x3 - 3x2 - 2x4 + 6x3 - x2 - 9
What we've done here is simply say "all the values in the second polynomial are being subtracted and we all know what to do with a - and a + or a - and another -". We've effectively subtracted the two expressions here by reversing the signs in the second expression.
You might want to look back at that again.
OK. Now we can collect the like terms.
4x4 - 2x4 + 5x3 + 6x3 - 3x2 - x2 - 9
= 2x4 + 11x3 - 4x2 - 9
You can only add or subtract values with the same powers of x.
Be careful: this means that you CANNOT add 3x4 and 3x3 because they have different powers of x and are therefore completely different values.
You also CANNOT add 3x2 and 3r2 because they are different variables and so different values.
Here is a fairly basic example: add 3x3 + 5x to 2x3 -4x
It helps to collect the two expressions into brackets first:
(3x3 + 5x) + (2x3 -4x)
Now you can sort out the signs more easily:
3x3 + 5x + 2x3 -4x
As it happens there is no sort out needed here!
Now collect the like terms:
(3x3 + 2x3) + (5x - 4x)
= 5x3 + x
It does get a lot more fiddly than this sometimes:
(4x4 + 5x3 - 3x2) - (2x4 - 6x3 + x2 + 9)
But use the same principles. Here the two expressions are already in brackets so we can SORT THE SIGNS OUT:
4x4 + 5x3 - 3x2 - 2x4 + 6x3 - x2 - 9
What we've done here is simply say "all the values in the second polynomial are being subtracted and we all know what to do with a - and a + or a - and another -". We've effectively subtracted the two expressions here by reversing the signs in the second expression.
You might want to look back at that again.
OK. Now we can collect the like terms.
4x4 - 2x4 + 5x3 + 6x3 - 3x2 - x2 - 9
= 2x4 + 11x3 - 4x2 - 9
Polynomials - The Basics
I've got fed up with doing straight line geometry, as it needs diagrams, and I am having real difficulty drawing & importing them.
So to polynomials. Firstly, what is a polynomial? Well it is an expression involving numbers and powers of x. Such as:
2x + 1
3x2 + 4x - 6
2x4 - 4x3 + 2x2 +2
These are all polynomials. They are usually written, as here, in descending powers of x, from highest to lowest.
Sometimes they are called polynomials of degree x, depending on the highest power of x. So a cubic expression is a polynomial of degree 3, a quadratic of degree 2, a linear of degree 1 and a constant (ie just a number on its own) of degree 0 (because the highest power of x here is 0, ie x0 - ie 1).
Polynomials DO NOT have fractional or negative powers.
2x-2 + 1
is not a polynomial.
That's the basics of it.
So to polynomials. Firstly, what is a polynomial? Well it is an expression involving numbers and powers of x. Such as:
2x + 1
3x2 + 4x - 6
2x4 - 4x3 + 2x2 +2
These are all polynomials. They are usually written, as here, in descending powers of x, from highest to lowest.
Sometimes they are called polynomials of degree x, depending on the highest power of x. So a cubic expression is a polynomial of degree 3, a quadratic of degree 2, a linear of degree 1 and a constant (ie just a number on its own) of degree 0 (because the highest power of x here is 0, ie x0 - ie 1).
Polynomials DO NOT have fractional or negative powers.
2x-2 + 1
is not a polynomial.
That's the basics of it.
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