Without it, even linear equations become extremely difficult to re-arrange.
Take 6xy + 3x - 2y = 7
Re-arrange this for x.
MOVE the non x bit.
6xy + 3x = 7- 2y.
SPOT the common factor
x(6y + 3) = 7 - 2y.
DIVIDE by (6y+3)
x = 7-2y
----
6y+3
PIECE of the proverbial.
Showing posts with label factorising polynomials. Show all posts
Showing posts with label factorising polynomials. Show all posts
Thursday, 22 January 2009
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
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.
Subscribe to:
Posts (Atom)
