Factorising of Polynomials... desperate help needed!!!
Hello! I desperately need help with factorising polynomials. I dont know how to do it as i was away when it was done in class at school; and my yearly assessment task is due tomorrrow.
Anyway.. the question is:
Fully factorise the polynomial so that it is written as a product of its linear factors.
y= x^3 - 2x^2 - x + 2
THANKYOU IN ADVANCE!!!
Anyway.. the question is:
Fully factorise the polynomial so that it is written as a product of its linear factors.
y= x^3 - 2x^2 - x + 2
THANKYOU IN ADVANCE!!!
Original post by Morgan00
Hello! I desperately need help with factorising polynomials. I dont know how to do it as i was away when it was done in class at school; and my yearly assessment task is due tomorrrow.
Anyway.. the question is:
Fully factorise the polynomial so that it is written as a product of its linear factors.
y= x^3 - 2x^2 - x + 2
THANKYOU IN ADVANCE!!!
Anyway.. the question is:
Fully factorise the polynomial so that it is written as a product of its linear factors.
y= x^3 - 2x^2 - x + 2
THANKYOU IN ADVANCE!!!
Guess
You literally say let f(x)=x^3 - 2x^2 - x + 2
and say f(x)=0
then solve for 0
The method is to guess. So start with small positive numbers i'd recommend going to about 3 then go with some negative numbers if you still don't get it.
After you have a factor use abgebraic long division to find the a quadratic which should factorise into 2 other factors
answer to one of the factors below
Spoiler
There may be a preferred method. (This is all ignoring complex roots).
A cubic (with 1 as the coefficient of x^3) will always have at least two factor (x-a)(x^2+bx+c). This is because a cubic must cross the x axis somewhere and so that gives you (x-a). It may cross the x axis in 2 other places or just touch the x axis at one place or not touch at all.
If it crosses in two other places you get (x-a)(x-b)(x-c). If only one then you get (x-a)(x-b)^2 and if none, you will find that the quadratic in
(x-a)(x^2+bx+c) has no roots.
At A-level, you generally find one root by trial and improvement (find a value of x where the polynomial is zero). You may then be invited to divide the (x-a) you just found into the original polynomial to get the quadratic (x^2+bx+c) from (x-a)(x^2+bx+c).
You then factorise the quadratic to get the other two roots if the quadratic factorises.
The division process can be by long division or equating coefficients. (The latter is often very easy)
Of course you may use a graphical calculator to see the answer immediately. If you have a graphical calculator you would be a bit silly not to learn how to use it properly.
A cubic (with 1 as the coefficient of x^3) will always have at least two factor (x-a)(x^2+bx+c). This is because a cubic must cross the x axis somewhere and so that gives you (x-a). It may cross the x axis in 2 other places or just touch the x axis at one place or not touch at all.
If it crosses in two other places you get (x-a)(x-b)(x-c). If only one then you get (x-a)(x-b)^2 and if none, you will find that the quadratic in
(x-a)(x^2+bx+c) has no roots.
At A-level, you generally find one root by trial and improvement (find a value of x where the polynomial is zero). You may then be invited to divide the (x-a) you just found into the original polynomial to get the quadratic (x^2+bx+c) from (x-a)(x^2+bx+c).
You then factorise the quadratic to get the other two roots if the quadratic factorises.
The division process can be by long division or equating coefficients. (The latter is often very easy)
Of course you may use a graphical calculator to see the answer immediately. If you have a graphical calculator you would be a bit silly not to learn how to use it properly.
Original post by Morgan00
Hello! I desperately need help with factorising polynomials. I dont know how to do it as i was away when it was done in class at school; and my yearly assessment task is due tomorrrow.
Anyway.. the question is:
Fully factorise the polynomial so that it is written as a product of its linear factors.
y= x^3 - 2x^2 - x + 2
THANKYOU IN ADVANCE!!!
Anyway.. the question is:
Fully factorise the polynomial so that it is written as a product of its linear factors.
y= x^3 - 2x^2 - x + 2
THANKYOU IN ADVANCE!!!
Original post by will'o'wisp
Guess
You literally say let f(x)=x^3 - 2x^2 - x + 2
and say f(x)=0
then solve for 0
The method is to guess. So start with small positive numbers i'd recommend going to about 3 then go with some negative numbers if you still don't get it.
After you have a factor use abgebraic long division to find the a quadratic which should factorise into 2 other factors
answer to one of the factors below
You literally say let f(x)=x^3 - 2x^2 - x + 2
and say f(x)=0
then solve for 0
The method is to guess. So start with small positive numbers i'd recommend going to about 3 then go with some negative numbers if you still don't get it.
After you have a factor use abgebraic long division to find the a quadratic which should factorise into 2 other factors
answer to one of the factors below
Spoiler
Not really guessing blindly, but you're making an educated guess as you should know to try all the factors of the constant term first.
Just plug in factors of the last term and see which one will make it equal to 0. Once you have one solution you can use long division to divide the polynomial by and factorise (if possible) your leftover quadratic.
Original post by RDKGames
Not really guessing blindly, but you're making an educated guess as you should know to try all the factors of the constant term first.
Just plug in factors of the last term and see which one will make it equal to 0. Once you have one solution you can use long division to divide the polynomial by and factorise (if possible) your leftover quadratic.
Just plug in factors of the last term and see which one will make it equal to 0. Once you have one solution you can use long division to divide the polynomial by and factorise (if possible) your leftover quadratic.
Above your post there is one which states the factor theorem where you have a general formula exactly equal to the cubic and thus you can equate coefficients, however if not given how do we get then general formula?
Original post by will'o'wisp
Above your post there is one which states the factor theorem where you have a general formula exactly equal to the cubic and thus you can equate coefficients, however if not given how do we get then general formula?
Can you rephrase this? I don't see what you are getting at.
@RDK games is just saying that if there is a factor (x-a) with a being simple, then in all likelihood, a will be a factor of the constant term in the original cubic. Now, I guess you might see a Q where a was a simple fraction, a half say or 3/2 but that is unlikely and so trying those factors first is a good way to go.
Original post by will'o'wisp
Above your post there is one which states the factor theorem where you have a general formula exactly equal to the cubic and thus you can equate coefficients, however if not given how do we get then general formula?
There is a general formula for the roots of a cubic, but that is not what is posted. The general formula is horrible to remember/use and you will never need to use it at A-level.
If you have a cubic equation with integer coefficients:
, then
if the equation is solvable using A-level methods(*), there will be at least one rational root. If you write this root as a fraction p/q in lowest terms, then
p must divide the constant coefficient (i.e. d)
q must divide the coefficient of x^3 (i.e. a).
This gives a finite (and hopefully short) list of values to try using trial and error.
So for example in the case here with x^3 - 2x^2 - x + 2
we know that any root p/q must have q | 1 (so WLOG q = 1), and p | 2, so p must be one of -2, -1, 1, 2.
Once you have one root, you can either keep guessing to see if you can pick up all the roots, or you can factor out the one root and use the quadratic formula.
(*) Excluding numerical methods which only give an approximation to the root. Since at A-level you shouldn't ever get asked a question that you can't do using A-level methods, you will either be able to find 1 root by guessing rationals, or you will be expected to use numerical methods to get an approximate root.
(edited 7 years ago)
Original post by nerak99
Can you rephrase this? I don't see what you are getting at.
@RDK games is just saying that if there is a factor (x-a) with a being simple, then in all likelihood, a will be a factor of the constant term in the original cubic. Now, I guess you might see a Q where a was a simple fraction, a half say or 3/2 but that is unlikely and so trying those factors first is a good way to go.
@RDK games is just saying that if there is a factor (x-a) with a being simple, then in all likelihood, a will be a factor of the constant term in the original cubic. Now, I guess you might see a Q where a was a simple fraction, a half say or 3/2 but that is unlikely and so trying those factors first is a good way to go.
Let's say that
how do we figure out the part by yourself? Or is it better to "guess" a value correctly and work things out from there?
Original post by nerak99
Can you rephrase this? I don't see what you are getting at.
@RDK games is just saying that if there is a factor (x-a) with a being simple, then in all likelihood, a will be a factor of the constant term in the original cubic. Now, I guess you might see a Q where a was a simple fraction, a half say or 3/2 but that is unlikely and so trying those factors first is a good way to go.
@RDK games is just saying that if there is a factor (x-a) with a being simple, then in all likelihood, a will be a factor of the constant term in the original cubic. Now, I guess you might see a Q where a was a simple fraction, a half say or 3/2 but that is unlikely and so trying those factors first is a good way to go.
if you are allowed to use a graphics calculator you could
i) draw the graph and see where it crosses the x axis. pick one of the whole number roots & use it to show the working to find the other roots. if a root was 4 then ( x - 4 ) will be a factor, and then you can use that to find the other factors
ii) use the equation solver. pick one of the answers & continue as above
i) draw the graph and see where it crosses the x axis. pick one of the whole number roots & use it to show the working to find the other roots. if a root was 4 then ( x - 4 ) will be a factor, and then you can use that to find the other factors
ii) use the equation solver. pick one of the answers & continue as above
Original post by will'o'wisp
Let's say that
how do we figure out the part by yourself? Or is it better to "guess" a value correctly and work things out from there?
how do we figure out the part by yourself? Or is it better to "guess" a value correctly and work things out from there?
(The polynomial you have given does not have a guessable root, however).
Original post by DFranklin
Assuming there is a guessable root, it's better to guess. In an A-level exam, the cubic will have been chosen so there is a guessable root.
(The polynomial you have given does not have a guessable root, however).
(The polynomial you have given does not have a guessable root, however).
Well that makes things a lot simpler
Thanks!Quick Reply
Related discussions
- MAT question help
- Points of inflection on a function of 9 degree
- Advanced difference of two squares
- Highers
- Knowing which particular integral to use for ordinary differential equations
- How many marks roughly would this question be worth (AQA A level Maths)?
- Extra MAT 2023
- Factorising quadratics when a isn't 1
- PSA: Do NOT buy the ti-84 for A-Levels
- Maths graph help
- A level Maths question
- What resources and tools would be good to get for my choices?
- Factorising question help
- cubic function, different ways of solving it?
- Mathematics Quadratics
- Simple differentiation question
- Binomial Expansion Question - A Level Maths
- stuck on a rearranging an expression question
- Rational canonical form
- Trigonometry alevel maths q
Latest
Trending
Last reply 2 days ago
Did Cambridge maths students find maths and further maths a level very easy?Last reply 2 weeks ago
Edexcel A Level Mathematics Paper 2 unofficial mark scheme correct me if wrongMaths
71
Trending
Last reply 2 days ago
Did Cambridge maths students find maths and further maths a level very easy?Last reply 2 weeks ago
Edexcel A Level Mathematics Paper 2 unofficial mark scheme correct me if wrongMaths
71
Articles for you
A guide to free resources for GCSE and A-level exam revision
Halls vs home: should I stay at home and commute to university or move out into halls or other student accommodation?
What is an LLM and how could it benefit your career?
Powerful questions business students forget to ask their career advisers