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# Factorising Cubic Polynomials watch

1. Let’s use this as an example:
f(x)=2x^3 -5x^2 -23x -10.

So we want the last 2 numbers in our brackets to multiply to -10. We can use (plus or minus) 10,1,5,2. What I’m wondering is if there is any strategical methods/ways of spotting which value could be substituted into f(x) to get 0 or is it just luck of the draw? Thanks
2. (Original post by Y11_Maths)
Let’s use this as an example:
f(x)=2x^3 -5x^2 -23x -10.

So we want the last 2 numbers in our brackets to multiply to -10. We can use (plus or minus) 10,1,5,2. What I’m wondering is if there is any strategical methods/ways of spotting which value could be substituted into f(x) to get 0 or is it just luck of the draw? Thanks
With this particular example, you can rule out + or - 10 immediately.

You can also rule out + or - 1 as 2-5-23-10 is not zero and -2-5+23-10 is also not zero.

Another option, if you're feeling lazy, is to use the table mode on your calculator.
3. (Original post by Y11_Maths)
Let’s use this as an example:
f(x)=2x^3 -5x^2 -23x -10.

So we want the last 2 numbers in our brackets to multiply to -10. We can use (plus or minus) 10,1,5,2. What I’m wondering is if there is any strategical methods/ways of spotting which value could be substituted into f(x) to get 0 or is it just luck of the draw? Thanks
Cubics are not as straight-forward to factorise like quadratics are. At the end of the day, you can always use the general formula but no sane person would use the cubic one.

Also, it's a cubic. It will have at most 3 real roots whose product must be equal to NOT -10, but in fact (*). So try the +ve and -ve factors of 5.

Once you have at least one real root, say it's , then you can just divide by and obtain a quadratic. This should be much simpler to factorise into where are its roots.
Hence you can say that

(*) This is quite simply because if we choose to denote the roots of as then we know that

Since the constant term we got is , we get that , hence the product is 5.
4. (Original post by BuryMathsTutor)
With this particular example, you can rule out + or - 10 immediately.

You can also rule out + or - 1 as 2-5-23-10 is not zero and -2-5+23-10 is also not zero.

Another option, if you're feeling lazy, is to use the table mode on your calculator.
I didn’t think of that! Can I use a calculator in both papers at as?
5. (Original post by Y11_Maths)
Let’s use this as an example:
f(x)=2x^3 -5x^2 -23x -10.

So we want the last 2 numbers in our brackets to multiply to -10. We can use (plus or minus) 10,1,5,2. What I’m wondering is if there is any strategical methods/ways of spotting which value could be substituted into f(x) to get 0 or is it just luck of the draw? Thanks
You want the last 3 numbers need to multiply to give -(-10/2) = 5.
So I basically notice that since there's odd and even coefficients, it may be worth trying an even number that is a factor of 10.
Also, the 3 numbers also need to add up to -(-5/2) = 5/2, so there must be a fractional root, so try x = ±1/2
That should start you off with a few ideas.
6. hence the product is 5.
Thank you for explaining this so clearly!
7. (Original post by 3pointonefour)
You want the last 3 numbers need to multiply to give -(-10/2) = 5.
So I basically notice that since there's odd and even coefficients, it may be worth trying an even number that is a factor of 10.
Also, the 3 numbers also need to add up to -(-5/2) = 5/2, so there must be a fractional root, so try x = ±1/2
That should start you off with a few ideas.
Ok thank you
8. (Original post by Y11_Maths)
I didn’t think of that! Can I use a calculator in both papers at as?
You can.
9. (Original post by BuryMathsTutor)
You can.
Are you required to have one or can the papers be done in the absence of a calculator?
10. (Original post by 3pointonefour)
You want the last 3 numbers need to multiply to give -(-10/2) = 5.
So I basically notice that since there's odd and even coefficients, it may be worth trying an even number that is a factor of 10.
Also, the 3 numbers also need to add up to -(-5/2) = 5/2, so there must be a fractional root, so try x = ±1/2
That should start you off with a few ideas.
How would they want me to set out my answer?
1) f(x) = (x-5)(2x+1)(x+2)
2) f(x) = 2(x-5)(x+1/2)(x+2)
11. (Original post by Y11_Maths)
How would they want me to set out my answer?
1) f(x) = (x-5)(2x+1)(x+2)
2) f(x) = 2(x-5)(x+1/2)(x+2)
The first way is how you'd usually set it out since it's much tidier and standard notation to leave linear factors like that
12. (Original post by 3pointonefour)
The first way is how you'd usually set it out since it's much tidier and standard notation to leave linear factors like that
Ok thanks
13. (Original post by Y11_Maths)
Are you required to have one or can the papers be done in the absence of a calculator?
You will need one.
14. Are you going to rename yourself Y12_Maths??
15. (Original post by Y11_Maths)
Are you required to have one or can the papers be done in the absence of a calculator?
At A Level you’ll often find that one factor is easy to find e.g. (x-1) and the others can be found by division.
16. (Original post by BuryMathsTutor)
You will need one.
Ok cheers
17. (Original post by Your Local Cat)
Are you going to rename yourself Y12_Maths??
I can? How do I change my name?
18. (Original post by Your Local Cat)
Are you going to rename yourself Y12_Maths??
19. (Original post by Notnek)
At A Level you’ll often find that one factor is easy to find e.g. (x-1) and the others can be found by division.
Yeah, that takes longer imo so I prefer to use (x-1)(ax^2 +bx + c) and fill in my a and c values to which I can easily find out b by comparing to my original f(x)
20. (Original post by 3pointonefour)
I hope you can 😂

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