How to solve cubic polynomials

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#1
Hi there,

I’m currently working on bridging work for yr 11-12 for a level maths and I don’t know how to factorise polynomials that have a x^3 or x^4 in them. Could someone please help me with the following questions - any advice on how to tackle these types of questions is also appreciated

Factorise fully:

X^3+x^2-2x

X^4+3x^3-18x^2

Thankyou !
0
1 month ago
#2
(Original post by LWalsh78)
Hi there,

I’m currently working on bridging work for yr 11-12 for a level maths and I don’t know how to factorise polynomials that have a x^3 or x^4 in them. Could someone please help me with the following questions - any advice on how to tackle these types of questions is also appreciated

Factorise fully:

X^3+x^2-2x

X^4+3x^3-18x^2

Thankyou !
First one; note how every term has at least one x factor.

Factor this out and you get x(x^2+x-2) … now this second factor is a quadratic that factorises …

So by factoring out this x you are left with factorising a quadratic.

Similar story with second cubic
1
#3
(Original post by RDKGames)
First one; note how every term has at least one x factor.

Factor this out and you get x(x^2+x-2) … now this second factor is a quadratic that factorises …

So by factoring out this x you are left with factorising a quadratic.

Similar story with second cubic
Hi there,

Thankyou with your help for the first one. For the second it’s x^4+3x^3-18x^2, so would you just factorise out x^2 to be left with x^2+3x-18 and then factorise that quadratic ?
0
1 month ago
#4
(Original post by LWalsh78)
Thankyou with your help for the first one. For the second it’s x^4+3x^3-18x^2, so would you just factorise out x^2 to be left with x^2+3x-18 and then factorise that quadratic ?
Yes.
1
#5
(Original post by DFranklin)
Yes.
Hi there,

How would you solve this to find the values for x. Eg x^2 (x^2+3x-18)
This factors to: x^2 (x+6)(x-3)

If you made that equal to 0, then could you solve x^2 (x+6)(x-3)=0 ?

How would you then solve that if you can ?
Last edited by LWalsh78; 1 month ago
0
1 month ago
#6
(Original post by LWalsh78)
Hi there,

How would you solve this to find the values for x. Eg x^2 (x^2+3x-18)
This factors to: x^2 (x+6)(x-3)

If you made that equal to 0, then could you solve x^2 (x+6)(x-3)=0 ?

How would you then solve that if you can ?
So we have x^2(x+6)(x-3)=0
For this equation to be equal to 0, one of the factors must be 0 because then we have 0*something*something else which will be 0 like we want.
We hence get our 3 solutions by putting each factor equal to 0 and solving each of the much more simpler equations.
0
1 month ago
#7
(Original post by LWalsh78)
Hi there,

I’m currently working on bridging work for yr 11-12 for a level maths and I don’t know how to factorise polynomials that have a x^3 or x^4 in them. Could someone please help me with the following questions - any advice on how to tackle these types of questions is also appreciated

Factorise fully:

X^3+x^2-2x

X^4+3x^3-18x^2

Thankyou !
In year 12 you learn will factor theorem, but usually, looking at the constant at the end (or if there isn't a constant just factor out x) eg. x^3+3x^2+3x+1, since the constant is 1, you plug in all factors of the constant, in this case 1, and see if the equation =0, if it is then its a factor. Then you can just divide the cubic to give you a quadratic.
Last edited by BlackH0mbs; 1 month ago
1
1 month ago
#8
(Original post by BlackH0mbs)
In year 12 you will factor theorem, but usually, looking at the constant at the end (or if there isn't a constant just factor out x) eg. x^3+3x^2+3x+1, since the constant is 1, you plug in all factor the constant, in this case 1, and see if the equation =0, if it is then its a factor. Then you can just divide the cubic to give you a quadratic.

Some cubics which have a special pattern to them can also be factorised more easily. Specifically, there's a trick for those where the ratio between the coefficients on the x3 and x2 terms is the same as the ratio between the x term and the constant. For example:

x3 + 4x2 - 7x - 28 (Note that the ratio between 1 and 4 is the same as the ratio between -7 and -28)
x2(x+4) - 7(x+4)
(x2 - 7)(x+4)
(x + sqrt7)(x - sqrt7)(x+4)

So in this case the roots are -4 and ±sqrt7.
Last edited by anosmianAcrimony; 1 month ago
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