# How to solve cubic polynomials

Watch
Announcements

Page 1 of 1

Go to first unread

Skip to page:

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 !

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

reply

Report

#2

(Original post by

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 !

**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 !

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

reply

(Original post by

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

**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

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

reply

Report

#4

(Original post by

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 ?

**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 ?

1

reply

(Original post by

Yes.

**DFranklin**)Yes.

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

reply

Report

#6

(Original post by

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 ?

**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 ?

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

reply

Report

#7

**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 !

Last edited by BlackH0mbs; 1 month ago

1

reply

Report

#8

(Original post by

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.

**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 x

^{3}and x

^{2}terms is the same as the ratio between the x term and the constant. For example:

x

^{3}+ 4x

^{2}- 7x - 28 (Note that the ratio between 1 and 4 is the same as the ratio between -7 and -28)

x

^{2}(x+4) - 7(x+4)

(x

^{2}- 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

0

reply

X

Page 1 of 1

Go to first unread

Skip to page:

### Quick Reply

Back

to top

to top