LWalsh78
Badges: 5
Rep:
?
#1
Report Thread starter 1 month ago
#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
reply
RDKGames
Badges: 20
Rep:
?
#2
Report 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
reply
LWalsh78
Badges: 5
Rep:
?
#3
Report Thread starter 1 month ago
#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
reply
DFranklin
Badges: 18
Rep:
?
#4
Report 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
reply
LWalsh78
Badges: 5
Rep:
?
#5
Report Thread starter 1 month ago
#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
reply
JustSomeGuy:/
Badges: 14
Rep:
?
#6
Report 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
reply
BlackH0mbs
Badges: 3
Rep:
?
#7
Report 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
reply
anosmianAcrimony
Badges: 20
Rep:
?
#8
Report 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.
^ Good advice.

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
0
reply
X

Quick Reply

Attached files
Write a reply...
Reply
new posts
Back
to top
Latest
My Feed

See more of what you like on
The Student Room

You can personalise what you see on TSR. Tell us a little about yourself to get started.

Personalise

How would you feel if uni students needed to be double vaccinated to start in Autumn?

I'd feel reassured about my own health (40)
14.76%
I'd feel reassured my learning may be less disrupted by isolations/lockdowns (88)
32.47%
I'd feel less anxious about being around large groups (33)
12.18%
I don't mind if others are vaccinated or not (23)
8.49%
I'm concerned it may disadvantage some students (14)
5.17%
I think it's an unfair expectation (70)
25.83%
Something else (tell us in the thread) (3)
1.11%

Watched Threads

View All