C4 Factor Theorem help...

Original post by Branny101

Could someone help me with the comparing coefficients method of solving a cubic equation?
Let's take the example:

x^3-5x^2+2x+8

Help would be appreciated :awesome:

Thaaanks

Posted from TSR Mobile

Firstly, comparing co-efficients is not used for solving equations

Secondly, that is not an equation

Reply 3

the bear

Original post by Branny101

Could someone help me with the comparing coefficients method of solving a cubic equation?
Let's take the example:

x^3-5x^2+2x+8

Help would be appreciated :awesome:

Thaaanks

Posted from TSR Mobile

x³ - 5x² + 2x + 8 = (x + a)(x + b)(x + c)

times out the RHS then compare the various powers of x

Reply 4

Kvothe the Arcane

Original post by Branny101

Could someone help me with the comparing coefficients method of solving a cubic equation?
Let's take the example:

x^3-5x^2+2x+8

Help would be appreciated :awesome:

Thaaanks

Posted from TSR Mobile

Assuming that the cubic does factorize. Say that

f(x)=x^3-5x^2+2x+8

. Find a number

\lambda

for which

f( \lambda )=0

. Then you can say that

f(x)=x^3-5x^2+2x+8=(x- \lambda)(ax^2+bx+c)

Comparing coefficients tells you what a, b and c are.

Reply 5

Original post by TenOfThem

Firstly, comparing co-efficients is not used for solving equations

Secondly, that is not an equation

Don't be so pedantic :smile:

you know I meant to factorise the expression f(x), thanks for the help anyways :ahee:

Reply 6

Original post by Branny101

Don't be so pedantic :smile:

you know I meant to factorise the expression f(x), thanks for the help anyways :ahee:

I am afraid that the number of students that I come across who do not know the difference between equation, expression, and formula frightens me

I assumed that you were one of them and needed correction

It is good to know that you are not :biggrin:

Reply 7

Original post by majmuh24

I would use synthetic division, major time saver :tongue:

First, use factor theorem to find a divisor and then use synthetic division to factorise to a quadratic and a linear that can easily be solved :smile:

Do you mean to set this equal to 0, so you want to solve

x^3-5x^2+2x+8=0

?

Posted from TSR Mobile

Sorry I should've made it clearer - to factorise :facepalm:

Reply 8

Original post by keromedic

Assuming that the cubic does factorize. Say that

f(x)=x^3-5x^2+2x+8

. Find a number

\lambda

for which

f( \lambda )=0

. Then you can say that

f(x)=x^3-5x^2+2x+8=(x- \lambda)(ax^2+bx+c)

Comparing coefficients tells you what a, b and c are.

But how do you compare the coefficients, my book says you say that as a=1 then b+c=-5 thus b=-6? It doesn't work this way though :frown:

Reply 9

Original post by Branny101

But how do you compare the coefficients, my book says you say that as a=1 then b+c=-5 thus b=-6? It doesn't work this way though :frown:

Are you setting up an equation like Bear has in Post 4?

Do you already have a factor?

Reply 10

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Original post by Branny101

Sorry I should've made it clearer - to factorise :facepalm:

Yeah, I know, you still use synthetic division. It's a lot easier.

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Reply 11

Original post by TenOfThem

Are you setting up an equation like Bear has in Post 4?

Do you already have a factor?

Yes, and no the question is to factorise out fully by means of comparing coefficients :frown:

( well it doesn't say specifically to compare coefficients but I wanted to see which way suits me)

Posted from TSR Mobile

(edited 10 years ago)

Reply 12

Original post by the bear

x³ - 5x² + 2x + 8 = (x + a)(x + b)(x + c)

times out the RHS then compare the various powers of x

Original post by Branny101

Yes, and no the question is to factorise out fully by means of comparing coefficients :frown:

So - what did you get when you multiplied out the RHS

Reply 13

Original post by majmuh24

Yeah, I know, you still use synthetic division. It's a lot easier.

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That's what I thought but would you lose marks for not using the correct method?

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Reply 14

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Original post by Branny101

That's what I thought but would you lose marks for not using the correct method?

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I don't think so, they both give the right answer

Anyway, I would start by using the factor theorem with factors of the constant term (8) and then compare coefficients to get the other terms, so you would get

(x+a)(x^2+bx+c) which can easily be factorized to linear factors.

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Reply 15

Original post by TenOfThem

So - what did you get when you multiplied out the RHS

ax^3+bx^2+cx+d

EDIT: scratch that there isn't a 'd' term - x^3+cx^2+bx^2+ax^2+acx+abx+abc
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(edited 10 years ago)

Reply 16