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Simplify into a single factorised expression.

(x – 3)^2 + 5(x – 3)^3

and

4x(2x + 1)^3 + 5(2x + 1)^4

(x – 3)^2 + 5(x – 3)^3

and

4x(2x + 1)^3 + 5(2x + 1)^4

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#2

(Original post by

Simplify into a single factorised expression.

(x – 3)^2 + 5(x – 3)^3

and

4x(2x + 1)^3 + 5(2x + 1)^4

**BenHarrow**)Simplify into a single factorised expression.

(x – 3)^2 + 5(x – 3)^3

and

4x(2x + 1)^3 + 5(2x + 1)^4

Post your workings and what you've tried.

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#3

**BenHarrow**)

Simplify into a single factorised expression.

(x – 3)^2 + 5(x – 3)^3

and

4x(2x + 1)^3 + 5(2x + 1)^4

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(Original post by

the first one, factorise (x+3)^2 out and the second one factorise (2x+1)^3 oit

**brainmaster**)the first one, factorise (x+3)^2 out and the second one factorise (2x+1)^3 oit

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#5

(Original post by

THanks man but the thing i am confused about is how you factorise that out as the answer for the first one is (x – 3)^2(5x – 14) but if you factorise (x-3)^2 surely it will be gone from equation, i just dont know how to factorise that step by step

**BenHarrow**)THanks man but the thing i am confused about is how you factorise that out as the answer for the first one is (x – 3)^2(5x – 14) but if you factorise (x-3)^2 surely it will be gone from equation, i just dont know how to factorise that step by step

(x - 3)^2 {......... + ........)

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(Original post by

Factorise does not mean remove it completely.

(x - 3)^2 {......... + ........)

**Muttley79**)Factorise does not mean remove it completely.

(x - 3)^2 {......... + ........)

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#7

**BenHarrow**)

THanks man but the thing i am confused about is how you factorise that out as the answer for the first one is (x – 3)^2(5x – 14) but if you factorise (x-3)^2 surely it will be gone from equation, i just dont know how to factorise that step by step

does it make sense now?

Last edited by brainmaster; 1 month ago

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#8

**Muttley79**)

Factorise does not mean remove it completely.

(x - 3)^2 {......... + ........)

so you have y^2+5y^3

y^2(1+5y)

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(Original post by

(x-3)^2(1+5(x-3))

(x-3)^2(1+5x-15)

(x-3)^2(5x-14)

does it make sense now?

**brainmaster**)(x-3)^2(1+5(x-3))

(x-3)^2(1+5x-15)

(x-3)^2(5x-14)

does it make sense now?

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#11

(Original post by

It is against the rules to post a solution - please edit.

**Muttley79**)It is against the rules to post a solution - please edit.

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(Original post by

a simple way to understand this is to let (x-3) = y

so you have y^2+5y^3

y^2(1+5y)

**brainmaster**)a simple way to understand this is to let (x-3) = y

so you have y^2+5y^3

y^2(1+5y)

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#13

(Original post by

how does it work then i know how to factorise normally but i dont understand this

**BenHarrow**)how does it work then i know how to factorise normally but i dont understand this

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#14

(Original post by

sorry for that. is it fine now?

**brainmaster**)sorry for that. is it fine now?

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#15

(Original post by

I understand that yes but i just dont understand how its the same for this equation

**BenHarrow**)I understand that yes but i just dont understand how its the same for this equation

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#16

(Original post by

I think you've lost a sign but please don't do the second one for the OP.

**Muttley79**)I think you've lost a sign but please don't do the second one for the OP.

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#17

(Original post by

can you do second one so i can fully understand as i am confused as to how the 1 just moves into the brackets

**BenHarrow**)can you do second one so i can fully understand as i am confused as to how the 1 just moves into the brackets

SImpler example: Factorise x^2 + x

x^2 + x = x( x + 1) x is a common factor here so I've taken it outside the bracket then inside the bracket I need an 'x' to get the x^2 and a '+1' to get the x.

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(Original post by

No, you need to try it for yourself. The '1' is what you need in the bracket to get the original term when it is multiplied out again.

SImpler example: Factorise x^2 + x

x^2 + x = x( x + 1) x is a common factor here so I've taken it outside the bracket then inside the bracket I need an 'x' to get the x^2 and a '+1' to get the x.

**Muttley79**)No, you need to try it for yourself. The '1' is what you need in the bracket to get the original term when it is multiplied out again.

SImpler example: Factorise x^2 + x

x^2 + x = x( x + 1) x is a common factor here so I've taken it outside the bracket then inside the bracket I need an 'x' to get the x^2 and a '+1' to get the x.

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#19

(Original post by

Oh ok i understand that thanks both of you for the help i will try secomd example now

**BenHarrow**)Oh ok i understand that thanks both of you for the help i will try secomd example now

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I got right answer through method thanks, final question i have is this

simplify the following

(2πh^2/rb)/4/3πhr^2

i have some current working but struggling to get to answer

simplify the following

(2πh^2/rb)/4/3πhr^2

i have some current working but struggling to get to answer

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