BenHarrow
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#1
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#1
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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MediocreSince01
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#2
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(Original post by BenHarrow)
Simplify into a single factorised expression.
(x – 3)^2 + 5(x – 3)^3

and
4x(2x + 1)^3 + 5(2x + 1)^4
You can factorise out the (x-3)^2 in the first one.

Post your workings and what you've tried.
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brainmaster
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(Original post by BenHarrow)
Simplify into a single factorised expression.
(x – 3)^2 + 5(x – 3)^3

and
4x(2x + 1)^3 + 5(2x + 1)^4
the first one, factorise (x+3)^2 out and the second one factorise (2x+1)^3 oit
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BenHarrow
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(Original post by brainmaster)
the first one, factorise (x+3)^2 out and the second one factorise (2x+1)^3 oit
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
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Muttley79
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(Original post by 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
Factorise does not mean remove it completely.

(x - 3)^2 {......... + ........)
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BenHarrow
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(Original post by Muttley79)
Factorise does not mean remove it completely.

(x - 3)^2 {......... + ........)
how does it work then i know how to factorise normally but i dont understand this
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brainmaster
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(Original post by 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(1 5(x-3))
does it make sense now?
Last edited by brainmaster; 1 month ago
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brainmaster
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(Original post by Muttley79)
Factorise does not mean remove it completely.

(x - 3)^2 {......... + ........)
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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Muttley79
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(Original post by brainmaster)

does it make sense now?
It is against the rules to post a solution - please edit.
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BenHarrow
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#10
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(Original post by brainmaster)
(x-3)^2(1+5(x-3))
(x-3)^2(1+5x-15)
(x-3)^2(5x-14)

does it make sense now?
can you do second one so i can fully understand as i am confused as to how the 1 just moves into the brackets
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brainmaster
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(Original post by Muttley79)
It is against the rules to post a solution - please edit.
sorry for that. is it fine now?
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BenHarrow
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#12
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(Original post by brainmaster)
a simple way to understand this is to let (x-3) = y
so you have y^2+5y^3
y^2(1+5y)
I understand that yes but i just dont understand how its the same for this equation
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Muttley79
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(Original post by BenHarrow)
how does it work then i know how to factorise normally but i dont understand this
(x – 3)^2 + 5(x – 3)^3= (x - 3)^2 [ what do you need here to multiply to get the first term + what do you need to get the second term]
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Muttley79
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(Original post by brainmaster)
sorry for that. is it fine now?
I think you've lost a sign but please don't do the second one for the OP.
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brainmaster
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(Original post by BenHarrow)
I understand that yes but i just dont understand how its the same for this equation
try this, let y=(x-3) and then factorise the equation fully and they replace y=x-3 back and see what you get. the substitute makes it simpler to see and understand
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brainmaster
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(Original post by Muttley79)
I think you've lost a sign but please don't do the second one for the OP.
won't do it
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Muttley79
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(Original post by 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
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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BenHarrow
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(Original post by 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.
Oh ok i understand that thanks both of you for the help i will try secomd example now
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Muttley79
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(Original post by BenHarrow)
Oh ok i understand that thanks both of you for the help i will try secomd example now
Post your working and we can help
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BenHarrow
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#20
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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
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