# help with very tricky expanding and factorising questions

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#1
hi, im really struggling with expanding and factorising (unit 2.2 in the edexcel textbook), i now know the answer but do not understand how to reach it. I would be very grateful for a step by step guide on how to reach the answer so i am able to fully understand the topic.
thank you

factorise completely

a) 14(p + 1)^2 + 21(p + 1) --> answer = 7(p + 1)(2p + 5)

b) 5(c + 1)^2 - 10(c + 1) --> answer = 5(c + 1)(c - 1)

c) 12(y + 4)^2 - 8(y + 4) --> answer = 4(y + 4)(3y + 10)

d) (a + 3b)^2 - 2(a + 3b) --> answer = (a + 3b)(a + 3b - 2)

e) 5(f +5) + 10f( f + 5) --> answer = 5(f + 1)(1 + 2f)

f) 5(a + b) - 10(a + b) --> answer = 5(a + b)(a + b - 2)
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4 weeks ago
#2
(Original post by kyefrankie)
hi, im really struggling with expanding and factorising (unit 2.2 in the edexcel textbook), i now know the answer but do not understand how to reach it. I would be very grateful for a step by step guide on how to reach the answer so i am able to fully understand the topic.
Thank you

factorise completely

a) 14(p + 1)^2 + 21(p + 1) --> answer = 7(p + 1)(2p + 5)

b) 5(c + 1)^2 - 10(c + 1) --> answer = 5(c + 1)(c - 1)

c) 12(y + 4)^2 - 8(y + 4) --> answer = 4(y + 4)(3y + 10)

d) (a + 3b)^2 - 2(a + 3b) --> answer = (a + 3b)(a + 3b - 2)

e) 5(f +5) + 10f( f + 5) --> answer = 5(f + 1)(1 + 2f)

f) 5(a + b) - 10(a + b) --> answer = 5(a + b)(a + b - 2)
i've never seen such easy questions in my life!!!!!
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4 weeks ago
#3
(Original post by kyefrankie)
hi, im really struggling with expanding and factorising (unit 2.2 in the edexcel textbook), i now know the answer but do not understand how to reach it. I would be very grateful for a step by step guide on how to reach the answer so i am able to fully understand the topic.
thank you

factorise completely

a) 14(p + 1)^2 + 21(p + 1) --> answer = 7(p + 1)(2p + 5)

b) 5(c + 1)^2 - 10(c + 1) --> answer = 5(c + 1)(c - 1)

c) 12(y + 4)^2 - 8(y + 4) --> answer = 4(y + 4)(3y + 10)

d) (a + 3b)^2 - 2(a + 3b) --> answer = (a + 3b)(a + 3b - 2)

e) 5(f +5) + 10f( f + 5) --> answer = 5(f + 1)(1 + 2f)

f) 5(a + b) - 10(a + b) --> answer = 5(a + b)(a + b - 2)
For a) what common expression(s), or factors, are there in the two terms?
1
#4
(Original post by mqb2766)
For a) what common expression(s), or factors, are there in the two terms?
i think 7( p + 1)
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4 weeks ago
#5
So what multiplies that in each of the two terms?
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#6
(Original post by mqb2766)
So what multiplies that in each of the two terms?
i think its 7
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4 weeks ago
#7
(Original post by kyefrankie)
i think its 7
For the first term, what multiplied 7(p+1) to get
14(p+1)^2

Then same question for the second term
21(p+1)?
Last edited by mqb2766; 4 weeks ago
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#8
(Original post by mqb2766)
For the first term, what multiplied 7(p+1) to get
14(p+1)^2

Then same question for the second term
21(p+1)?
2 for 14
3 for 21
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4 weeks ago
#9
(Original post by kyefrankie)
2 for 14
3 for 21
But for the first term
7(p+1)(2)
Is not
14(p+1)^2
There is an extra term needed.

The second is correct.
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#10
(Original post by mqb2766)
But for the first term
7(p+1)(2)
Is not
14(p+1)^2
There is an extra term needed.

The second is correct
im a bit confused
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4 weeks ago
#11
(Original post by kyefrankie)
im a bit confused
0
4 weeks ago
#12
(Original post by kyefrankie)
hi, im really struggling with expanding and factorising (unit 2.2 in the edexcel textbook), i now know the answer but do not understand how to reach it. I would be very grateful for a step by step guide on how to reach the answer so i am able to fully understand the topic.
thank you

factorise completely

a) 14(p + 1)^2 + 21(p + 1) --> answer = 7(p + 1)(2p + 5)

b) 5(c + 1)^2 - 10(c + 1) --> answer = 5(c + 1)(c - 1)

c) 12(y + 4)^2 - 8(y + 4) --> answer = 4(y + 4)(3y + 10)

d) (a + 3b)^2 - 2(a + 3b) --> answer = (a + 3b)(a + 3b - 2)

e) 5(f +5) + 10f( f + 5) --> answer = 5(f + 1)(1 + 2f)

f) 5(a + b) - 10(a + b) --> answer = 5(a + b)(a + b - 2)
f) 5a+5b-10a+10b
-5a+15b
5(-a+3b)
then you gotta factorise it again, you can use a calculator
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#13
(Original post by mqb2766)
(Original post by mqb2766)
But for the first term
7(p+1)(2)
Is not
14(p+1)^2
There is an extra term needed.

The second is correct.
this part
0
4 weeks ago
#14
(Original post by kyefrankie)
this part
14(p+1)^2
and you have to split it up into two terms which multiply together to give the original expression. One is
7(p+1)
If you just multiplied it by 2, you'd get
14(p+1)
Which isn't the original expression. As well as 2, what else do you need to multip!y it by?
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#15
(Original post by mqb2766)
14(p+1)^2
and you have to split it up into two terms which multiply together to give the original expression. One is
7(p+1)
If you just multiplied it by 2, you'd get
14(p+1)
Which isn't the original expression. As well as 2, what else do you need to multip!y it by?
(p + 1)
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4 weeks ago
#16
Yes, so the first term is written as
7(p+1) * 2(p+1)
The second is
7(p+1) * 3

Now factorize 7(p+1) out of both terms and group together the other parts.
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#17
(Original post by mqb2766)
Yes, so the first term is written as
7(p+1) * 2(p+1)
The second is
7(p+1) * 3

Now factorize 7(p+1) out of both terms and group together the other parts.
is it 7p + 7
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4 weeks ago
#18
(Original post by kyefrankie)
is it 7p + 7
You've got half an eye on the answer? 7(p+1) is part of it? We have
ab + ac = a(b+c)
where a is common expression and b and c are the other two factors you've just worked out. ab and ac are the two original terms. So put it in that form.
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#19
i think its
a = 7(p+1)
b = 7p
c = 7
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4 weeks ago
#20
(Original post by kyefrankie)
i think its
a = 7(p+1)
b = 7p
c = 7
b = 2(p+1)
c = 3

Not sure of your background, but I'd recommend going over something like
https://www.bbc.co.uk/bitesize/guide...d2p/revision/1
Before going any further. That has the very basics, try doing the simple exercises first.
Last edited by mqb2766; 4 weeks ago
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