# help with very tricky expanding and factorising questions

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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)

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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#2

(Original post by

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)

**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)

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#3

(Original post by

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)

**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)

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(Original post by

For a) what common expression(s), or factors, are there in the two terms?

**mqb2766**)For a) what common expression(s), or factors, are there in the two terms?

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(Original post by

So what multiplies that in each of the two terms?

**mqb2766**)So what multiplies that in each of the two terms?

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#7

(Original post by

i think its 7

**kyefrankie**)i think its 7

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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(Original post by

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)?

**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)?

3 for 21

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#9

7(p+1)(2)

Is not

14(p+1)^2

There is an extra term needed.

The second is correct.

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(Original post by

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

**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

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#12

**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)

-5a+15b

5(-a+3b)

then you gotta factorise it again, you can use a calculator

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(Original post by

About what?

**mqb2766**)About what?

(Original post by

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.

**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.

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#14

(Original post by

this part

**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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(Original post by

You start with

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?

**mqb2766**)You start with

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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#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.

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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(Original post by

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.

**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.

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#18

(Original post by

is it 7p + 7

**kyefrankie**)is it 7p + 7

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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#20

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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