help with very tricky expanding and factorising questions

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)

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?

So what multiplies that in each of the two terms?

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)?

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

im a bit confused

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)

About what?

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

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?

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

i think its
a = 7(p+1)
b = 7p
c = 7