# Thread: Several algebra questions regarding factorisation

1. ## Several algebra questions regarding factorisation

A very good day to all.
I have several questions all relating to algebra, and I will be very thankful if you can help me solve them. Please also show your workings clearly so that I may understand wholly. I am sorry if these questions take a lot a time.

Factorise the following:

1. $(x + y)(a + b) - (y + z)(a + b)$

2. $(2x + y)^2 -3(2x + y)$

3. $5(m - 2n) - (m -2n)^2$

4. $p^2 -9q^2 +6qr -r^2$

5. $4a^2 -b^2 -2bc -c^2$

2. Originally Posted by PythagorasNeophyte
A very good day to all.
I have several questions all relating to algebra, and I will be very thankful if you can help me solve them. Please also show your workings clearly so that I may understand wholly. I am sorry if these questions take a lot a time.

Factorise the following:

1. $(x + y)(a + b) - (y + z)(a + b)$
Observe that $(a+b)$ is the common factor. So $(x+y)(a+b)-(y+z)(a+b)=(a+b)(\ldots)$

2. $(2x + y)^2 -3(2x + y)$
Observe that $(2x+y)$ is the common factor. So $(2x+y)^2-3(2x+y)=(2x+y)(\ldots)$

3. $5(m - 2n) - (m -2n)^2$
Observe that $(m-2n)$ is the common factor. So $5(m-2n)-(m-2n)^2=(m-2n)(\ldots)$

4. $p^2 -9q^2 +6qr -r^2$
This one is slightly more complicated. First observe that $p^2-9q^2+6rq-r^2=p^2-(9q^2-6qr+r^2)$.

Now, $9q^2-6qr+r^2=(3q-r)^2$. So now we are left with $p^2-(3q-r)^2$, which can be factored using the difference of squares formula.

5. $4a^2 -b^2 -2bc -c^2$
This is similar to #4. Observe that $4a^2-b^2-2bc-c^2=(2a)^2-(b^2+2bc+c^2)$.

Now, $b^2+2bc+c^2=(b+c)^2$. So now we are left with $(2a)^2-(b+c)^2$, which can be factored using the difference of squares formula.

Does this make sense? Can you try these problems now?

3. Thank you so much for your help Chris!

However, I still could not work out questions 1, 2 and 3 myself. I know the common factors but I am not sure about the other terms.

Also, I would appreciate if you could verify my answers for question 4 and 5.
The answer for question 4 might be---

$p^2 -(3q -r)^2$
$=[p^2 +(3q-r)][p^2 -(3q -r)]$
$=(p^2 +3q -r)(p^2 -3q +r)$

Whereas for question 5---

$(2a)^2 -(b+c)^2$
$=[2a +(b+c)][2a -(b +c)$
$= (2a +b +c)(2a -b -c)$

Is my answers and workings correct?

4. Originally Posted by PythagorasNeophyte
However, I still could not work out questions 1, 2 and 3 myself. I know the common factors but I am not sure about the other terms.

$(a+b)(x+y-y-z)=(a+b)(x-z)$

$(2x+y)^2-3(2x+y)=(2x+y)(2x+y)-3(2x+y)=(2x+y)(2x+y-3)$

same as second
$(m-2n)(5-m+2n)$