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.\(\displaystyle (x + y)(a + b) - (y + z)(a + b)\)

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

2. \(\displaystyle (2x + y)^2 -3(2x + y)\)

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

3. \(\displaystyle 5(m - 2n) - (m -2n)^2\)

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

4. \(\displaystyle p^2 -9q^2 +6qr -r^2\)

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

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

5. \(\displaystyle 4a^2 -b^2 -2bc -c^2\)

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

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

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