# Several algebra questions regarding factorisation

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

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

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

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

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

#### Chris L T521

MHF Hall of Fame
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?

• HallsofIvy and PythagorasNeophyte

#### PythagorasNeophyte

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---

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

Whereas for question 5---

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

Is my answers and workings correct?

#### yeKciM

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.

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

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

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