# Math Help - More Complex Factorisation

1. ## More Complex Factorisation

How can I factorise the following? I am new to more complex forms of factorisation and am teaching myself at home.

1. 4p^5q^3 + 6p^2q^4
2. 3(x+6)^2 - 9(x-6)
3. 12a^4 - 3
4. 36p^3 - 25pq^2
5. 2x^2 + x -21

How can I factorise the following? I am new to more complex forms of factorisation and am teaching myself at home.

1. 4p^5q^3 + 6p^2q^4
2. 3(x+6)^2 - 9(x-6)
3. 12a^4 - 3
4. 36p^3 - 25pq^2
5. 2x^2 + x -21
the first one is
$4p^5q^3 + 6p^2q^4$
$2p^2q^3(2p^3 + 3q)$

not sure if the 2nd one is typed ok

the 3rd one is

$12a^4 - 3$
$3 (2 a^2-1) (2 a^2+1)$

do you see what is happening???

3. ## Thank you very much, bigwave!

By the way, the 2nd one is in its original form from the textbook.

Ah! This is definitely an aid in my understanding. But I just have to become accustomed to putting into practice by myself.

So, correct me if I am wrong:
(6p^2-5pq)(6p+5q)

No, it's not right. I am baffled by this one.

4. ## as for the 4th

$
36p^3 - 25pq^2

$

first factor out what has exponents
so factor out p

$p\left(36p^2 - 25q^2\right)$

this will factor into a difference of squares

$
p\left(6p - 5q\right)\left(6p + 5q\right)
$

has the fog cleared

5. the last one,
$2x^2 + x -21=(x-3)(2x+7).$.

6. looks good to me

7. Any ideas about number 2?

8. Originally Posted by Raoh
the last one,
$2x^2 + x -21=(x-3)(2x+7).$.
Yes, I did manage to work this one out by myself. All except number 2.

Yes, I did manage to work this one out by myself. All except number 2.

given:
$3(x+6)^2 - 9(x-6)$

first expand the brackets

$3(x^2 +12x +36) - 9x +54$

$3x^2 +36x +108 - 9x + 54$

combine like terms

$

3x^2 + 27x + 162
$

factor out 3

$3(x^2 + 9x +54)$

hope this helps..