Difficult factorisation? (1 question!)?

Question

Hi,
I'm doing h/w on factorising quadratics and I really can't work out how I'd go about factorising this expression:
p^2 - 8pq + 15q^2
Any ideas (+ explanations)?
Thanks!
Lorna

It's alright - I've done it now but if you still want to answer, I'll give you 10 points!

Answer 1

(p-3q)(p-5q)

Answer 2

p^2 - 8pq + 15q^2
= (p - 5q)(p - 3q)

Equate each binomial factor to zero:

p - 5q = 0
p = 5q Answer

p - 3q = 0
p = 3q Answer

teddyboy

Answer 3

p^2 - 8pq + 15q^2 ;
Lert's try to complete a square:
(p -4q)^2 = p^2 -8pq +16q^2
if we add and sustract (16 q^2) to (p^2 -8pq) , we get:

(p -4q)^2 -16q^2 + 15q^2 = (p -4q)^2 - q^2

use the identity (a^2 - b^2) = (a - b) ( a + b)

we get: (p -4q -q)(p -4q +q) = ( p -5q)(p -3q)

Answer 4

p^2 - 8pq + 15q^2

=>p^2 - 5pq - 3pq + 15q^2

=>p(p - 5q) - 3q(p - 5q)

=>(p-5q)(p-3q)

Answer 5

(a-b)^2 = a^2 - 2ab + b^2
a^2 - b^2 = (a-b)(a+b)

p^2 - 2*4pq + (4q)^2 - q^2
=(p-4q)^2 - q^2
=(p-3q)(p-5q)