I ahve got an algebra question and it is really difficult for me...?

Question

Factorise
1 - p2- p3q + pq

note: the 2 and 3 in the question represents the power of so p2 is p square.

Thx for looking.

Answer 1

(1-p)2 -pq(p2+1)=(1-p)2-pq(p+1)2

Answer 2

First of all, most mathematicians and math students write powers using "^". So your p2 is p^2 in normal syntax. Also, it is preferable for you to write the p^3q as (p^3)q or qp^3 so we know you aren't talking about p^(3q) (p to the power of 3q).
Anyways, look at the second two terms alone:
-q(p^3) + pq factors as qp(-p^2+1) = qp(1-p^2).
Now note what the first two terms are: 1-p^2.
So your polynomial is (1-p^2) + qp(1 - p^2).
Now you can factor out the (1-p^2) to get:
(1-p^2)(1+qp).
The key to these problems is to think flexibly, and realize exactly what you're doing. Basically, you're going to do one or two easy factorizations, such as bringing out the qp from the last two terms. You might also have needed to factor out something from the first two terms to get your common factor. Once you do this (if you need to) it should be obvious what is the common factor between the factored pairs (like (1-p^2) above). Then you just factor this and you're done. However, in order to get that far, you need to realize which pairs you need to factor. It's not always the last two and the first two. You need to consider every pair and whether factoring it would help you to get a common factor with the rest of the problem.

Answer 3

Expression
-qp^3 - p^2 + qp + 1


Result
-(p-1) (p+1) (pq+1)

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Answer 4

Given expression
=(1-p^2)-(p^3q-pq)
={(1)^2-(p)^2}-pq{(p)^2-(1)^2}
=(1+p)(1-p)-pq(p+1)(p-1)
= -(p+1)(p-1)-pq(p+1)(p-1)
= -(p+1) (p-1)(1+pq)

Answer 5

p(p-p^2q+q)