Help with Algebra?

Question

Factor completely:
1-s^4p^4

and solve usind the principle of zero..
r^2+9=6r

Answer 1

Hi,

Factor completely. this is the difference of perfect squares:

1-s^4p^4

1 is really 1² and s^4p^4 is really (s²p²)² so both factors start with 1 and end with s²p². One factor has a "+" between them and the other has a "-" between them. The factors you get are:

(1 - s²p²)(1 + s²p²)

The first factor is still the difference of perfect squares, so it factors again. 1 is still 1² and s²p² is really (sp)², so both factors start with 1 and end with sp. One factor has a "+" between them and the other has a "-" between them. The factors you get are:

(1 - sp)(1 + sp)(1 + s²p²)

For the second problem, solve using the principle of zero..
r² + 9 = 6r

Move all terms to one side of the equation, set equal to zero.
r² - 6r + 9 = 0

Factor it.
(r - 3)(r - 3) = 0

Both factors are the same, so just set one of them equal to zero and solve.

r - 3 = 0
r = 3 This is the answer for both factors.

I hope that helps!! :-)

Answer 2

3

Answer 3

First one is a difference of 2 squares, so

1-s^4p^4 =

(1-s^2p^2)(1+s^2p^2)

and out of that, the first factor is also a difference of 2 squares...

=(1-sp)(1+sp)(1+s^2p^2)

For the second problem, subtract 6r from both sides and rewrite as

r^2 - 6r + 9 = 0

which factors to

(r-3)(r-3) = 0

so if r=3, your factors equal zero and the equation is satisfied.

Answer 4

Question 1
(1 - s²p²) (1 + s²p²)
(1 - sp) (1 + sp) (1 + s²p²)

Question 2
r² - 6r + 9 = 0
(r - 3) (r - 3) = 0
r = 3

Answer 5

1-s^4p^4
=1-(sˆ2pˆ2)ˆ2
= (1+s^2p^2) (1-s^2p^2)
= (1+s^2p^2) [1-(sp)^2]
= (1+s^2p^2) (1+sp)(1-sp)

remember the basic principle:
xˆ2-yˆ2
=(x+y)(x-y)


r^2+9=6r
rˆ2 - 6r + 9 = 0
(r-3)(r-3) = 0
r=3

Answer 6

I don't understand what you are asking in the first question.

second one is

(r - 3)^2

r = 3

Answer 7

1^2-(s^2.p^2)^2
(1-s^2.p^2)(1+s^2.p^2)
(1-s.p)(1+s.p)(1+s^2.p^2)

(r-3)^2=0
r=3 ans