Question

@mathmate

Answer 1

@mathmate

Answer 2

factor 25-y^2

Answer 3

I would just square root both right? :P

Answer 4

difference of two squares!

Answer 5

(x+5)(x-5) ?

Answer 6

Yep!

Answer 7

Thanks :)

Answer 8

Now,
Simplify \(\Large \frac{3x^3}{6x^2}\)

Answer 9

\[\frac{ 3*x*x*x }{ 3*2*x*x }\]
\[\frac{ x }{ 2 }\]

Answer 10

What you should do \(whenever\) you divide out anything from the denominator is to make sure the factor cannot be zero.
3 is not zero, so nothing needs to be done! But...

Answer 11

-_-

Answer 12

You would specify that the answer is \(\frac{x}{2}\) for x\(\ne\)0

Answer 13

is that ok?

Answer 14

Remember whenever we cancel a factor from the denominator, we are not allowed to cancel unless the factor is \(not\) equal to zero.

Answer 15

Most teachers will take half the points away if you forgot that, or some of these conditions.

Answer 16

O_O

Answer 17

That was over copied!
Simplify \(\Large \frac{x(x^2+6)}{x^2}\)

Answer 18

(x^2+6) / x

Answer 19

Condition?

Answer 20

\[x \neq1\]

Answer 21

You cancelled x, so the condition is x\(\ne\)0

Answer 22

It's what you cancelled should not equal to zero.

Answer 23

x\(\ne\)1 is for cancelling (x-1)

Answer 24

are we good?

Answer 25

Yes

Answer 26

ok, ready for the next one!

Answer 27

mhm

Answer 28

Simplify \(\Large \frac{p^2-2p+1}{p^2-1}\)

Answer 29

Remember to specify the conditions if you cancelled any factor.

Answer 30

\[\frac{ (p−1)(p−1) }{ p^2-1}\]

Answer 31

Continue! There are factors for the denominator.

Answer 32

diff of two squares!

Answer 33

In the worst case, use the quadratic formula!

Answer 34

\[\frac{ (p-1)(p-1) }{ (p+1)(p-1) }\]

Answer 35

Excellent, you can now finish off. Tag on the conditions whenever you cancel!

Answer 36

\[\frac{ p-1 }{ p+1 }~~~~~~~ x \neq 1\]

Answer 37

** -1

Answer 38

x\(\ne\)1 is correct, because it is the same as (x-1)\(\ne\)0

Answer 39

is that clear how we got x\(\ne\)1 ?

Answer 40

yes

Answer 41

ready for another?

Answer 42

no more of these ;-;

Answer 43

This is going to be all of these for the rest of 11!
Just one more of these simple ones.
Once you can do the simple ones, you'll even like the more complicated ones! lol

Simplify \(\Large \frac{3(4-m)}{6(m-4)}\)

Answer 44

2

Answer 45

you get the idea, but a little too fast ! :(

Answer 46

Remember m-4 = -(4-m), and 3/6=1/2

Answer 47

... and the conditions!

Answer 48

x /= 1/3 , 3/6

Answer 49

Well, we have
\(\Large \frac{3(4-m)}{6(m-4)}\)
= \(\Large \frac{-3(m-4)}{6(m-4)}\)
=\(\Large \frac{-3(m-4)}{3*2(m-4)}\) so we cancel 3 and (m-4)
but we know that 3\(\ne\)0, so we don't have to specify as a condition, the other one is
m-4\(\ne\)0. If we add 4 to each side, we have m\(\ne\)4

Answer 50

So the final answer is
-1/2, where m\(\ne\)4

Answer 51

are we good, especially with conditions?
We'll be working with conditions and factorization throughout ch. 11

Answer 52

yessss

Answer 53

34.
simplify \(\Large \frac{x^2-9}{x^2-5x-6}\)

Answer 54

\[\frac{ (x+3)(x-3) }{ (x-6)(x+1)}\]

Answer 55

anything else to do or to write?

Answer 56

(note: the factoring is correct! Well-done!)

Answer 57

nope thats all : P

Answer 58

If you did not cancel anything, then you make a note: the expression is already in its simplest form. (and no conditions if nothing cancelled)

Answer 59

ok :)

Answer 60

Simplify if possible \(\Large \frac{3x-5}{25-30x+9x^2}\)

Answer 61

i need help ;-;

Answer 62

Remember steps for factoring:
1. take out GCF (common factors)
2. Check if it is difference of 2 squares (Must be a difference, and only has two terms)
3. Check if it is a perfect square:
First and last terms are perfect squares, and middle term is twice the product of the square-roots of the two end terms.
Example: a^2+2ab+b^2 : End terms are perfect squares, middle = 2(a)(b), so yes, this is a perfect square, equal to (a+b)^2.
4. Try factoring.

Answer 63

1. any common factors (in the denominator)?

Answer 64

no

Answer 65

2. Is it diff. of 2 squares?

Answer 66

no

Answer 67

3. Is it a perfect square?

Answer 68

no

Answer 69

Hmm, let's see.
Square-root of first term = sqrt(25) =5
square-root of last term =sqrt(9x^2)= 3x
(so far so good, both terms are perfect squares)

Answer 70

Now do a foil on (5-3x) and see if the middle term fits.
(5-3x)(5-3x)=5^2-15x-15x+9x^2=5^2-30x+9x^2 Yay, so it's a perfect, equal to
(5-3x)^2

Answer 71

So now can you complete it?

Answer 72

Shall we go back to perfect squares?

Answer 73

\[\frac{ 1 }{ 3x-5 }\]

Answer 74

Watch out: 5-3x = -1 (3x-5) and also since we cancelled 3x-5, we specify the condition x\(\ne\)5/3

Answer 75

>.<

Answer 76

So the answer is:

Answer 77

1/3x-5
x/= 5/3

Answer 78

Yep, that's correct!

Answer 79

For perfect squares, it has to satisfy 3 conditions
1. end terms are perfect squares, i.e. you can find the square roots without leaving the radical.
2. middle term equals twice the square-roots of the end terms.
3. the middle term has the same sign of the factored terms.

Answer 80

We check that with an example
Factor 81x^2- xy +64y^2.

Answer 81

Note that condition one means that both end terms must be positive.
Here
first term = 81x^2, square root = 9x

Answer 82

Last term = 64 y^2, square-root = ?

Answer 83

@pooja195 What is the square-root of the last term?

Answer 84

3 :/ ?

Answer 85

Last term = 64 y^2, square-root = ?

Answer 86

8y

Answer 87

Good.
What's twice the product of the square-roots?

Answer 88

huh ?

Answer 89

We need to multiply together the square-roots (of the first and third terms), and double it.

Answer 90

512? :/

Answer 91

im so confused :/

Answer 92

the square roots are 9x and 8y,so the product is 72xy, and doubling it shoud give 144xy.
This should match the middle term (if the expression is a perfect square).
It doesn't because I forgot to fill in the 144 in the middle.
So we conclude that (9x+8y)^2=81x^2+144xy+64y^2=(9x+8y)^2.