Question

See Attached; hey, is it possible if anyone can walk me through this review problem by any chance?

Answer 1

#6 is the one that I need assistance with, thanks again!

Answer 2

Hint:

\(\Large \frac{1}{\sqrt{4+5x}} = (4+5x)^{-\frac{1}{2}}\)

Answer 3

thanks so much!

Answer 4

I'm working it out right now..

Answer 5

I need help

Answer 6

If you had x^2, what must you multiply it by to get x^0 = 1 ?

Answer 7

i have no idea

Answer 8

it would be x^(-2), agreed?

Answer 9

x^2 times x^(-2) = x^(2+(-2)) = x^0 = 1

Answer 10

yes, that makes sense

Answer 11

ok so we're going to use this idea to factor

as @TheSmartOne already stated, \[\Large \frac{1}{\sqrt{4+5x}} = (4+5x)^{-\frac{1}{2}}\]

Answer 12

The exponential notation with the -1/2 is easier to deal with since fractions are a pain

Answer 13

What must we multiply (4+5x)^(-1/2) with to get 1 as the result?

ie fill in the blank

(4+5x)^(-1/2) times _________ = 1

Answer 14

hints
1) The bases must stay the same for the exponents to be added
2) The exponents must add to 0

Answer 15

(4+5x)^(1/2)

Answer 16

yes

Answer 17

So if we multiplied that first term by (4+5x)^(1/2), then the (4+5x)^(-1/2) would pair up with it and they would multiply to 1

Answer 18

yes

Answer 19

This idea will help us factor. Let's digress to a similar and simpler example

Let's say we had 2x+10

Naturally you can see that 2 factors out to get 2(x+5)

Let's do a more rigorous form of factoring.

---------------------------------------

Step 1) Start with 2x+10

Step 2) Multiply the expression by 1 to get 1*(2x+10). Leave the 1 there

Step 3) Transform the '1' into 2/2 or equivalently 2^1*2^(-1)

Step 4) In step 3, the numerator '2' will be outside the parenthesis. The denominator '2' will be divided by EVERY term inside. So 2x/2 = 1x = x and 10/2 = 5

So that's a more detailed process of going from 2x+10 to 2(x+5)

Let me know if these steps are something you agree with

Answer 20

yup

Answer 21

In step 4, instead of dividing by 2, we can multiply by 2^(-1) which leads to the same result

2^1 times 2^(-1) = 2^(1+(-1)) = 2^0 = 1

Answer 22

ok..

Answer 23

makes sense, just tricky and a bit creative

Answer 24

So what we'll do to factor problem 6 is multiply the entire expression by \[\Large (4+5x)^{1/2}*(4+5x)^{-1/2}\]
which is a fancy form of "1"

Answer 25

Since we want to factor out the (4+5x)^(-1/2), this means the "(4+5x)^(-1/2)" will stay outside the parenthesis

The (4+5x)^(1/2) will be distributed through to both terms

If you multiply (4+5x)^(1/2) with the first term, you would find that those radical terms go away (because they multiply to 1)

What happens when you multiply (4+5x)^(1/2) with the second term?

Answer 26

hold on..

Answer 27

1/2sqrt(4+5x)

Answer 28

You multiplied \[\Large (150x-80)(4+5x)^{1/2}\] with \[\Large (4+5x)^{1/2}\] right?

Answer 29

yup

Answer 30

What is \(\Large (4+5x)^{1/2}\) times \(\Large (4+5x)^{1/2}\) equal to?

Answer 31

hint: x^3 times x^3 = x^(3+3)

Answer 32

hey, I have to get going, but thanks for the tips, very helpful; I just have to complete this problem later.. Thanks again!!

Answer 33

Ok I'll show you a slightly different way to factor 2x+10


\[\Large 2x+10 = 1*(2x+10)\]
\[\Large 2x+10 = {\color{red}{1}}*(2x+10)\]
\[\Large 2x+10 = {\color{red}{2^1*2^{-1}}}*(2x+10)\]
\[\Large 2x+10 = 2^1*2^{-1}*(2x+10)\]
\[\Large 2x+10 = 2^1*{\color{blue}{2^{-1}}}*(2x+10)\]
\[\Large 2x+10 = 2^1*({\color{blue}{2^{-1}}}*2x+{\color{blue}{2^{-1}}}*10)\]
\[\Large 2x+10 = 2*(1x+5)\]
\[\Large 2x+10 = 2*(x+5)\]

This is the same idea used in problem 6, just a bit more complicated of course

Answer 34

ok thanks again!