1. ## Factoring...1

What is the first step in factoring the following problem?

(ax + b)^(-1/2) - root {(ax + b)/b}

2. ## Re: Factoring...1

Seems you'd start by using this rule: a^(-p) = 1 / a^p

1 / (ax + b)^(1/2) - {(ax + b)/b}^(1/2)

Remember that root n = n^(1/2)

Also, if you continued the factoring process, I suggest
setting ax + b = k, to get something simpler to work with:
1 / k^(1/2) - (k/b)^(1/2)

3. ## Re: Factoring...1

Originally Posted by DenisB
Seems you'd start by using this rule: a^(-p) = 1 / a^p

1 / (ax + b)^(1/2) - {(ax + b)/b}^(1/2)

Remember that root n = n^(1/2)

Also, if you continued the factoring process, I suggest
setting ax + b = k, to get something simpler to work with:
1 / k^(1/2) - (k/b)^(1/2)
I'll work on this problem using k as a substitute for
(ax + b) as you suggested and post my work this weekend.

4. ## Re: Factoring...1

I tried but couldn't factor completely. Can someone work this out for me?

5. ## Re: Factoring...1

1 / k^(1/2) - (k/b)^(1/2)

right side: (k/b)^(1/2) = k^(1/2) / b^(1/2)
so we now have:
1 / k^(1/2) - k^(1/2) / b^(1/2)

which is of style: 1/u - u/v

I'll let you continue...

6. ## Re: Factoring...1

Originally Posted by DenisB

1 / k^(1/2) - (k/b)^(1/2)

right side: (k/b)^(1/2) = k^(1/2) / b^(1/2)
so we now have:
1 / k^(1/2) - k^(1/2) / b^(1/2)

which is of style: 1/u - u/v

I'll let you continue...
In place of v, you meant b.

1/u - u/v = 1/u - u/b, right?

This looks like a subtraction of two fractions.
Do I proceed in terms of subtracting two fractions and then simplify some more?

7. ## Re: Factoring...1

In place of v, you meant b.

1/u - u/v = 1/u - u/b, right?
No. I replaced k^(1/2) with u and b^(1/2 with v, to get 1/u - u/v

Simplify that by multiplying by uv to get one fraction only.

Then you're finished. You can substitute back in if you wish.

8. ## Re: Factoring...1

Originally Posted by DenisB
No. I replaced k^(1/2) with u and b^(1/2 with v, to get 1/u - u/v

Simplify that by multiplying by uv to get one fraction only.

Then you're finished. You can substitute back in if you wish.
Great. I can take it from here. I will post my work tomorrow.

9. ## Re: Factoring...1

1/u - 1/v becomes (v - u)/uv.

What value for u and v must I substitute back in?

10. ## Re: Factoring...1

1/u - 1/v becomes (v - u)/uv.

What value for u and v must I substitute back in? Denis answered this in the quote box below, which was a couple of posts prior to this.
That last denominator needs grouping symbols around it, such as:

(v - u)/(uv)

Originally Posted by DenisB
No. I replaced k^(1/2) with u and b^(1/2)with v, to get 1/u - u/v
$\dfrac{b^{1/2} - k^{1/2}}{k^{1/2}b^{1/2}} \ =$

11. ## Re: Factoring...1

Originally Posted by greg1313
That last denominator needs grouping symbols around it, such as:

(v - u)/(uv)

$\dfrac{b^{1/2} - k^{1/2}}{k^{1/2}b^{1/2}} \ =$
So, do I now factor or is this the final answer?

12. ## Re: Factoring...1

So, do I now factor or is this the final answer?
Although k = ax + b in this method, I don't see a direct/handy route for showing a factor being taken out.

Starting over:

(ax + b)^(-1/2) - root {(ax + b)/b}

$(ax + b)^{-1/2} - \dfrac{(ax + b)^{1/2}}{b^{1/2}} =$

$(ax + b)^{-1/2} - (ax + b)^{1/2}b^{-1/2} =$

$(ax + b)^{-1/2}{b^{-1/2}}[b^{1/2} - (ax + b)] =$

$[b(ax + b)]^{-1/2}(b^{1/2} - ax - b)$

After this post, I don't expect to post further on this thread.

13. ## Re: Factoring...1

Originally Posted by greg1313
Although k = ax + b in this method, I don't see a direct/handy route for showing a factor being taken out.

Starting over:

(ax + b)^(-1/2) - root {(ax + b)/b}

$(ax + b)^{-1/2} - \dfrac{(ax + b)^{1/2}}{b^{1/2}} =$

$(ax + b)^{-1/2} - (ax + b)^{1/2}b^{-1/2} =$

$(ax + b)^{-1/2}{b^{-1/2}}[b^{1/2} - (ax + b)] =$

$[b(ax + b)]^{-1/2}(b^{1/2} - ax - b)$

After this post, I don't expect to post further on this thread.
I appreciate the breakdown and quick replies.