1. ## Factoring...2

What is the first step in factoring the following problem?

64 (x - a)^4 - x + a

2. ## Re: Factoring...2

rewrite as
$64(x-a)^4-1(x-a)$
and then you can factor out the x - a
$=(x-a)[64(x-a)^3-1]$ and then complete the factor using the difference of two cubes

3. ## Re: Factoring...2

(x - a)[64 (x - a)^3 - 1]

We use a^3 – b^3 = (a – b)(a^2 + ab + b^2).

Let a = (x - a)

Let b = 1

(x - a - 1)[(x - a)^2 + (x - a)(1) + (1)^2]

(x - a - 1)[64 (x - a)^2 + (x - a) + 1

Putting it all together I got the following:

(x - a)[(x - a - 1)64 (x - a)^2 + (x - a) + 1]

Is this right?

4. ## Re: Factoring...2

(x - a)[64 (x - a)^3 - 1]

We use a^3 – b^3 = (a – b)(a^2 + ab + b^2).

Let a = (x - a)

Let b = 1

(x - a - 1)[(x - a)^2 + (x - a)(1) + (1)^2]

(x - a - 1)[64 (x - a)^2 + (x - a) + 1

Putting it all together I got the following:

(x - a)[(x - a - 1)64 (x - a)^2 + (x - a) + 1]

Is this right?
Not quite. Note that:
$64 (x - a)^3 - 1 = 4^3*(x-a)^3 -1^3 = [4(x-a)]^3 - 1^3$ and now factorise the diff of 2 cubes.

5. ## Re: Factoring...2

Originally Posted by Debsta
Not quite. Note that:
$64 (x - a)^3 - 1 = 4^3*(x-a)^3 -1^3 = [4(x-a)]^3 - 1^3$ and now factorise the diff of 2 cubes.
Ok. I will work on it later or tomorrow.

6. ## Re: Factoring...2

Ok. I will work on it later or tomorrow.
[4(x-a)-1]([4 (x-a)]^2 + 4 (x-a) + 1)

(4x - 4a - 1)[(4x-4a)^2 + (4x - 4a) - 1]

Yes or no?

7. ## Re: Factoring...2

Originally Posted by Debsta
Not quite. Note that:
$64 (x - a)^3 - 1 = 4^3*(x-a)^3 -1^3 = [4(x-a)]^3 - 1^3$ and now factorise the diff of 2 cubes.
[4(x-a)-1]([4 (x-a)]^2 + 4 (x-a) + 1]

(4x - 4a - 1)[(4x-4a)^2 + (4x - 4a) + 1]

Yes or no?

8. ## Re: Factoring...2

[4(x-a)-1]([4 (x-a)]^2 + 4 (x-a) + 1]

(4x - 4a - 1)[(4x-4a)^2 + (4x - 4a) + 1]

Yes or no?
Yes

9. ## Re: Factoring...2

Originally Posted by Debsta
Yes
Good.

10. ## Re: Factoring...2

It feels good to be right.