# stuck on some problems

• May 2nd 2010, 01:53 PM
mmaquick
stuck on some problems
I have been out of math for years and I am stuck on 3-4 problems of my first struggle since returning to school

10x(to the 9th power) = 2x( ) I think it's 5x to the 8th power but am I forgetting some rules or something?

x(to the 2nd power) + 27 + 28 I believe it's prime but I may be wrong. none of the other answers work out for me or is it (x+28)(x-1)

how do I work out something like x(to the 4th power) -1
• May 2nd 2010, 02:04 PM
stapel
Quote:

Originally Posted by mmaquick
10x(to the 9th power) = 2x( ) I think it's 5x to the 8th power but am I forgetting some rules or something?

Is it (10x)^9, or 10x^9?

Quote:

Originally Posted by mmaquick
x(to the 2nd power) + 27 + 28 I believe it's prime but I may be wrong. none of the other answers work out for me or is it (x+28)(x-1)

What value have they given you for x, that you have concluded that the value of the expression is prime? What "other answers" are you talking about?

Quote:

Originally Posted by mmaquick
how do I work out something like x(to the 4th power) -1

What are you supposed to be doing with x^4 - 1? What is the "working out" that you want to do?
• May 2nd 2010, 02:14 PM
mmaquick
Quote:

Originally Posted by stapel
Is it (10x)^9, or 10x^9?

What value have they given you for x, that you have concluded that the value of the expression is prime? What "other answers" are you talking about?

What are you supposed to be doing with x^4 - 1? What is the "working out" that you want to do?

10x^9

2nd question I am factoring completly, and if the polynomial is prime I choose that answer my other options are
(x+7)(x-4) - is option A.
(x+28)(x-1) - opttion B
prime - option C
(x+28)(x+1) option D.

with x^4 - 1 these are my options
(x^2+1) (x-1)(x+1) option A
(x+2)(x^2-2x+4) -B
(x-2)(x^2 +2x +4) -C
(x+8)(x^2-1) -D
• May 2nd 2010, 02:23 PM
piglet
Quote:

Originally Posted by mmaquick

with x^4 - 1 these are my options
(x^2+1) (x-1)(x+1) option A
(x+2)(x^2-2x+4) -B
(x-2)(x^2 +2x +4) -C
(x+8)(x^2-1) -D

Right, i haven't a clue what your trying to do in the first 2 questions but here's how you do the question above

Use the difference of two squares formula twice i.e. \$\displaystyle x^{4} - 1 = (x^{2})^{2} - (1)^{2} = (x^{2}+1)(x^{2}-1) = (x^{2}+1)((x)^{2} - (1)^{2}) = (x^{2}+1)(x+1)(x-1) \$

Difference of two squares formula \$\displaystyle x^{2} - y^{2} = (x+y)(x-y) \$
• May 2nd 2010, 02:26 PM
mmaquick
Quote:

Originally Posted by piglet
Right, i haven't a clue what your trying to do in the first 2 questions but here's how you do the question above

Use the difference of two squares formula twice i.e. \$\displaystyle x^{4} - 1 = (x^{2})^{2} - (1)^{2} = (x^{2}+1)(x^{2}-1) = (x^{2}+1)((x)^{2} - (1)^{2}) = (x^{2}+1)(x+1)(x-1) \$

Difference of two squares formula \$\displaystyle x^{2} - y^{2} = (x+y)(x-y) \$

so would (x^2 + 1) (x-1) (x+1) work ? the - & + are switched
• May 2nd 2010, 02:38 PM
piglet
Quote:

Originally Posted by mmaquick
so would (x^2 + 1) (x-1) (x+1) work ? the - & + are switched

Yes of course it would. Think of \$\displaystyle x \$ as just a number. It's also useful to think of an easy example like the following; \$\displaystyle 2*3 = 3*2 = 6 \$
• May 2nd 2010, 02:59 PM
mmaquick
Quote:

Originally Posted by piglet
Right, i haven't a clue what your trying to do in the first 2 questions but here's how you do the question above

Use the difference of two squares formula twice i.e. \$\displaystyle x^{4} - 1 = (x^{2})^{2} - (1)^{2} = (x^{2}+1)(x^{2}-1) = (x^{2}+1)((x)^{2} - (1)^{2}) = (x^{2}+1)(x+1)(x-1) \$

Difference of two squares formula \$\displaystyle x^{2} - y^{2} = (x+y)(x-y) \$

I worked out that x^2 + 27 + 28 is prime so I no longer need assistance on that one

I am having more problems on the ones you seem good at
does x^6+1 work out this way
(x^3) + (1)^2 = (x^3-1)(x^3+1) = (x^3 +1) (x^3-1) did I do that right or is it something like (x+1)(x-1)(x^4-x^2+1)
• May 2nd 2010, 04:39 PM
mmaquick
Quote:

Originally Posted by mmaquick
I worked out that x^2 + 27 + 28 is prime so I no longer need assistance on that one

I am having more problems on the ones you seem good at
does x^6+1 work out this way
(x^3) + (1)^2 = (x^3-1)(x^3+1) = (x^3 +1) (x^3-1) did I do that right or is it something like (x+1)(x-1)(x^4-x^2+1)

still stuck on this one...
• May 3rd 2010, 01:48 AM
piglet
Quote:

Originally Posted by mmaquick
still stuck on this one...

Yes this question is very similar but it has a little twist. First you use the difference of two square formula that i mentioned above so
\$\displaystyle x^{6}+1 = (x^{3})^{2} + (1)^{2} = (x^{3}+1)(x^{3}-1) \$

Now you need to use the difference of two CUBES on \$\displaystyle (x^{3}-1)\$

Formula = \$\displaystyle x^{3 }- y^{3} = (x-y)(x^{2}+xy+y^{2}) \$

so \$\displaystyle (x^{3}-1) = (x-1)(x^{2} +x + 1) \$

You can also tidy up \$\displaystyle (x^{3}+1) \$ using the SUM of two CUBES formula.

Formula = \$\displaystyle x^{3}+y^{3} = (x+y)(x^{2}-xy+y^{2}) \$

so \$\displaystyle (x^{3}+1) = (x+1)(x^{2} -x +1) \$

=> your overall answer \$\displaystyle x^{6}+1 = (x-1)(x^{2} +x + 1)(x+1)(x^{2} -x +1) \$

Hope that helps