Two questions involving factoring

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In summary, the conversation involves two questions on factoring: 1) 5x^1/2 - 15x^3/2, and 2) 3(x-6)^2 + 2(x-6)^4 + 3/(x-6). The first question has been correctly factored to 5x^1/2 (1-3x), but the asker is unsure if it can be simplified further. The second question has also been partially factored, but the asker forgot to divide by (x-6) after multiplying by it. The correct final answer is (x-6)^2 (3+2(x-6)^2) + 3. The asker
  • #1
erik05
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Hello all. I have two questions involving factoring and I seem to be stuck.

1) [tex] 5x^\frac {1}{2} - 15x^\frac{3}{2} [/tex]


I tried it and got: [tex] 5x^\frac {1}{2} (1-3x) [/tex]
I'm not too sure if you could go any further than this or if there's another way to approach this. The answer seems too simple and knowing my teacher, there's probably a more complicated one. Any thoughts?

2) [tex] 3(x-6)^2 + 2(x-6)^4 + \frac {3}{x-6} [/tex]

so far I got: [tex] 3(x-6)^3 + 2(x-6)^5 + 3 [/tex] and taking out a common factor of [tex] (x-6)^3 [/tex] I got: [tex] (x-6)^3 (3+2(x-6)^2) + 3 [/tex]
I don't think this can be the simplest form so any suggestions or ideas? Thanks for the help.
 
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  • #2
erik05 said:
Hello all. I have two questions involving factoring and I seem to be stuck.

1) [tex] 5x^\frac {1}{2} - 15x^\frac{3}{2} [/tex]


I tried it and got: [tex] 5x^\frac {1}{2} (1-3x) [/tex]
I'm not too sure if you could go any further than this.

2) [tex] 3(x-6)^2 + 2(x-6)^4 + \frac {3}{x-6} [/tex]

so far I got: [tex] 3(x-6)^3 + 2(x-6)^5 + 3 [/tex] and taking out a common factor of [tex] (x-6)^3 [/tex] I got: [tex] (x-6)^3 (3+2(x-6)^2) + 3 [/tex]
I don't think this can be the simplest form so any suggestions or ideas? Thanks for the help.

#1) CORRECT

#2) You forgot to divide thru by (x - 6) after multiplying by (x - 6) in your first step, so your final answer should be:

[tex] \frac {(x-6)^3 (3+2(x-6)^2) + 3} {x - 6} [/tex]


~~
 
  • #3
I have a question. Would it be correct then to simplify it even further by dividing the top [tex] (x-6)^3[/tex] with the bottom [tex] (x-6) [/tex] to get an answer of [tex] (x-6)^2 (3+2(x-6)^2)+3 [/tex] or no?
 
  • #4
erik05 said:
I have a question. Would it be correct then to simplify it even further by dividing the top [tex] (x-6)^3[/tex] with the bottom [tex] (x-6) [/tex] to get an answer of [tex] (x-6)^2 (3+2(x-6)^2)+3 [/tex] or no?
It would not be correct to do that because of the +3 on the end.
 
  • #5
I should learn to look at the question more carefully. Thanks man.
 

FAQ: Two questions involving factoring

What is factoring?

Factoring is the process of finding the factors, or numbers that can be multiplied together to equal a given number, of a given polynomial or expression.

Why is factoring important?

Factoring is important because it allows us to simplify complicated expressions and solve equations more easily. It also helps us identify patterns and relationships within mathematical concepts.

How do you factor a polynomial?

To factor a polynomial, you need to look for common factors and use algebraic techniques such as grouping, difference of squares, trinomial factoring, or the quadratic formula.

What is the difference between factoring and solving?

Factoring involves breaking down an expression into its factors, while solving involves finding the value(s) of the variable that make the equation true.

Can all polynomials be factored?

No, not all polynomials can be factored. Some polynomials, such as prime polynomials, cannot be factored any further because they only have one set of factors.

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