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anonymous
 5 years ago
Factoring is used to solve quadratic equations in order to turn each term into zero to get a solution set. Using the form ax^2+bx+c=0, you need to figure out what numbers would equal c when they are multiplied while the same numbers will total the amount of b when they're added. An example would be 5x^25x10=0.
5x^25x10
5x(x2)(x+1)=0
Because the a in ax^2 is equal to zero, 5x=0. Since x2=0 and x+1=0 the solution sets are x=0, x=2, x=1
How do I comment on this for a discussion question to the person that wrote this?
anonymous
 5 years ago
Factoring is used to solve quadratic equations in order to turn each term into zero to get a solution set. Using the form ax^2+bx+c=0, you need to figure out what numbers would equal c when they are multiplied while the same numbers will total the amount of b when they're added. An example would be 5x^25x10=0. 5x^25x10 5x(x2)(x+1)=0 Because the a in ax^2 is equal to zero, 5x=0. Since x2=0 and x+1=0 the solution sets are x=0, x=2, x=1 How do I comment on this for a discussion question to the person that wrote this?

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Did the instructor or a student write this?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0a student....just not sure what to write here...to agree or not and to go into detail...I am not good with this stuff...

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0The student explained the concept somewhat well. They said that when looking to factor a quadratic, you should find two numbers whose product is c and whose sum is b. Were you all supposed to mention if you have an a that doesn't equal 1? Also, the student factored 5x^2  5x  10 to 5x(x  2)(x + 1) when it SHOULD be 5(x  2)(x + 1). Therefore, the solution set should be x = 1 or x = 2, not x = 0. These are ideas, I'm not sure what else you could extrapolate from that answer.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0You could also point out that the a in ax^2 is not actually zero... it's five.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0ok...thank you so much!
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