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Math Help - Basic Factoring Question

  1. #1
    Member cmf0106's Avatar
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    Basic Factoring Question

    Hello I am just starting factoring, the book lists the correct method for solving the following problem as.
    8x+8= 8*x+8*1=8(x+1)

    but what if you were to do it like this?
    8x+8=4*2x + 4*2=4(2x+2)

    Very confusing since they both give the correct answer once distributed.
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by cmf0106 View Post
    Hello I am just starting factoring, the book lists the correct method for solving the following problem as.
    8x+8= 8*x+8*1=8(x+1)

    but what if you were to do it like this?
    8x+8=4*2x + 4*2=4(2x+2)

    Very confusing since they both give the correct answer once distributed.
    the last is not fully factored, since there is still a common factor of 2 among the terms in the brackets. so yes, it is still correct, but as far as factoring goes, it is incomplete
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  3. #3
    Member cmf0106's Avatar
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    Thanks & one last thing for now.
    In 12xyz(4x^4y^2z^5+5x^3z^2+3x^5y)
    x^2 can be factored out because in the term 5x^3z^2 x^2 is the MOST that can removed and still remain an x value. And since factoring is "divided into" we are subtracting the exponent values because \frac{a^m}{a^n}=a^{m-n}. Thus x^2 can be factored out, does that logic sound correct? Also this procedure is not done for the z and y variables because they do not show up in all the terms right?
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by cmf0106 View Post
    Thanks & one last thing for now.
    In 12xyz(4x^4y^2z^5+5x^3z^2+3x^5y)
    x^2 can be factored out because in the term 5x^3z^2 x^2 is the MOST that can removed and still remain an x value. And since factoring is "divided into" we are subtracting the exponent values because \frac{a^m}{a^n}=a^{m-n}. Thus x^2 can be factored out, does that logic sound correct? Also this procedure is not done for the z and y variables because they do not show up in all the terms right?
    basically, you can factor out the lowest power of any term common to all terms. in the case of x, it is x^3

    your logic seems about right, but i would call it "dividing out of" but maybe that makes no sense. you are correct with y and z. but yes, you watch the powers in the same way you stated. and you can double check yourself by seeing that when you are multiplying out again, the powers add up to give you what was in the original
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  5. #5
    Member cmf0106's Avatar
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    Same type of question for this one 28xy^2-14x
    The book has the answer -7x(-4y^2+2 but I solved it
    14x * 2y^2 - 14x * 1= 14x(2y^2-1). It seems like I picked the correct number combination, because in the original problem 14x is the largest number that divides into both 14x and 28xy
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by cmf0106 View Post
    Same type of question for this one 28xy^2-14x
    The book has the answer -7x(-4y^2+2 but I solved it
    14x * 2y^2 - 14x * 1= 14x(2y^2-1). It seems like I picked the correct number combination, because in the original problem 14x is the largest number that divides into both 14x and 28xy
    your answer is better. good job. the book still left a factor of 2 to be taken out
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