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Math Help - General Factoring

  1. #1
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    Question General Factoring

    can anyone please help me with this question:

    Xwith a exponent of 3 -4x with exponent 2+5x-20..I dont know how to make the exponents small maybe someone can help me with that to.
    Last edited by jay123; September 22nd 2007 at 05:01 PM. Reason: forgot exponent
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    Quote Originally Posted by jay123 View Post
    can anyone please help me with this question:

    Xwith a exponent of 3 -4x with exponent 2+5x-20..I dont know how to make the exponents small maybe someone can help me with that to.
    Here we can factor by groups. note that there is a common x^2 among the first two terms, and a common 5 among the last two, so pull those out.

    x^3 - 4x^2 + 5x - 20 = x^2 (x - 4) + 5(x - 4)

    now what can we do with that?


    by the way, write powers using "^"

    so you should write x^3 - 4x^2 + 5x - 20
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  3. #3
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    I believe we can do this:

    (x-4)^2 (x^2+5) right? wait a second can I make (x-4)^2?
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    Quote Originally Posted by jay123 View Post
    I believe we can do this:

    (x-4)^2 (x^2+5) right? wait a second can I make (x-4)^2?
    no, you factor out the common (x - 4). how did you make it squared? it is simply (x - 4) \left( x^2 + 5 \right)
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  5. #5
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    i c. ok but what about when there is X^4-81? I know that since it is ^4 that 3 goes into 81 4 times. How would i write this ouT?
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    Quote Originally Posted by jay123 View Post
    i c. ok but what about when there is X^4-81? I know that since it is ^4 that 3 goes into 81 4 times. How would i write this ouT?
    no. this is the difference of two squares (do you see it?)

    we factor the difference of two squares as follows:

    a^2 - b^2 = (a + b)(a - b)

    now, how do you think this question should be done
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  7. #7
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    sorry i dont understand....difference of squares yes...can i separate
    x^4-81

    (x-3) (x+3)

    x^2+3x-3x-9= X^2-9
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    Quote Originally Posted by jay123 View Post
    sorry i dont understand....difference of squares yes...can i separate
    x^4-81

    (x-3) (x+3)

    x^2+3x-3x-9= X^2-9
    note that: x^4 - 81 = \left( x^2 \right)^2 - 81

    now what?
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  9. #9
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    can you help me with this problem

    (2y+3)^2-5(2y+3)+6
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  10. #10
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    If we call u=2y+3, it remains to factorise u^2-5u+6

    Factorise that, then back substitute.
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    Quote Originally Posted by jay123 View Post
    can you help me with this problem

    (2y+3)^2-5(2y+3)+6
    did you get the last one? the difference of two squares?

    note that for this question, we have a quadratic in 2y + 3. if you can't see that right away, try replacing 2y + 3 with a variable, say x, then you get:

    x^2 - 5x + 6

    now continue...
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  12. #12
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    yes, i did.
    I still don't understand...
    x-5x+6 how do i solve this?

    what rule should i follow?
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  13. #13
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    lets see if i X^2-5x+6 = (x-2)(x-3)...then what????do i multiply

    (X-2)(X-3)(2y+3) but i still have the (2y+3)^2
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    Quote Originally Posted by jay123 View Post
    lets see if i X^2-5x+6 = (x-2)(x-3)...then what????do i multiply

    (X-2)(X-3)(2y+3) but i still have the (2y+3)^2
    no, just replace the x with 2y + 3, that's what you used x to represent, remember?
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  15. #15
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    lol...im still lost.

    do i
    X^2-5(2y+3)+6
    X^2-10y-15+6
    X^2-10y+1
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