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Math Help - Factoring polynomials

  1. #1
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    Factoring polynomials

    Hey, I have a question, you can factor using guess and check or grouping, here it is

    -13a + 12 - 4a^2

    the answer is -(a+4)(4a-3)

    Not sure how to get it using either one, please help! Thanks!
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    Quote Originally Posted by MathMack View Post
    Hey, I have a question, you can factor using guess and check or grouping, here it is

     -13a + 12 - 4a^2

    the answer is -(a+4)(4a-3)

    Not sure how to get it using either one, please help! Thanks!
    First re-arrange:

     -4a^2 - 13a + 12

    Now factor out the negative sign

     -(4a^2 + 13a - 12)

    Now we should take the  4a^2 term and factorise that, keeping in mind that it is not yet complete. (The   x_{1} and  x_{2} terms represent the constants)

     -(4a + x_{1})(a + x_{2})

    Now we should see which constants give us a product of 12.
    (1 x 12 ; 2 x 6 ; 3 x 4)

    Now from those options we should find which also satisfies the middle term.

    Can you take it from here? Just do it step by step.
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  3. #3
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    I also have 4 more, any help is appreciated!

    2.) n^3-n^2+n-1

    3.) 4a(a+2) - 8 (a+2)

    4.) -6x^3 +10x^2 - 4x^4

    5.) a rectangle has area described by the polynomial 6x^2 + x -1 suare centimeters. Find the length and width of the rectangle. Thanks again
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  4. #4
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    kinda still stuck on that first one. Thanks for your help
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  5. #5
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    Quote Originally Posted by MathMack View Post
    I also have 4 more, any help is appreciated!

    2.) n^3-n^2+n-1

    3.) 4a(a+2) - 8 (a+2)

    4.) -6x^3 +10x^2 - 4x^4

    5.) a rectangle has area described by the polynomial 6x^2 + x -1 suare centimeters. Find the length and width of the rectangle. Thanks again
    2)  n^2(n - 1) + 1(n - 1)
    = (n^2 + 1)(n - 1)

    3) (4a - 8)(a + 2)

    4) -2(3x^3 - 5x^2 + 2x^4)
    = -2x^2(3x - 5 + 2x^2)
    = -2x(2x^2 + 3x - 5)
    = -2x(2x + 5)(x - 1)

    5) 6x^2 + x - 1

    We know that this value is equal to 2 values multiplied by each other. So let's factorise it:

    (3x - 1)(2x + 1)
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  6. #6
    Bar0n janvdl's Avatar
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    Quote Originally Posted by MathMack View Post
    kinda still stuck on that first one. Thanks for your help
    Which part do you not understand? Give me the exact part where you got stuck and why you do not understand it.
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  7. #7
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    I got the part where you take the negative out, but after that i'm lost, I know it's (4a something)(a something) but lost to go after that, I know you must find factors but not sure
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  8. #8
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    Quote Originally Posted by MathMack View Post
    I got the part where you take the negative out, but after that i'm lost, I know it's (4a something)(a something) but lost to go after that, I know you must find factors but not sure
    You must find the factors of the 12 at the end. Then from all those factors, you must find the pair which satisfies the middle term too.

    In other words: When you expand the brackets again, you should end up with the same expression that you started with.
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  9. #9
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    and the answers in the back of the book for the problems you answered are a little different, thanks for helping me, i'm just confused!

    2.) n^3-n^2+n-1
    (n-1) (n^2 +1)
    3.) 4a(a+2) - 8 (a+2)
    4(a+2)(a-2)
    4.) -6x^3 +10x^2 - 4x^4
    -2x^2(2x+5)(x-1)
    5.) a rectangle has area described by the polynomial 6x^2 + x -1 suare centimeters. Find the length and width of the rectangle.
    2x-5
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  10. #10
    Bar0n janvdl's Avatar
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    Quote Originally Posted by MathMack View Post
    and the answers in the back of the book for the problems you answered are a little different, thanks for helping me, i'm just confused!

    2.) n^3-n^2+n-1
    (n-1) (n^2 +1)
    3.) 4a(a+2) - 8 (a+2)
    4(a+2)(a-2)
    4.) -6x^3 +10x^2 - 4x^4
    -2x^2(2x+5)(x-1)
    5.) a rectangle has area described by the polynomial 6x^2 + x -1 suare centimeters. Find the length and width of the rectangle.
    2x-5
    3) In number 3 they merely factorised (4a - 8) to 4(a - 2)

    5) Is the answer (2x - 5) ??
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  11. #11
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    yes and on the first one we worked on they have the answer -(a+4) (4a-3)

    Still lost on that one
    THANKS JANVDLL!
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  12. #12
    Bar0n janvdl's Avatar
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    Quote Originally Posted by MathMack View Post
    yes and on the first one we worked on they have the answer -(a+4) (4a-3)

    Still lost on that one
    THANKS JANVDLL!
    You're welcome but I'm more interested in making you understand.

    Tell me why 4 and -3 are the constants in the brackets?
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  13. #13
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    that's what i'm not sure on either, the part that we should find two numbers for is positive 12
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  14. #14
    Bar0n janvdl's Avatar
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    Quote Originally Posted by MathMack View Post
    that's what i'm not sure on either, the part that we should find two numbers for is positive 12
    But remember we factored a negative out...

    So then -(4a^2 + 13a - 12)

    Now just put that negative sign aside(But don't throw it away!).

    Imagine the equation would be only  4a^2 + 13a - 12

    Now can you see why we use 4 and -3 ?
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  15. #15
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    okay, I finally got that one all figured out completly, just took awhile, The next one I can't figure out is the n^3-n^2 +n -1
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