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Math Help - Factoring equation...

  1. #1
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    Factoring equation...

    How would I factor these equations?

    1. 6x + 3x = 0

    2. -2x + 5x + 3 = 0

    I know how to factor equations without the number before
    x. But when the number is there I don't know how to factor it.
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by lp_144 View Post
    How would I factor these equations?

    1. 6x + 3x = 0

    2. -2x + 5x + 3 = 0

    I know how to factor equations without the number before
    x. But when the number is there I don't know how to factor it.
    for the first, 3x is a common term. just factor it out

    for the second, in general we use what we call the AC-method to factor, if possible, a quadratic in which the coefficient of the squared term is not 1. but for small numbers like 2, we can use trial and error, it would generally be a lot simpler.

    so -2x^2 + 5x + 3 = 0

    \Rightarrow 2x^2 - 5x - 3 = 0

    \Rightarrow (2x ~~a)(x ~~b) = 0

    now in the place of a and b, you want numbers that multiply to give you -3, but a and 2b will add to give -5
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    Quote Originally Posted by Jhevon View Post
    for the first, 3x is a common term. just factor it out

    for the second, in general we use what we call the AC-method to factor, if possible, a quadratic in which the coefficient of the squared term is not 1. but for small numbers like 2, we can use trial and error, it would generally be a lot simpler.

    so -2x^2 + 5x + 3 = 0

    \Rightarrow 2x^2 - 5x - 3 = 0

    \Rightarrow (2x ~~a)(x ~~b) = 0

    now in the place of a and b, you want numbers that multiply to give you -3, but a and 2b will add to give -5
    So it would be (2x+1)(x-3)?
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by lp_144 View Post
    So it would be (2x+1)(x-3)?
    yes, very good

    now finish the problem. what about the first problem?
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    Quote Originally Posted by Jhevon View Post
    yes, very good

    now finish the problem. what about the first problem?
    I'm still unsure about the first one. Can you please explain?
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    Quote Originally Posted by lp_144 View Post
    I'm still unsure about the first one. Can you please explain?
    when you have a common term, you can factor it out

    example:

    factorize 3xy + 5xy^2

    we want to factor out the lowest powers of all things common. i have x common to both terms, and i have y common to both terms. there is a y^2 in the second term, but i can't factor that out, because i only have y in the first term. so x and y are common to both terms, both terms have them, so i will factor out xy, to get:

    xy(3 + 5y)

    so i took out the xy from the 3, and i have none left in the brackets. as for the 5xy^2, i took out the x and one of the y's, so i am left with one y on the inside, hence the 5y

    now you have,

    6x^2 + 3x = 0

    i told you that 3x is common to both, so factor it out
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  7. #7
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    Quote Originally Posted by Jhevon View Post
    now you have,

    6x^2 + 3x = 0

    i told you that 3x is common to both, so factor it out
    When I factored 3x out, I got 3x(3x).
    What do I do from here?
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by lp_144 View Post
    When I factored 3x out, I got 3x(3x).
    What do I do from here?
    6x^2 + 3x = 0

    \Rightarrow 3x(2x + 1) = 0

    step 1: look carefully at what i did and make sure you understand it.

    step 2: finish the problem. how would we solve for x?
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    Quote Originally Posted by Jhevon View Post
    6x^2 + 3x = 0

    \Rightarrow 3x(2x + 1) = 0

    step 1: look carefully at what i did and make sure you understand it.

    step 2: finish the problem. how would we solve for x?
    I think I got it.

    3x(2x+1)=0

    3x=0
    x=0

    2x+1=0
    2x=-1
    x= -\frac {1}{2}
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  10. #10
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    Quote Originally Posted by Jhevon View Post
    for the first, 3x is a common term. just factor it out

    for the second, in general we use what we call the AC-method to factor, if possible, a quadratic in which the coefficient of the squared term is not 1. but for small numbers like 2, we can use trial and error, it would generally be a lot simpler.

    so -2x^2 + 5x + 3 = 0

    \Rightarrow 2x^2 - 5x - 3 = 0

    \Rightarrow (2x ~~a)(x ~~b) = 0

    now in the place of a and b, you want numbers that multiply to give you -3, but a and 2b will add to give -5
    No, the numbers a and b multiply to give you -6 (not -3) because 2*(-3) = -6, and their sum will be -5

    See here

    \Rightarrow 2x^2 - 5x - 3 = 0

    \Rightarrow 2x^2 - 6x +1x - 3 = 0\;(here, \;a=-6 \; and \;b=+1)

    \Rightarrow 2x(x-3) +1(x - 3) = 0

    \Rightarrow (2x + 1)(x-3) = 0

    \Rightarrow x=\frac{-1}{2}\;and\;x=3
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  11. #11
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Shyam View Post
    No, the numbers a and b multiply to give you -6 (not -3) because 2*(-3) = -6, and their sum will be -5

    See here

    \Rightarrow 2x^2 - 5x - 3 = 0

    \Rightarrow 2x^2 - 6x +1x - 3 = 0\;(here, \;a=-6 \; and \;b=+1)

    \Rightarrow 2x(x-3) +1(x - 3) = 0

    \Rightarrow (2x + 1)(x-3) = 0

    \Rightarrow x=\frac{-1}{2}\;and\;x=3
    note that your a and b multiply to give -6 as i said. i was not doing the AC-method here.
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