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How do you factor out 3x^2 - 3x - 18 = 0?

I really need help on this! Please explain step by step!

8 個解答

相關度
  • 1 十年前
    最愛解答

    3(x^2-x-6) = 3(x-3)(x+2) = 0

    x = 3, -2

  • 匿名
    1 十年前

    Hello,

    The first thing you can do is GCF factoring, or finding the Greatest Common Factor.

    In this case, it is 3, because all parts of the equation are divisible by three.

    The result is 3(x^2 - x - 6) = 0 because we just took out 3 from every part.

    Next, we move on to "fast factoring."

    We must find two numbers that add up to -1 and multiply to -6

    Two numbers that work are positive 2 and negative 3. (2 * -3 = -6 and -3 + 2 = -1)

    so we end up with 3(x + 2)(x - 3) = 0 and that is the simplest it can get.

    Answer: 3(x+2)(x-3)

    (don't forget about the three you factored out in the beginning)

    And to be more specific, the solutions are x = -2 and 3

  • 1 十年前

    First you factor out the 3 by dividing each term by 3. This leaves you with 3(x^2-x-6)=0. Factor inside the parentheses and you get 3(x-3)(x+2)=0. To solve set each factor equal to zero.

    x-3=0------x=3

    x+2=0-----x=-2

  • 匿名
    1 十年前

    3x^2 - 3x - 18 = 0

    Divide both sides by 3.

    x^2 - x - 6 = 0

    (x + 2)(x - 3) = 0

    x = -2, 3

  • 匿名
    1 十年前

    (3x-9)(x+2)= 3x^2-3x-18 =0

    then each factor = to 0

    3x-9 = 0

    3x=9

    x=3 this is one solution

    x+2= 0

    x= -2

    this is the second solution

    to prove if these are the correct solution

    you have to plug them in in the original equation and if the result is the same in tthis case zero

    you can take both of them as your solutions

  • 1 十年前

    3 (x² - x - 18) = 0

    x² - x - 6 = 0

    As the formula, ax² + bx + c = 0, find 2 numbers which their combination is b, and multiplication is c

    b here is -1

    c here is -6

    So, 2 numbers we are finding are 2 and -3 (because 2+(-3)=-1, 2x(-3)=-6)

    So, we have (x+2)(x-3)=0

    And, in order to make that equation 0, we need (x+2) or (x-3) to be 0

    x+2=0

    x= -2

    OR

    x-3=0

    x=3

    So, x has 2 values, x= -2, 3

  • 1 十年前

    There are two solutions (x1 and x2)

    lets say a is the number multiplying x^2 (in your case it 3), b is the number multiplying x (your case -3) and c is the free coefficient (in your case -18)

    the regular expression goes like this

    x1= [-b+sqrt(b^2-4*a*c)]/2

    x2= [-b-sqrt(b^2-4*a*c)]/2

    If you use this expression to calculate lengths the negative one doesn't exist.

  • 1 十年前

    3(x^2-x-6) = 3(x-3)(x+2) = 0

    x = 3, -2

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