Anonymous
Anonymous asked in Science & MathematicsMathematics · 1 decade ago

Is this factorable: 4x^2 - 13x -12 ?????

I think that there is no solution...but I am not sure.

10 Answers

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  • 1 decade ago
    Favorite Answer

    4x² - 13x - 12

    4x² - 16x + 3x - 12

    4x(x - 4) + 3(x - 4)

    (4x + 3)(x - 4)

    - - - - - - - - - - -s-

  • Ray
    Lv 5
    1 decade ago

    the way that i do it is multiply the leading coefficient by the constant

    4 * -12 = - 48

    then list the factors of 48

    1 and 48, 2 and 24, 3 and 16, 4 and 12, 6 and 8

    then do any of these factors subtract and give me 13?

    that would be 3 and 16, but we need a negative 13.....so 16 would be negative. I now know that this can be factored. To find the factors you make fractions using the leading coefficient as the denominator and the two factors as the numerators,

    3/4 and -16 / 4 and simplify when possible but leave in rational form

    3 / 4 and -4 / 1 the numerators are the constants in my factors and the denominators are the coefficients of the x's

    (4x + 3 ) ( x - 4 ) .... this process should work for all quadratic trinomials.

  • hume
    Lv 4
    4 years ago

    4x^2 - 13x = 12 circulate each and every little thing to the left hand ingredient, 4x^2 - 13x - 12 = 0 we can clean up this utilising the "ac" technique. Multiply a and c, and locate 2 components of ac which upload to b. In our case, a = 4 and c = -12, so ac = -40 8. What 2 components of -40 8 upload as much as -13? the respond is -sixteen and +3. chop up -13x into -16x and +3x. 4x^2 - 16x + 3x - 12 = 0 ingredient the 1st 2 words and the 2nd 2 words. 4x(x - 4) + 3(x - 4) = 0 look at how we've (x - 4) in straightforward now; ingredient (x - 4) from the full expression. (x - 4)(4x + 3) = 0 meaning x = { 4, -3/4 }

  • 1 decade ago

    Since everybody else has shown you lots of methods, here is yet another one:

    4x² - 13x - 12

    = 4{x² - 13/4x - 3}

    = 4{x² - 13/4x +(13/8)² - (13/8)² - 3} the 13/8 is just half the middle term i.e. half 13/4

    = 4{ (x - 13/8)² - 361/64 }

    Note. -(13/8)² - 3 = -169/64 - 3 x 64/64 = -169/64 - 192/64 = -361/64

    = 4{ (x - 13/8)² - (19/8)² } . . . . it's now a diffrence of two squares inside the curly brackets

    = 4{ (x - 13/8 - 19/8)(x - 13/8 + 19/8) }

    = 4(x - 32/8)(x + 6/8)

    = 4(x - 4)(x + 3/4)

    = (x - 4)(4x + 3)

    This method is often called "completing the square" - it is the method used to derive the quadratic formula used for solving quadratic equations. I would not recommend using it unless you are really strong in fractions, or are allowed to use a fraction-capable calculator.

    You only asked if it was factorisable !

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  • Anonymous
    1 decade ago

    Yes, it factors into (4x+3)(x-4).

    In order to do this, think "factors of -48 that add (or subtract) to -13." The only pair that works is -16 and +3. The outside terms multiply to -16 and the inside terms multiply to +3.

  • 1 decade ago

    try the divisors of 12 : 1, 2 , 3, 4,6,12

    1 no

    2 no

    3 no

    4 yes : 64 - 52 - 12

    so 4 a a zero

    rest is easy

  • 1 decade ago

    All quadratic equations are factorable even if the factors involve complex numbers,

    Determine the roots of the equation which in your case are real ,say they are r1 and r2, then the expression is (x - r1)(x - r2)

  • 1 decade ago

    (4x+3)(x-4)

    x=3/4 x=4

  • 1 decade ago

    yes it is,

    (4x+3)(x-4)=4x^2-13x-12

    so x=-3/4 or +4

  • xoxox
    Lv 5
    1 decade ago

    no, it's not factorable

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