Factorization of Quadratic Expressions

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Factoring Trinomials.wmv

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1 of 2 videos by BRIAN M.

Key Questions

  • Factorization of a quadratic expression is the opposite of expansion, and is the process of putting the brackets back into the expression rather than taking them out.

    To factorize a quadratic expression of the form #ax^2+bx+c# you must find two numbers that add together to give the first coefficient of #x# and multiply to give the second coefficient of #x#.

    An example of this would be the equation #x^2 + 5x + 6#, which factorizes to give the expression #(x+6)(x-1)#

    Now, one might expect the solution to include the numbers 2 and 3, as these two numbers both add together to give 5 and multiply to give 6. However, as the signs differ in the factorized equation, then the solution to the equation must be #(x+6)(x-1)#, as #+6 -1# gives #5#, and #6 times 1# yields a solution of 6.

    The equation can be checked by multiplying the solutions back into the equation to give the original quadratic of #x^2 + 5x + 6#.

  • Good question .

    basically what are asking me to factor is :

    #ax^2 + bx +c#

    here is an eg.

    #4x^2 – 15x + 9.#

    First you need to make column like this

    SUM = COEFFICIENT OF x =( -15)
    product = (COEFFICIENT of x^2 #*# COEFFICIENT of constant)= 36
    Now you have to find factors which have a sum of -15 and a product of 36

    THE FACTORS ARE (-12), (-3)

    So now what we do is split the middle term in the equation and factor it out

    #4x^2 – 12x– 3x +9.#
    #4x(x – 3)–3 (x-3).#

    #(4x–3 )(x – 3).#

    HENCE FACTORED

    #f(4x^2 – 15x + 9.) = (4x–3 )(x – 3)#

  • Example 1

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


    Example 2

    #2x^2-9x-5=(2x+1)(x-5)#


    Example 3

    #x^2-9=(x+3)(x-3)#


    I hope that this was helpful.

  • Answer:

    If both the roots satisfy the quadratic equation, then you have factorised it correctly.

    Explanation:

    I will try explaining with an example.
    #x^2+4x-12=0#
    The roots are #2# and #-6#.
    Now simply put the values in the quadratic equation.

    First, I will try #-6#
    #(-6)^2+4(-6)-12=0#
    #36-24-12=0#
    #cancel36-cancel36=0#
    #0=0#
    which is true.

    Now, I will try #2#
    #2^2+4*2-12=0#
    #4+8-12=0#
    #cancel12-cancel12=0#
    #0=0#
    which is also true.

    #:.# Our answer was correct.

    Hope you got it :)

  • Extension to factoring, when the trinomials do not factor into a square (it also works with squares).

    Sum-product-method
    Say you have an expression like #x^2+15x+36#
    Then you try to write #36# as the product of two numbers, and #15# as the sum (or difference) of the same two numbers. In this case (with both being positive) it's not so hard. You take the sum.

    You can write #36=1*36=2*18=3*12=4*9=6*6#
    Sums of these are #37,20,15,13,12# respectively
    Differences are #35,16,9,5,0# respectively
    #15=+3+12# will do. So the factoring becomes:
    #(x+3)(x+12)#
    Check your answer! #=x^2+12x+3x+36#

    It's a bit harder when one or two of the numbers are negative, let's take #x^2-15x+36#
    Same as the first, only now both factors are negative
    #(x-3)(x-12)=x^2-12x-3x+36=# the original

    Extra
    If the last number (#36#) is negative, you will have to work with the difference of the factors. Check the next one yourself:
    #x^2+5x-36=(x+9)(x-4)=?#

    And now try: #x^2-5x-36=?#

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