Anonymous
Anonymous asked in Science & MathematicsMathematics · 1 month ago

How did they get x^2 - 4x + 3 = 0 => (x-3)(x-1)? Do you do it by intuition or is there something to memorize?

How did they get (x-3)(x-1) 

How did they factor it? Do you just do it by intuition or is there something to memorize?

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  • 1 month ago
    Favourite answer

    x² - 4x + 3 = 0

    Usually they'll give you problems that are easy enough to figure out with integer factors. 

    You are looking for two numbers:

    They should *multiply* to be the last number (3)

    They should *add* to be the middle coefficient (-4)

    Clearly 3 is positive. You either need two positive numbers or two negative numbers to get a positive product. The only possible products are:

    1 * 3 = 3

    -1 * -3 = 3

    But obviously since the middle coefficient (-4) is negative, we want the second pair. You can't get a negative sum from two positive numbers.

    -1 + -3 = -4

    From there, you just write it as:

    (x - 1)(x - 3) = 0

    Later on you'll learn more methods for doing this that don't rely on factoring such as completing the square or the quadratic formula.

    Summary:

    Think of two numbers that multiply to be the last number. 

    - You'll need two numbers of the same sign if the number is positive. You'll need two numbers of opposite signs if the number is negative.

    - Those same numbers need to *add* to be the middle coefficient.

  • 1 month ago

    x^2 - 4x + 3 = 0 

    x^2 - x - 3x + 3 = 0 

    x(x - 1) - 3(x - 1) = 0

    (x - 1)(x - 3) = 0

  • Alan
    Lv 7
    1 month ago

    For quadratic equation, 

    there is always 

    the quadratic formula 

    (-b +/- sqrt( b^2 -4ac) )/  2a 

    where form is  ax^2 + bx +c  = 0  

    so a = 1 , b = -4 ,  c= 3  then 

    plug into formula

    (+4  +/- sqrt( 16 -12) ) / 2  = 2 +/-  sqrt(4)/2

    = 2 +/- 1  = 1 or 3  

    root of 1 and 3 

    means factor 

    (x-1)(x-3)     

    otherwise, you a few things to help 

    1st rational roots theorem 

    so all possible roots  of al

    are all factor of p/q where p is the lowest coefficient 

    and q the highest

     

    +/- 3 , +/- 1   

    ax^2 + bx + c   = (x+ f)(x+g)  = x^2 + (f+g)x + fg  

    b = (f+g)  

    c = fg  

    so you have try the possibilites 

    or +/- 3, +/- 1  which  

    -4 = f + g  

    and 

    3 = fg   

    just check the values.  

    There is a 3rd method using 

    a completing of squares. 

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