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?
- PuzzlingLv 71 month agoFavourite 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.
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.
- KrishnamurthyLv 71 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
- AlanLv 71 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
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
3 = fg
just check the values.
There is a 3rd method using
a completing of squares.