Solving Equation: x² - 3x - 40 = 0

[Gringo123]

My Maths textbook gives the following equation:
x squared - 3x - 40 = 0

.. and it gives the following as a model answer: (where I've wriiten right or wrong there is a tick or a cross)

try 2 x 20x, 4 x 10x, 5 x 8 (right)
(x-5) (x + 8) (wrong)
(x+5) (x-8) (right)
x = -5 or 8

I have no idea what this method is and why it has been used here. Can anyone please expain it to me?
Thanks

 

[tiny-tim]

Hi Gringo123!

This is the rational root theorem …

basically, if the roots are whole numbers, then they have to be factors of 40 (times ±1).

So first they tried ±2 and ±20 (by putting eg 2 into the equation to see if it comes out 0, as it should), then they tried ± 4 and ±10, then they tried ±5 and ±8 …

woohoo! it's -5 and 8 !

See http://en.wikipedia.org/wiki/Rational_roots" for more details. ​

 

[HallsofIvy]

My Maths textbook gives the following equation:
x squared - 3x - 40 = 0

.. and it gives the following as a model answer: (where I've wriiten right or wrong there is a tick or a cross)

try 2 x 20x, 4 x 10x, 5 x 8 (right)
(x-5) (x + 8) (wrong)
(x+5) (x-8) (right)
x = -5 or 8

I have no idea what this method is and why it has been used here. Can anyone please expain it to me?
Thanks

Essentially, they are using a "trial and error" method of factoring the left side. I presume the textbook has already show that (x+ a)(x+ b)= x²+ (a+b)x+ ab. In order that you have (x+a)(x+b)= x<sup>+ (a+b)x+ ab= x2- 3x- 40, you must have a+ b= -3 and ab= -40. So start looking for integers factors of 40: 1* 40, 2*20, 4*40, 5*8, 8*5, 10*4, 20*2, and 40*1. Since it doesn't matter which you call "a" and which "b", you don't need to look at "8*5, 10*4, 20*2, and 40*1", they are the same as "1* 40, 2*20, 4*40, and 5*8". I don't know why they did not include "1*40"- perhaps they thought that was too obviously wrong.

Since the product of two positive or two negative numbers is positive, in order to get ab= 40, one factor must be positive and the other negative. Now we check each of those to see if they add to -3: 1- 40= -39 not -3; 2- 20= -17, not -3; 4- 10= -6, not -3; 5- 8= -3. success!

Okay, we now know the numbers are a= 5, b= -8: check: (x+ 5)(x- 8)= x2+ (5-8)+ 5(-8)= x2- 3x- 40.
Now we know that x2- 3x- 40= (x+ 5)(x- 8)= 0. Since we also know that if the product of two numbers is 0, at least one of the numbers must be 0 (that is a special property of "0" and does not work if the product is any number except 0), we know that either x+ 5= 0 or x- 8= 0. From the first we see that x= -5 is a solution to the equation, from the other, we know that x= 8 is a solution.

By the way, it might happen that NONE of the integer factors of ab, the constant term in the polynomial, add to the coefficient of x. In that case, it cannot be factored with integer coefficients. Fortunately, for quadratic equations, there are other methods, such as "completing the square" or the "quadratic formula", to solve them.</sup>