find a postive value of the given variable so that the trinonmial is factorable

Question

x²+bx-11; b= ?

Answer 1

To find a positive number b so x2+bx-11 is factorable without guessing, let's write the generic factors:

 

(x ± n₁)(x ± n₂)

 

What do we know about n₁ and n₂?

 

The product of n₁ and n₂ must be -11 and the sum of n₁ and n₂ must be b, where b is positive.

 

Okay, what 2 numbers multiply to get -11?  Well, either 11 and -1 or -11 and 1.

 

If we use -11 and 1, we get:

 

(x - 11)(x + 1) = x² + 1x - 11x - 11 = x² - 10x - 11.

 

Here b = -10, but a requirement states b must be positive, so lets use 11 and -1 then:

 

(x + 11)(x - 1) = x² - 1x + 11x - 11 = x² + 10x - 11.

 

Here b = 10 so this is the solution without guessing.

Answer 2

If b = 10 then x² + 10x -11 = (x+11)(x-1). 

 

There's another possibility: b = -10: then x² - 10x -11 =(x-11)(x+1), but this case has been ruled out because the question asked for a positive value of b.