Question

S t are positive integers

Answer 1

Given:

The expanded form of $(x+s)(x-t)$ is $x^2+Kx-40$.

Where, s, t, K are positive integers.

To find:

The smallest possible value of K.

Solution:

The expanded form of $(x+s)(x-t)$ is $x^2+Kx-40$. It means,

$(x+s)(x-t)=x^2+Kx-40$

$x^2-tx+sx-st=x^2+Kx-40$

$x^2+(s-t)x-st=x^2+Kx-40$

On comping both sides, we get

$K=s-t$              ...(i)

K is a positive integer if s>t.

And

$st=40$

The factor pairs of 40 are (1,40), (2,20), (4,10), (5,8), (8,5), (10,4), (20,2) and (40,1).

Since s>t, therefore the possible values for (s,t) are (8,5), (10,4), (20,2) and (40,1).

Using (i), find the value of K for each factor pair.

$K=8-5=3$  

$K=10-4=6$      

$K=20-2=18$          

$K=40-1=39$          

Therefore, the smallest possible value of K is 3.