Question

S t are positive integers

(X +S) (x-t) is expanded - Simplified

The

answer

is x² + Kx-40 where K is a positive integer

work out the

out the smallest possible value of K

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.