Question

s and t are positive integers.
(x+s)(x-t) is expanded and simplified
The answer is x^2+kx-40 where k is a positive integer
Work out the smallest possibe value of k
(4 marks)
ANSWE CAREFULLY, WITH WORKING OUT!
(question from: may 2020 paper 1 maths gcse mocks)
(NO LINK OR I WILL REPORT)

Answer 1

Given :  (x+s)(x-t) =  x^2+kx-40

s and t are positive integers.   k is a positive integer

To Find :  smallest possible value of k

Solution:

(x+s)(x-t) =  x²+kx-40

=> x²  + x(s - t) - st  = x²  + kx  - 40

=> s-t = k

st = 40

as s and t  are positive integers.

and s -t = k is also positive integer hence s > k

s           t        st        s - t = k

40        1        40        39

20        2       40        18

10         4        40       6

8          5        40       3

smallest possible value of k is 3

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