Question

18x²-3x-36=3(3x+a)(2x-b) for all values of x, where a and b are positive integers. What are the values of the positive integers a and b?

Answer 1

18x^2 - 3x - 36 = 3 ( 3x + a ) ( 2x + b )

simplifying right hand side

foiling ( 3x + a ) ( 2x - b )

6x^2 - 3bx + 2ax - ab

3 ( 6x^2 - 3bx + 2ax - ab )

18 x^2 - 9bx + 6ax - 3ab

now comparing left hand side to right hand side

-9b + 6a = - 3

-3ab = - 36

ab = 12

a = 12/b

plugging this value into equation 1

- 9b + 6( 12/b) = - 3

multiplying both sides by b

- 9b^2 + 72 = - 3b

adding 9b^2 and subtrcating 72 from both sides

9b^2 - 3b - 72 = 0

dividing both sides by 3

3b^2 - b - 24 = 0

( 3b + 8 ) ( b - 3) = 0

b = 3

a = 12/3

a = 4

so, values of a = 4 , b = 3