The factorization of 8x³ - 125 is (2x - 5)(jx² + kx + 25). What are the values of j and k?

Solution:

Given: 8x³ - 125 = (2x - 5)( jx² + kx + 25)

But from algebraic identity,

a³ - b³= (a - b)(a² - ab +b²)

Therefore, 8x³ - 125 = (2x)³ - (5)³

⇒(2x)³ - (5)³ = (2x - 5) ( (2x)² + 2x × 5 + 5²)

⇒(2x)³ - (5)³ = (2x - 5) (4x² + 10x +25) -----(1)

But given

8x³ - 125 = (2x - 5) (jx² + kx + 25)-------(2)

Comparing equations (1) and (2)

(2x - 5) (4x² + 10x +25) = (2x - 5) (jx² + kx + 25)

Equating the like terms on both the sides

⇒ j = 4 and k = 10.

The factorization of 8x³ - 125 is (2x - 5)(jx² + kx + 25). What are the values of j and k?

Summary :

If 8x³ - 125 = (2x - 5)(jx² + kx + 25) then j = 4 and k = 10. Factorization is when you break a number down into smaller numbers that, multiplied together, give you that original number. Example 12 can be factorized as 2×2×3 which on multiplying gives 12.