Question

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

Answer 1

Answer:

$j=4$

$k=10$

Step-by-step explanation:

we have that

$8x^{3}-125=(2x-5)(jx^{2} +kx+25)$

Applying the distributive property to the right side

$(2x-5)(jx^{2} +kx+25)=2jx^{3}+2kx^{2} +50x-5jx^{2}-5kx-125$

Combine like terms

$=2jx^{3}+(2k-5j)x^{2} +(50-5k)x-125$

Compare with $8x^{3}-125$

so

$2j=8$ -------> $j=4$

$(2k-5j)=0$ ------> substitute the value of j and solve for k

$2k=5(4)$ ------> $k=10$

Answer 2

What number is multiplied by 2 in left parentheses to get 8 in original expression? That is j, the coefficient for x^2. Follow the same pattern to find k, the middle coefficient.

j= x^2 coefficient
k= x coefficient


= 8x^3 – 125

= (2x−5)(4x^2+10x+25)


ANSWER: j= 4; k= 10

Hope this helps! :)