Question

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

Answer 1

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! :)

Answer 2

`j=4`

`k=10`

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`