Question

Factor completely.

49t^6 - 4k^8

A.) (7t³ + 2k^4)(7t³ - 2k^4)
B.) (7t³ - 2k^4)(7t³ - 2k^4)
C.)(7t³ + 2k^4)(7t³ + 2k^4)

Answer 1

A.) (7t³ + 2k^4)(7t³ - 2k^4)

Explanation:

Factor the following:

49 t^6 - 4 k^8

49 t^6 - 4 k^8 = (7 t^3)^2 - (2 k^4)^2:

(7 t^3)^2 - (2 k^4)^2

Factor the difference of two squares. (7 t^3)^2 - (2 k^4)^2 = (7 t^3 - 2 k^4) (7 t^3 + 2 k^4):

Answer: (7 t^3 - 2 k^4) (7 t^3 + 2 k^4)

Answer 2

A.) (7t³ + 2k^4)(7t³ - 2k^4)

Explanation:

This is what we call a conjugate binomial which is two binomials with similar terms but with one of those terms with different signs, this shows a result of the first term squared minus the second term squared.

The formula for conjugate binomials are:

(a+b)(a-b)=
`a^(2)-b^(2)`

In this case the multiplication of

(7t³ +
`2k^(8)`)(7t³ -
`2k^(8)` )= 49
`t^(6)` + 14
`k^(4)t^(3)` - 14
`k^(4)t^(3)` -
`4k^(8)`

After substracting the second and the third term you are left with:

`49t^(6) - 4k^(8)`