Yes, 4x + 5 is a factor of 64x³ + 125.
The technique used to find this out is by extracting the cube roots of the perfect cube binomial. The cuberoot of 64x³ is 4x, and for 125 it's 5 (As 4x • 4x • 4x = 64x³, and 5 • 5 • 5 = 125). We use the a³ + b³ = (a+b)(a²-ab+b²) formula as it is 64x³ + 125.
1. Rewrite in exponential form.
[tex]64 {x}^{3} + 125 = {4}^{3} {x}^{3} + 5 {}^{3} [/tex]
2. Remove the exponents.
[tex]64 {x}^{3} + 125 = 4x + 5[/tex]
3. Use the formula: a³ + b³ = (a+b)(a²-ab+b²), wherein a and b are the cube roots of the variable (64x³ = 4x) and coefficients (125 = 5).
[tex]64 {x}^{3} + 125 = (4x + 5)((4x) {}^{2} - (4x)(5) + 5 {}^{2} )[/tex]
4. Simplify.
[tex]64 {x}^{3} + 125 = (4x + 5)(16x {}^{2} - 20x + 25)[/tex]
Thus, 4x + 5 is a factor of 64x³ + 125.