Final answer:
The correct factorization of the polynomial -y^6 + 4y^5 + 16y^4 - 64y^3 is -y^3(y - 4)(y + 4), which is option (b). This involves factoring out y^3 and applying the difference of cubes and difference of squares formulas.
Step-by-step explanation:
The student is asking how to factor the polynomial -y^6 + 4y^5 + 16y^4–64y^3 completely. To factor this expression, we can first take out the common factor y^3, which gives us:
-y^3(y^3 - 4y^2 - 16y + 64)
Looking at the cubic polynomial inside the parenthesis, we can recognize it as a difference of cubes since 4y^2 is (2y)^2 and 64 is 4^3:
-y^3((y^2)^3 - (4)^3)
Now, we can factor it using the difference of cubes formula, a^3 - b^3 = (a - b)(a^2 + ab + b^2), which applies here with a = y^2 and b = 4:
-y^3((y^2 - 4)(y^4 + 4y^2 + 16))
The quadratic y^2 - 4 can be further factored as it is a difference of squares:
-y^3((y - 4)(y + 4)(y^4 + 4y^2 + 16))
The remaining factor, y^4 + 4y^2 + 16, cannot be factored further over the real numbers, so the completely factored expression is:
-y^3(y - 4)(y + 4)
Thus, the correct answer is option (b) -(y^3)(y - 4)(y + 4).