16x^5y^4 + 6x^3y - 2x^2y - 24x^2y^4
= 2x^2y (8x^3y^3 + 3x - 1 - 12y^3)
= -2x^2y (-8x^3y^3 - 3x + 1 + 12y^3)
Naturally, if the divisor had only an x instead of x^2, all the terms in the quotient would have an extra x in them.
12y^3 + 1 - 3x - 8x^3y^3 is of degree 6, with integer coefficients
Part A: Divide (16x^5y^4 + 6x^3y - 2x^2y - 24x^2y^4) by -2x^2y.
Part B: How would your answer in Part A be affected if the x2 variable in the denominator was just an x?
Part C: What is the degree and classification of the polynomial you got in Part A?
Is this first part right?
(x^2 y (16 x^5 y^4+6 x^3 y-2 x^2 y-24 x^2 y^4))/(-2)
Factor (2 x^2 y (8 x^3 y^3+3 x-1-12 y^3) x^2 y)/(-2)
(2 x^2 y (8 x^3 y^3+3 x-1-12 y^3) x^2 y)/(-2).
Add powers together (2^1-1 x^2+2 y^1+1 (8 x^3 y^3+3 x-1-12 y^3))/(-1)
simplify
-x^4 y^2 (8 x^3 y^3+3 x-1-12 y^3)
1 answer
1 answer