Please help ASAP. Thanks!

in Calculus Answers by Level 1 User (940 points)

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Let’s simplify the function:

(x²(x²-2x-3))/(x²(x²-5x+6))=((x-3)(x+1))/((x-3)(x-2)), except when x=0;

(x+1)/(x-2) except when x-3=0, that is, when x=3. x-2=0, that is, x=2, means we will be dividing by zero.

So we can cancel out the factors x² and x-3, except when x=0 and 3, and we can evaluate the function, except when x=2. So the boxes must contain 0, 3 and 2.

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