Asked by Anonymous
Divide.
[(X^2 +6+9)/(X-1)]/(X^2-9)/(x^2-2x+1)
• [(x+3)(x-1)]/(x-3)
• [(x-3)(x+1)]/(x+3)
• [(x+3)(x+1)/(x-3)
• [(x-3)(x-1)]/x+3)
How?
[(X^2 +6+9)/(X-1)]/(X^2-9)/(x^2-2x+1)
• [(x+3)(x-1)]/(x-3)
• [(x-3)(x+1)]/(x+3)
• [(x+3)(x+1)/(x-3)
• [(x-3)(x-1)]/x+3)
How?
Answers
Answered by
bobpursley
assuming you meant 6x in the first term.
that first term factors to (x+3)(x+3)
now in the nenominator, (x-3)(x+3)/((x-1)^2
I see in the numerator
(x+3)^2/(x-1)*(x-1)^2/(x-3)(x+3)
then
(x+3)(x-1)/(x-3)
check that.
that first term factors to (x+3)(x+3)
now in the nenominator, (x-3)(x+3)/((x-1)^2
I see in the numerator
(x+3)^2/(x-1)*(x-1)^2/(x-3)(x+3)
then
(x+3)(x-1)/(x-3)
check that.
Answered by
Anonymous
That helps
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