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# (x^2-3x+2)/(x^2-4x+4) / (X^2-2x+1)/(3x^2-12)

Dividing Radical expressions, i have to divide and simplify. I know i have to multiply so have to do the reciprocal, but im stuck after that.

HELP!

x2-3x+2 can be factored into (x-2)(x-1) (numerator of 1st fraction)

x2-4x+4 can be factored into (x-2)(x-2) (denominator of 1st fraction)

so the first fraction reduces to (x-1)/(x-2)

x2-2x+1 can be factored into (x-1)(x-1) (numerator of 2nd fraction)

3x2-12 can be factored into (3) (x-2)(x+2) (denominator of 2nd fraction)

so the fraction is [(x-1)(x-1) / (3)(x-2)(x+2)]

Now we have to remember that  (a/b) / (c/d) = "extreems / means"  or a*d/b*c

or multiply the means together and multiply the extreems together then divide the extreems by the means.

For our problem, we would have [3(x-1)(x-2)(x+2)] / (x-2)(x-1)(x-1)

NOTICE: THE (X-1) CAN BE CANCELLED FROM TOP AND BOTTOM!

NOTICE: THE (X-2) CAN BE CANCELLED FROM BOTH TOP AND BOTTOM!

We are now left with 3(x+2) /(x-1)  the ANSWER

YOU DO NOT HAVE TO DO THE RECIPROCAL!

For rational expressions the "means/extreems" method is much easier and should have been introduced to you long long ago.

x2 - 3x + 2   *  3x2 - 12

x- 4x + 4      x2 - 2x + 1

(x - 1)(x - 2) * 3(x - 2)(x + 2)

(x - 2)(x - 2)    (x - 1)(x - 1)

3(x + 2)

(x - 1)