Asked by mark
                what is the quotient in simplified form? State any restrictions on the variable. (6-x)/x^2+3x-28 divided x^2-36/x^2+5x-36
            
            
        Answers
                    Answered by
            scott
            
    {(6-x)/[(x+7)(x-4)]} / {[(x+6)(x-6)]/[(x+9)(x-4)]} = [(6-x)(x+9)(x-4)] / [(x+7)(x-4)(x+6)(x-6)] =
-(x+9) / [(x+7)(x+6)]
x cannot equal -6 or -7 ... division by zero is a no-no
    
-(x+9) / [(x+7)(x+6)]
x cannot equal -6 or -7 ... division by zero is a no-no
                    Answered by
            Reiny
            
    (6-x)/(x^2+3x-28) ÷ ( (x^2-36)/(x^2+5x-36) )
= (6-x)/(x^2+3x-28) * (x^2+5x-36)/(x^2 - 36)
= (6-x)/( (x+7)(x-4) ) * (x+9)(x-4)/((x-6)(x+6) )
= -(x+9)/( (x+7)(x+6) ) , x ≠ 4,6
The restriction of x ≠ 4,6 <b>must be stated</b>, otherwise we actually did divide by zero.
The restriction of x ≠ -7, -6 is implied, in the original problem as well as the answer, it doesn't really have to be stated
    
= (6-x)/(x^2+3x-28) * (x^2+5x-36)/(x^2 - 36)
= (6-x)/( (x+7)(x-4) ) * (x+9)(x-4)/((x-6)(x+6) )
= -(x+9)/( (x+7)(x+6) ) , x ≠ 4,6
The restriction of x ≠ 4,6 <b>must be stated</b>, otherwise we actually did divide by zero.
The restriction of x ≠ -7, -6 is implied, in the original problem as well as the answer, it doesn't really have to be stated
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