Question 188267
If zero was in the denominator the terms would be undefined.....

1 in the denominator lets factor the term... {{{x^2+3x}}}
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x(x+3).....the restriction is that the denominator cannot be zero...so set 
x(x+3)=0--->now set both of those factors equal to 0.x=0 and x+3=0,so the restriction is that   x cannot be 0 or -3
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2) factor the denominator (2x-5)(x+4)....now set each factor equal to zero.
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2x-5=0 and x+4=0....so the restriction is that x cannot be 5/2 or -4.
:
If zero was in the denominator the term would be undefined.....