Question
Simplify the expression
s32imt3−5imts2
Evaluate
((2×s3m)t3−(5×sm)t)i
Remove the parentheses
(2×s3m×t3−5×sm×t)i
Multiply the terms
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Multiply the terms
2×s3m×t3
Multiply the terms
s32m×t3
Multiply the terms
s32mt3
(s32mt3−5×sm×t)i
Multiply the terms
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Multiply the terms
5×sm×t
Multiply the terms
s5m×t
Multiply the terms
s5mt
(s32mt3−s5mt)i
Subtract the terms
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Simplify
s32mt3−s5mt
Reduce fractions to a common denominator
s32mt3−s×s25mts2
Multiply
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Evaluate
s×s2
Multiply the terms with the same base by adding their exponents
s1+2
Add the numbers
s3
s32mt3−s35mts2
Write all numerators above the common denominator
s32mt3−5mts2
s32mt3−5mts2i
Multiply the terms
s3(2mt3−5mts2)i
Multiply the terms
s3i(2mt3−5mts2)
Solution
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Evaluate
i(2mt3−5mts2)
Apply the distributive property
i×2mt3−i×5mts2
Multiply the numbers
2imt3−i×5mts2
Multiply the numbers
2imt3−5imts2
s32imt3−5imts2
Show Solution

Find the excluded values
s=0
Evaluate
((2×s3m)t3−(5×sm)t)i
To find the excluded values,set the denominators equal to 0
s3=0s=0
The only way a power can be 0 is when the base equals 0
s=0s=0
Solution
s=0
Show Solution
