SOLUTION: What is the remainder when 3t^2+5t-7 is divided by t-5?

Question 175211: What is the remainder when 3t^2+5t-7 is divided by t-5?
Answer by gonzo(654) (Show Source): You can put this solution on YOUR website!
3t^2 + 5t - 7 divided by t-5
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t goes into 3t^2 by a factor of 3t.
3t * (t-5) = 3t^2 - 15t
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3t^2 + 5t - 7 minus 3t^2 - 15t equals 20t - 7
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t goes into 20t by a factor of 20
20 * (t-5) = 20t - 100
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20t - 7 minus 20t - 100 equals 93
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the remainder is 93/(t-5)
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the complete answer is 3t + 20 + 93/(t-5)
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to prove this is true, then (t-5) * (3t + 20 + 93/(t-5)) should equal 3t^2 + 5t - 7
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when you multiply polynomials, each factor in one polynomial has to be multiplied by each factor in the other polynomial.
this means that:
(a + b) * (c + d + e) equals (ac + ad + ae) + (bc + bd + be) = first way
alternatively, this also means that:
(a + b) * (c + d + e) equals (ac + bc) + (ad + bd) + (ae + be) = second way
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the answer is the same either way.
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multiplying them out the first way, you get:
(t-5) * (3t + 20 + 93/(t-5)) = t*(3t + 20 + 93/(t-5)) - 5*(3t + 20 + 93/(t-5))
which equals
3t^2 + 20t + 93t/(t-5) - 15t - 100 - 5*93/(t-5)
which equals
3t^2 + 5t - 100 + 93t/(t-5) - 5*93/(t-5)
which equals
3t^2 + 5t - 100 + 93*(t-5)/(t-5)
which equals
3t^2 + 5t - 100 + 93
which equals
3t^2 + 5t - 7
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multiplying them out the second way, you get:
(t-5) * (3t + 20 + 93/(t-5)) = (t-5) * 3t) + ((t-5) * 20) + (t-5)*93/(t-5)
which equals
3t^2 - 15t + 20t - 100 + 93
which equals
3t^2 + 5t - 7
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the answer is the same.
the remainder is 93/(t-5)
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