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
---
t goes into 3t^2 by a factor of 3t.
3t * (t-5) = 3t^2 - 15t
---
3t^2 + 5t - 7 minus 3t^2 - 15t equals 20t - 7
---
t goes into 20t by a factor of 20
20 * (t-5) = 20t - 100
---
20t - 7 minus 20t - 100 equals 93
---
the remainder is 93/(t-5)
---
the complete answer is 3t + 20 + 93/(t-5)
---
to prove this is true, then (t-5) * (3t + 20 + 93/(t-5)) should equal 3t^2 + 5t - 7
---
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
---
the answer is the same either way.
---
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
---
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
---
the answer is the same.
the remainder is 93/(t-5)
---
|
|
|
