Asked by Brooklyne
I'm supposed to substitute (3+h) into the equation 3t^3-2^t+8. I keep getting 3h^3-h^2+80 but it says that's incorrect and the real answer is 3h^3+26h+75h+80. Can somebody explain how they got that?
Answers
Answered by
Steve
Your math is full of typos. If you want a polynomial, that is
3t^3-t^2+8
Plugging in (3+h) for t, you get
3(3+h)^3-(3+h)^2+8
= 3(27+27h+9h^2+h^3)-(9+6h+h^2)+8
= 81+81h+27h^2+3h^3-9-6h-h^2+8
= 3h^3+26h^2+75h+80
When you have trouble getting an answer, show your work, so we can see where you go wrong.
3t^3-t^2+8
Plugging in (3+h) for t, you get
3(3+h)^3-(3+h)^2+8
= 3(27+27h+9h^2+h^3)-(9+6h+h^2)+8
= 81+81h+27h^2+3h^3-9-6h-h^2+8
= 3h^3+26h^2+75h+80
When you have trouble getting an answer, show your work, so we can see where you go wrong.
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