SOLUTION: find the values of a, b,c, and d that make the equation true. (4t^3-at^2-2bt+5)-(ct^3+2t^2-6t+3) = t^3-2t+d

Algebra ->  Polynomials-and-rational-expressions -> SOLUTION: find the values of a, b,c, and d that make the equation true. (4t^3-at^2-2bt+5)-(ct^3+2t^2-6t+3) = t^3-2t+d      Log On


   

Question 198237This question is from textbook algebra and trigonometry
: find the values of a, b,c, and d that make the equation true.
(4t^3-at^2-2bt+5)-(ct^3+2t^2-6t+3) = t^3-2t+d This question is from textbook algebra and trigonometry

Answer by jim_thompson5910(35256) About Me  (Show Source):
You can put this solution on YOUR website!
%284t%5E3-at%5E2-2bt%2B5%29-%28ct%5E3%2B2t%5E2-6t%2B3%29+=+t%5E3-2t%2Bd Start with the given equation.


4t%5E3-at%5E2-2bt%2B5-ct%5E3-2t%5E2%2B6t-3+=+t%5E3-2t%2Bd Distribute


4t%5E3-at%5E2-2bt-ct%5E3-2t%5E2%2B6t-2+=+t%5E3-2t%2Bd Combine like terms.


%284t%5E3-ct%5E3%29%2B%28-at%5E2-2t%5E2%29%2B%28-2bt%2B6t%29%2B%282%29+=+t%5E3-2t%2Bd Group like terms.


%284-c%29t%5E3%28-a-2%29t%5E2%2B%28-2b%2B6%29t%2B2+=+t%5E3-2t%2Bd Factor out the respective GCFs from each group


Since the coefficient of t%5E3 on the left side is 4-c and on the right side is 1, this means that we know that 4-c=1


Because the coefficient for t%5E2 on the left is -a-2. However, there is no t%5E2 on the right. So this means that the coefficient of t%5E2 on the right is 0. So -a-2=0


Finally, the coefficients for the "t" terms on the left and right sides are -2b%2B6 and -2 respectively. So this means that -2b%2B6=-2

Take note that the only constants on the left and right sides are "2" and "d". So we know that 2=d or d=2


So all you need to do is solve the equations:

system%284-c=1%2C-a-2=0%2C-2b%2B6=-2%29


To find the values of 'a', 'b', and 'c'. I'll let you do that.