(z-2)(z^2+az+b)=z^3-4z^2+6z-4 a,b are an element of Z. Honestly have no idea how to solve this all help welcome. Thanks
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Originally Posted by khorne55 (z-2)(z^2+az+b)=z^3-4z^2+6z-4 a,b are an element of Z. Honestly have no idea how to solve this all help welcome. Thanks 1. Expand the LHS: $\displaystyle (z-2)(z^2+az+b) = z^3 + z^2(a - 2) + z(b - 2a) - 2b$ 2. Now compare the coefficients of the RHS: $\displaystyle \begin{array}{lcr}a-2&=&-4 \\ b-2a&=&6 \\ -2b&=&-4 \end{array}$ 3. Solve the system of equations for a and b.
Last edited by earboth; Jan 5th 2013 at 06:39 AM.
Thanks a lot ! a= -2 and b =2 That clears out the whole situation. I am really grateful for your help!
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