Given that (1+x/2)^8(1-x)^6)≡1+Ax+Bx^2... find the constants A and B how would I solve for them, other than completely expanding each binomial?
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Originally Posted by DemonX01 (1+x/2)^8(1-x)^6)≡1+Ax+Bx^2... find the constants A and B $\displaystyle \left(1+\frac{x}{2}\right)^8=1+4x+14x^2+\cdots$ $\displaystyle (1-x)^6=1-x+15x^2+\cdots$ Mulriply those and see what happens.
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