I am wondering why on step 2 they factored out (8A-8B + D) .... where did they g

Question

I am wondering why on step 2 they factored out (8A-8B + D) .... where did they get a negative? They got -8B . Everything points to a positive. Would it make any difference if I left it a positive number ? Thank you . Precalculus clear the fractions by multiplying each side by" x2 =(Ax+B)(x2 + 4)' +(Cr+D)(x2 + 4)-(Ex+F) The result is | Step 2 0 By expanding the identity in equation (2), we obtain x2=x'(4)-x"(B)+x'(C)+x"(8A-8B + D)-x(16A+4C+ E)+16B+4D + F We equate the coefficients of like powers of x to get, A=0 B=0 Chapter 11.5 Problem 42AYU

Explanation / Answer

You are correct it is supposed to be positive not negative.

The expansion done was wrong for coefficients of x^2.

It was supposed to be .

(8B+D)

A does not exist for x^2.

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