Question

Can someone please explain how to do this problem?

Answer 1

\[(49+16b+2b^2) \div (2b+4)\]

Answer 2

\[(2b^2+16b+49)\div (2b+4) \] Reordered.

Answer 3

OK.

The problem here is that -

2b^2 + 16b + 49.

a = 2, b = 16, c = 49.

b^2-4ac = 16^2 - 4.49.2

=> 256 - 392 = -136.

So u cannot cancel the denominator completely.(the solutions of numerator are imaginary hence factorising eqn is quite obscure.)

However u can make them as 3 separate fractions and evaluate -

2b^2/2(b+2) + 16b/2(b+2) + 49/2b+4

=> b^2/b+2 + 8b/b+2 + 49/2b+4.

Answer 4

Ok, thanks :)