Simplify:

*x*(

*x*(

*x*(2 –

*x*) + 1) +1 ) + 1 +

*x*(

*x*(

*x*(

*x*– 2) + 1) + 1) + 1

This is supposed to come out to be zero. Does it? If not, can you fix it up so that it does?

Oogh. Some students actually LIKE working these out, bless ’em. (And we need ’em!)

Here goes:

*x* (*x* (*x* (2 – *x*) + 1) + 1) + 1 + *x* (*x* (*x* (*x* – 2) + 1) + 1) + 1

= *x* (*x* (2*x* – *x*^{2} + 1) + 1) + 1 + *x* (*x* (*x*^{2} – 2*x* + 1) + 1) + 1

= *x* (2*x*^{2} – *x*^{3} + *x* + 1) + 1 + *x* (*x*^{3} – 2*x*^{2} + *x* + 1) + 1

= 2*x*^{3} – *x*^{4} + *x*^{2} + *x* + 1 + *x*^{4} – 2*x*^{3} + *x*^{2} + *x* + 1

= 2*x*^{2} + 2*x* + 2 which is not zero, not identically anyway.

Change the right half to:

*x* (*x* ( *x* (*x* – 2) – 1) – 1) – 1

= *x* (*x* (*x*^{2} – 2*x* – 1) – 1) – 1

= *x* (*x*^{3} – 2*x*^{2} – *x* – 1) – 1

= *x*^{4} – 2*x*^{3} – *x*^{2} – *x* – 1

Now the sum of the left half and right half will equal zero.