   SEARCH HOME Math Central Quandaries & Queries  Darya, a student: show that (x2-x+3)(2x2-3x-9) = Ax4+Bx3+C where A,B and C are constants to be found. Hi Darya,

You question is simply a situation of showing that when you multiply these two trinomials and then collect like terms, some of the terms will cancel out. The constants will be the coefficients left once you've collected like terms.

The problem with FOIL (first terms, outside terms, inside terms, last terms) is that it only works for binomials. It really just a trick to remind you to multiply all the terms of on binomial to all the terms of the other binomial. If you consider it to be distributing all the terms, it is a little easier to translate that skill to polynomials of any size.

For example:
(x-5)(9x-4) = x(9x-4) -5(9x-4) ⇒I distributed all the terms of the first binomial to the second binomial
= 9x2-4x-45x+20 ⇒I multiplied each monomial, x & -5, by 9x-4
=9x2-49x+20 ⇒ I collect like terms

It also works for trinomials:
(x2-5x+6)(x2-3x+1)=x2(x2-3x+1)-5x(x2-3x+1)+6(x2-3x+1) ⇒I multiplied x2, -5x & 6 by (x2-3x+1)
=x4-3x3+x2-5x3+15x2-5x+6x2-18x+6
=x4-8x3+22x2-23x+6

Multiplying trinomials can get a little messy but if you follow this method it will make it a little easier. Hope this helps,

Janice     Math Central is supported by the University of Regina and The Pacific Institute for the Mathematical Sciences.