The traditional way of teaching the multiplication of binomials is FOIL: First, Outside, Inside, Last. For instance: \[(x + 3)(2x – 5) = (x)(2x) + (x)(-5) + (3)(2x) + (3)(-5) \\ = 2x^2 -5x + 6x – 15 \\ = 2x^2 + x – 15\]
This works well enough for binomials, but for more complicated cases, such as multiplying trinomials, it’s easy to lose track of the various terms. One method that seems to be gaining traction of late is to use a grid. For instance, let’s say you want to multiply \(x^2 + 2x – 3\) by \(4x^2 – x + 1\). First, set it up by creating a 4×4 grid with the two polynomials on the top and the side:
| \(x^2\) | \(+2x\) | \(-3\) | |
| \(4x^2\) | |||
| \(-x\) | |||
| \(+1\) |
Next, in each cell, place the product of the column head and the row head. For instance, \((x^2)(4x^2) = 4x^4\) while \((+2x)(4x^2) = +8x^3\):
| \(x^2\) | \(+2x\) | \(-3\) | |
| \(4x^2\) | \(4x^4\) | \(+8x^3\) | \(-12x^2\) |
| \(-x\) | \(-x^3\) | \(-2x^2\) | \(+3x\) |
| \(+1\) | \(+x^2\) | \(+2x\) | \(-3\) |
Now collect up the like terms along the diagonals: \(4x^4 + 8x^3 – x^3 – 12x^2 – 2x^2 + x^2 + 3x + 2x – 3\).
Finally, combine the like terms: \((x^2 + 2x – 3) \cdot (4x^2 – x + 1) = 4x^4 + 7x^3 – 13x^2 + 5x – 3\).
The advantage of this system is that there’s no concern about losing a particular term, because it’s easy to see from the grid if something’s missing. The one bit of caution involves cases where not all terms are present in the original multiplicands. For instance, consider \((x^3 + 2x – 3) \cdot (4x^2 – x + 1)\). If you fill out the grid as above, the diagonals won’t match up with like terms. You can either sort the like terms afterwards or add a column headed \(0x^2\).