Multiplying and dividing polynomials extends the arithmetic of numbers you already know to letters. Splitting a rectangle's area into four pieces shows at once why the product of two linear expressions expands the way it does, and when dividing by a linear expression, synthetic division gives the quotient and remainder quickly from the coefficients alone. Just as 17 = 5 × 3 + 2, a polynomial follows the structure (dividend) = (divisor) × (quotient) + (remainder), and this is the thread running through the whole unit. In the first picture the pieces of the area model change with b; in the second, the last cell of the synthetic division table changes with the constant term.
Dividing 17 by 5 in whole numbers gives quotient 3 and remainder 2, which can be written as the single statement 17 = 5 × 3 + 2. Polynomial division is most useful when it is kept in the same form, one equation, rather than as a quotient and a remainder memorised separately. Multiplication is similar: if you only memorise the expanded result, it is easy to lose a sign or a coefficient. So this page shows multiplication as an area picture and division as a table of coefficients, following each term from where it comes from to where it goes.
The product of two polynomials can be seen as the area of a rectangle. Expanding (x + 3)(x + b) gives the sum of the areas of four smaller rectangles. In the picture, 3 stays fixed and only b changes.
The area picture is made of a square of side x and three pieces to its right, below it and in the bottom-right corner. The big square has area x², the right piece bx, the lower piece 3x and the corner piece 3b. Changing b from 1 to 7 changes only the width of the two right-hand pieces, and with it the coefficient 3 + b of the x-term and the constant 3b in the expansion underneath. For example, with b = 4, (x + 3)(x + 4) = x² + 7x + 12.
When dividing a polynomial by x − a, synthetic division gives the quotient and remainder quickly from the coefficients alone.
If you only memorise the order for filling in a synthetic division table, it is easy to miss why the last cell is the remainder. Dividing by a linear expression must leave a remainder of lower degree, that is, a constant. So the cells along the bottom row give the coefficients of the quotient, and only the very last one is the remainder. In the picture above the quotient is always x² + 3x − 2, and changing the constant term changes only the last cell, to (constant term) − 2. When the constant term is 2 the remainder is 0, which means x³ + 2x² − 5x + 2 is exactly divisible by x − 1.
For a question on the expansion formulas, first decide which formula's left-hand side the expression matches. If the same expression is multiplied by itself, as in (x + 3)², use the perfect-square formula; if only a sum and a product are given, use a rearrangement such as x² + y² = (x + y)² − 2xy. Before putting in numbers, picture which piece of the area model each term belongs to; that makes it harder to drop the 2 in 2ab. After writing the answer, substituting a simple value such as x = 1 into both sides is a quick check.
Put x = a into the division equation f(x) = (x − a)Q(x) + R and the first part vanishes, leaving f(a) = R. So the last cell of the table can be found directly as f(a) without doing the division; this is the remainder theorem of the next unit. If the remainder is 0, x − a is a factor, so factorising, which uses the expansion formulas in reverse, rests on the same equation. Keeping hold of the fact that the area-model expansion and the table-based division use the same letters and the same coefficients ties the next units together.
Last updated