Expanding Three Binomials

Two brackets, then the last

Expand any two of the brackets first: $\left(x+1\right)\left(x+2\right)=x^2+3x+2$(x + 1)(x + 2) = x² + 3x + 2. That leaves $\left(x^2+3x+2\right)\left(x+3\right)$(x² + 3x + 2)(x + 3).

Multiply every term of the first bracket by x, then every term by 3: $x^3+3x^2+2x$x³ + 3x² + 2x and $3x^2+9x+6$3x² + 9x + 6. Collecting like terms gives $x^3+6x^2+11x+6$x³ + 6x² + 11x + 6.

Expanded all at once, there are 2 × 2 × 2 = 8 products before collecting. Doing two brackets first collects some of them early.

Where each coefficient comes from

In (x + a)(x + b)(x + c), each product takes one term from every bracket. An $x^2$x² term takes x from two brackets and a number from the third, so the coefficient of $x^2$x² is a + b + c.

An x term takes x from one bracket and numbers from the other two: ab + bc + ac. The constant takes a number from all three: abc.

For ${\left(x+2\right)}^3$(x + 2)³ that gives $x^3+6x^2+12x+8$x³ + 6x² + 12x + 8: 2 + 2 + 2 = 6, three pairs of 2 × 2 = 12, and 2 × 2 × 2 = 8.

Negative numbers

The pattern holds with signs. (x − 1)(x + 2)(x − 3) has numbers −1, 2 and −3. The $x^2$x² coefficient is −1 + 2 − 3 = −2, the x coefficient is (−1)(2) + (2)(−3) + (−1)(−3) = −2 − 6 + 3 = −5, and the constant is (−1)(2)(−3) = 6.

So $\left(x-1\right)\left(x+2\right)\left(x-3\right)=x^3-2x^2-5x+6$(x − 1)(x + 2)(x − 3) = x³ − 2x² − 5x + 6.

When x has a coefficient

The shortcut for the coefficients needs x on its own in each bracket, so expand step by step instead. $\left(2x+1\right)\left(x+3\right)=2x^2+7x+3$(2x + 1)(x + 3) = 2x² + 7x + 3, and multiplying by (x − 1) gives $2x^3+5x^2-4x-3$2x³ + 5x² − 4x − 3.

Check with a number

Put the same value of x into the brackets and into the expansion; they must agree. At x = 1, (x + 1)(x + 2)(x + 3) = 2 × 3 × 4 = 24, and 1 + 6 + 11 + 6 = 24.

At x = 2, (x − 1)(x + 2)(x − 3) = 1 × 4 × (−1) = −4, and 8 − 8 − 10 + 6 = −4. Avoid a value that makes a bracket zero: then both sides are zero and the check proves less.

The usual slips

Mixing up the coefficients. Adding the numbers gives the $x^2$x² coefficient; multiplying all three gives the constant.

Cubing each term. ${\left(x+2\right)}^3$(x + 2)³ is not $x^3+8$x³ + 8: it is (x + 2)(x + 2)(x + 2), with $6x^2$6x² and 12x in the middle.

Losing a sign. With a negative number, write each product with its sign before collecting.