Math · College algebra · Worked example

Expand (x − 2)³ with the binomial theorem

Expand ${(x - 2)}^{3}$ .

{(x - 2)}^{3}

Read the coefficients from row 3

The power is $3$ , so the coefficients are $1$ , $3$ , $3$ , $1$ . The powers of a fall from $3$ to $0$ while the powers of $b$ rise from $0$ to $3$ .

{(a + b)}^{3} = a^{3} + 3 a^{2} b + 3 a b^{2} + b^{3}

Substitute $a = x$ and $b = - 2$

The binomial is $x + (- 2)$ , so $b$ is $- 2$ , sign included. Keep $- 2$ in parentheses so each power carries its sign.

x^{3} + 3 x^{2} (- 2) + 3 x {(- 2)}^{2} + {(- 2)}^{3}

Simplify each term

$3 x^{2} (- 2) = - 6 x^{2}$ , $3 x (4) = 12 x$ and ${(- 2)}^{3} = - 8$ . The odd powers of $- 2$ are negative, so the signs alternate.

x^{3} - 6 x^{2} + 12 x - 8

Check with a number

At $x = 3$ , ${(3 - 2)}^{3} = 1$ , and $27 - 54 + 36 - 8 = 1$ . Both sides agree. The coefficients’ sizes, $1 + 6 + 12 + 8 = 27$ , give $3^{3}$ , which is ${(x + 2)}^{3}$ at $x = 1$ .

Result

${(x - 2)}^{3}$ $= x^{3} - 6 x^{2} + 12 x - 8$ .

{(x - 2)}^{3} = x^{3} - 6 x^{2} + 12 x - 8

Your turn

1. One like the example

Expand ${(x + 3)}^{4}$ .

Show the answer and explanation

$x^{4} + 12 x^{3} + 54 x^{2} + 108 x + 81$ .

Row 4 is $1$ , $4$ , $6$ , $4$ , $1$ . The terms are $x^{4}$ , $4 x^{3} (3) = 12 x^{3}$ , $6 x^{2} (9) = 54 x^{2}$ , $4 x (27) = 108 x$ and $3^{4} = 81$ . At $x = 1$ the sum is $256 = 4^{4}$ , as ${(1 + 3)}^{4}$ requires.

\begin{array}{l} {(x + 3)}^{4} \\ x^{4} + 4 x^{3} (3) + 6 x^{2} {(3)}^{2} + 4 x {(3)}^{3} + 3^{4} \\ x^{4} + 12 x^{3} + 54 x^{2} + 108 x + 81 \end{array}

\begin{array}{l} {(x + 3)}^{4} \\ x^{4} + 4 x^{3} (3) + 6 x^{2} {(3)}^{2} \\ + 4 x {(3)}^{3} + 3^{4} \\ x^{4} + 12 x^{3} + 54 x^{2} \\ + 108 x + 81 \end{array}

2. A variation

Expand ${(2 x - 1)}^{3}$ . Here $a$ has a coefficient and $b$ is negative.

Show the answer and explanation

$8 x^{3} - 12 x^{2} + 6 x - 1$ .

With $a = 2 x$ and $b = - 1$ , the terms are ${(2 x)}^{3} = 8 x^{3}$ , $3 {(2 x)}^{2} (- 1) = - 12 x^{2}$ , $3 (2 x) {(- 1)}^{2} = 6 x$ and ${(- 1)}^{3} = - 1$ . At $x = 1$ , ${(2 - 1)}^{3} = 1$ and $8 - 12 + 6 - 1 = 1$ .

\begin{array}{l} {(2 x - 1)}^{3} \\ {(2 x)}^{3} + 3 {(2 x)}^{2} (- 1) + 3 (2 x) {(- 1)}^{2} + {(- 1)}^{3} \\ 8 x^{3} - 12 x^{2} + 6 x - 1 \end{array}

\begin{array}{l} {(2 x - 1)}^{3} \\ {(2 x)}^{3} + 3 {(2 x)}^{2} (- 1) \\ + 3 (2 x) {(- 1)}^{2} + {(- 1)}^{3} \\ 8 x^{3} - 12 x^{2} + 6 x - 1 \end{array}

3. Somewhere new

Find ${1.01}^{5}$ exactly by writing it as ${(1 + 0.01)}^{5}$ .

Show the answer and explanation

$1.0510100501$ .

Row 5 is $1$ , $5$ , $10$ , $10$ , $5$ , $1$ , and every power of $1$ is $1$ . The terms are $1$ , $0.05$ , $0.001$ , $0.00001$ , $0.00000005$ and $0.0000000001$ , which add to $1.0510100501$ . The first two terms alone give $1.05$ , already close.

\begin{array}{l} {(1 + 0.01)}^{5} \\ 1 + 5 (0.01) + 10 {(0.01)}^{2} + 10 {(0.01)}^{3} + 5 {(0.01)}^{4} + {(0.01)}^{5} \\ 1.0510100501 \end{array}

\begin{array}{l} {(1 + 0.01)}^{5} \\ 1 + 5 (0.01) + 10 {(0.01)}^{2} \\ + 10 {(0.01)}^{3} + 5 {(0.01)}^{4} \\ + {(0.01)}^{5} \\ 1.0510100501 \end{array}

Keep exploring

Open the steps in Math and expand ${(x - 2)}^{4}$ with row 1, $4$ , $6$ , $4$ , $1$ . The last row should read $x^{4} - 8 x^{3} + 24 x^{2} - 32 x + 16$ .

Return to the concept →

Sources and scope

Authored study material. Tool results depend on the stated inputs and model assumptions.

After reading

Study it, then try it

Practice the key terms as flashcards, then open the example in its tool, change a value and keep a useful result on your board.

Study the 5 key terms Explore Pascal’s Triangle Check the expansion in Math See row 3 as paths in Binomial & the Galton Board Open worked example on a board The binomial theorem in Math Reference

Your existing work stays on this device. Examples open as editable copies.

Read the coefficients from row 3

Substitute a=xa=x and b=−2b=−2

Simplify each term

Check with a number

Result

Your turn

1. One like the example

x4+12x3+54x2+108⁢x+81x4+12x3+54x2+108x+81.

2. A variation

8x3−12x2+6⁢x−18x3−12x2+6x−1.

3. Somewhere new

1.05101005011.0510100501.

Keep exploring

Study it, then try it

Substitute $a = x$ and $b = - 2$

$x^{4} + 12 x^{3} + 54 x^{2} + 108 x + 81$ .

$8 x^{3} - 12 x^{2} + 6 x - 1$ .

$1.0510100501$ .