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, and 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}

Keep −2 in parentheses so each power carries its sign.

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

3x²(−2) = −6x², 3x(4) = 12x and (−2)³ = −8. The odd powers of −2 are negative, so the signs alternate.

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

At x = 3, (3 − 2)³ = 1, and 27 − 54 + 36 − 8 = 1. Both sides agree.

(x − 2)³ = x³ − 6x² + 12x − 8.

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

Expand (x + 3)⁴.

Show the answer and explanation

Row 4 is 1, 4, 6, 4, 1. The terms are x⁴, 4x³(3) = 12x³, 6x²(9) = 54x², 4x(27) = 108x and 3⁴ = 81.

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

Open the steps in Math and expand (x − 2)⁴ with row 1, 4, 6, 4, 1: the checker confirms x⁴ − 8x³ + 24x² − 32x + 16.

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