Examples
(x-1)3(x−1)3
Step 1
Use the binomial expansion theorem to find each term. The binomial theorem states (a+b)n=n∑k=0nCk⋅(an-kbk)(a+b)n=n∑k=0nCk⋅(an−kbk).
3∑k=03!(3-k)!k!⋅(x)3-k⋅(-1)k3∑k=03!(3−k)!k!⋅(x)3−k⋅(−1)k
Step 2
Expand the summation.
3!(3-0)!0!(x)3-0⋅(-1)0+3!(3-1)!1!(x)3-1⋅(-1)1+3!(3-2)!2!(x)3-2⋅(-1)2+3!(3-3)!3!(x)3-3⋅(-1)33!(3−0)!0!(x)3−0⋅(−1)0+3!(3−1)!1!(x)3−1⋅(−1)1+3!(3−2)!2!(x)3−2⋅(−1)2+3!(3−3)!3!(x)3−3⋅(−1)3
Step 3
Simplify the exponents for each term of the expansion.
1⋅(x)3⋅(-1)0+3⋅(x)2⋅(-1)1+3⋅(x)1⋅(-1)2+1⋅(x)0⋅(-1)31⋅(x)3⋅(−1)0+3⋅(x)2⋅(−1)1+3⋅(x)1⋅(−1)2+1⋅(x)0⋅(−1)3
Step 4
Step 4.1
Multiply (x)3(x)3 by 11.
(x)3⋅(-1)0+3⋅(x)2⋅(-1)1+3⋅(x)1⋅(-1)2+1⋅(x)0⋅(-1)3(x)3⋅(−1)0+3⋅(x)2⋅(−1)1+3⋅(x)1⋅(−1)2+1⋅(x)0⋅(−1)3
Step 4.2
Anything raised to 00 is 11.
x3⋅1+3⋅(x)2⋅(-1)1+3⋅(x)1⋅(-1)2+1⋅(x)0⋅(-1)3x3⋅1+3⋅(x)2⋅(−1)1+3⋅(x)1⋅(−1)2+1⋅(x)0⋅(−1)3
Step 4.3
Multiply x3x3 by 11.
x3+3⋅(x)2⋅(-1)1+3⋅(x)1⋅(-1)2+1⋅(x)0⋅(-1)3x3+3⋅(x)2⋅(−1)1+3⋅(x)1⋅(−1)2+1⋅(x)0⋅(−1)3
Step 4.4
Evaluate the exponent.
x3+3x2⋅-1+3⋅(x)1⋅(-1)2+1⋅(x)0⋅(-1)3x3+3x2⋅−1+3⋅(x)1⋅(−1)2+1⋅(x)0⋅(−1)3
Step 4.5
Multiply -1−1 by 33.
x3-3x2+3⋅(x)1⋅(-1)2+1⋅(x)0⋅(-1)3
Step 4.6
Simplify.
x3-3x2+3⋅x⋅(-1)2+1⋅(x)0⋅(-1)3
Step 4.7
Raise -1 to the power of 2.
x3-3x2+3x⋅1+1⋅(x)0⋅(-1)3
Step 4.8
Multiply 3 by 1.
x3-3x2+3x+1⋅(x)0⋅(-1)3
Step 4.9
Multiply (x)0 by 1.
x3-3x2+3x+(x)0⋅(-1)3
Step 4.10
Anything raised to 0 is 1.
x3-3x2+3x+1⋅(-1)3
Step 4.11
Multiply (-1)3 by 1.
x3-3x2+3x+(-1)3
Step 4.12
Raise -1 to the power of 3.
x3-3x2+3x-1
x3-3x2+3x-1