Examples
(x-9)3(x−9)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⋅(-9)k3∑k=03!(3−k)!k!⋅(x)3−k⋅(−9)k
Step 2
Expand the summation.
3!(3-0)!0!(x)3-0⋅(-9)0+3!(3-1)!1!(x)3-1⋅(-9)1+3!(3-2)!2!(x)3-2⋅(-9)2+3!(3-3)!3!(x)3-3⋅(-9)33!(3−0)!0!(x)3−0⋅(−9)0+3!(3−1)!1!(x)3−1⋅(−9)1+3!(3−2)!2!(x)3−2⋅(−9)2+3!(3−3)!3!(x)3−3⋅(−9)3
Step 3
Simplify the exponents for each term of the expansion.
1⋅(x)3⋅(-9)0+3⋅(x)2⋅(-9)1+3⋅(x)1⋅(-9)2+1⋅(x)0⋅(-9)31⋅(x)3⋅(−9)0+3⋅(x)2⋅(−9)1+3⋅(x)1⋅(−9)2+1⋅(x)0⋅(−9)3
Step 4
Step 4.1
Multiply (x)3(x)3 by 11.
(x)3⋅(-9)0+3⋅(x)2⋅(-9)1+3⋅(x)1⋅(-9)2+1⋅(x)0⋅(-9)3(x)3⋅(−9)0+3⋅(x)2⋅(−9)1+3⋅(x)1⋅(−9)2+1⋅(x)0⋅(−9)3
Step 4.2
Anything raised to 00 is 11.
x3⋅1+3⋅(x)2⋅(-9)1+3⋅(x)1⋅(-9)2+1⋅(x)0⋅(-9)3x3⋅1+3⋅(x)2⋅(−9)1+3⋅(x)1⋅(−9)2+1⋅(x)0⋅(−9)3
Step 4.3
Multiply x3x3 by 11.
x3+3⋅(x)2⋅(-9)1+3⋅(x)1⋅(-9)2+1⋅(x)0⋅(-9)3x3+3⋅(x)2⋅(−9)1+3⋅(x)1⋅(−9)2+1⋅(x)0⋅(−9)3
Step 4.4
Evaluate the exponent.
x3+3x2⋅-9+3⋅(x)1⋅(-9)2+1⋅(x)0⋅(-9)3x3+3x2⋅−9+3⋅(x)1⋅(−9)2+1⋅(x)0⋅(−9)3
Step 4.5
Multiply -9−9 by 33.
x3-27x2+3⋅(x)1⋅(-9)2+1⋅(x)0⋅(-9)3x3−27x2+3⋅(x)1⋅(−9)2+1⋅(x)0⋅(−9)3
Step 4.6
Simplify.
x3-27x2+3⋅x⋅(-9)2+1⋅(x)0⋅(-9)3x3−27x2+3⋅x⋅(−9)2+1⋅(x)0⋅(−9)3
Step 4.7
Raise -9−9 to the power of 22.
x3-27x2+3x⋅81+1⋅(x)0⋅(-9)3x3−27x2+3x⋅81+1⋅(x)0⋅(−9)3
Step 4.8
Multiply 8181 by 33.
x3-27x2+243x+1⋅(x)0⋅(-9)3x3−27x2+243x+1⋅(x)0⋅(−9)3
Step 4.9
Multiply (x)0(x)0 by 11.
x3-27x2+243x+(x)0⋅(-9)3x3−27x2+243x+(x)0⋅(−9)3
Step 4.10
Anything raised to 00 is 11.
x3-27x2+243x+1⋅(-9)3x3−27x2+243x+1⋅(−9)3
Step 4.11
Multiply (-9)3(−9)3 by 11.
x3-27x2+243x+(-9)3x3−27x2+243x+(−9)3
Step 4.12
Raise -9−9 to the power of 33.
x3-27x2+243x-729x3−27x2+243x−729
x3-27x2+243x-729x3−27x2+243x−729