Algebra Examples
(x-3)2(x−3)2
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).
2∑k=02!(2-k)!k!⋅(x)2-k⋅(-3)k2∑k=02!(2−k)!k!⋅(x)2−k⋅(−3)k
Step 2
Expand the summation.
2!(2-0)!0!(x)2-0⋅(-3)0+2!(2-1)!1!(x)2-1⋅(-3)1+2!(2-2)!2!(x)2-2⋅(-3)22!(2−0)!0!(x)2−0⋅(−3)0+2!(2−1)!1!(x)2−1⋅(−3)1+2!(2−2)!2!(x)2−2⋅(−3)2
Step 3
Simplify the exponents for each term of the expansion.
1⋅(x)2⋅(-3)0+2⋅(x)1⋅(-3)1+1⋅(x)0⋅(-3)21⋅(x)2⋅(−3)0+2⋅(x)1⋅(−3)1+1⋅(x)0⋅(−3)2
Step 4
Step 4.1
Multiply (x)2 by 1.
(x)2⋅(-3)0+2⋅(x)1⋅(-3)1+1⋅(x)0⋅(-3)2
Step 4.2
Anything raised to 0 is 1.
x2⋅1+2⋅(x)1⋅(-3)1+1⋅(x)0⋅(-3)2
Step 4.3
Multiply x2 by 1.
x2+2⋅(x)1⋅(-3)1+1⋅(x)0⋅(-3)2
Step 4.4
Simplify.
x2+2⋅x⋅(-3)1+1⋅(x)0⋅(-3)2
Step 4.5
Evaluate the exponent.
x2+2x⋅-3+1⋅(x)0⋅(-3)2
Step 4.6
Multiply -3 by 2.
x2-6x+1⋅(x)0⋅(-3)2
Step 4.7
Multiply (x)0 by 1.
x2-6x+(x)0⋅(-3)2
Step 4.8
Anything raised to 0 is 1.
x2-6x+1⋅(-3)2
Step 4.9
Multiply (-3)2 by 1.
x2-6x+(-3)2
Step 4.10
Raise -3 to the power of 2.
x2-6x+9
x2-6x+9