Algebra Examples

Expand Using the Binomial Theorem
(x-3)2(x3)2
Step 1
Use the binomial expansion theorem to find each term. The binomial theorem states (a+b)n=nk=0nCk(an-kbk)(a+b)n=nk=0nCk(ankbk).
2k=02!(2-k)!k!(x)2-k(-3)k2k=02!(2k)!k!(x)2k(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!(20)!0!(x)20(3)0+2!(21)!1!(x)21(3)1+2!(22)!2!(x)22(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
Simplify each term.
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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.
x21+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+2x(-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
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