Algebra II: Evaluating higher powers of a monomial using Binomial Theorem
This is a free lesson from our course in Algebra II
 
   
In this lesson you'll learn mainly the fundamentals of the Binomial Theorem, which lead into the concept of finding a binomial coefficient that incorporates factorials into its formula. Then we will be going over to show how to compute the higher power of a binomial expression, without having to multiply the whole thing out. All this will be done using some examples with solution and with the help of audio, video presentation in instructor's own hand writing. Computing binomial expansion (x + a)n yields:
       xn + nC1xn-1 a1 + nC2xn-2 a2 + nC3xn-3 a3 +.....+ nCn-1x an-1 + nCnan .  (More text below video...)
<h2> Combinations and permutations to compute probability</h2> <p> linear, equations, video, variable, example, Algebra II, substitution, system of equation, solving equation, equations, coefficients, math help, equations in three variables, practice questions, quizzes</p> <p> Combinations and permutations to compute probability</p>
(Continued from above) This rule can be used to find out specific term as well. E.g. 7th term in the expansion of (x+5)10, is
3281250 x 4.
Remember about this theorem:
the top number of the binomial coefficient is always n, which is the  exponent  on your binomial.
the bottom number of the binomial coefficient starts with 0 and goes up 1 each time until you reach n, which is the exponent on your binomial.
the 1st term of the expansion has (first term of the binomial), raised to the n power i.e. exponent on the binomial.  From there first term's exponent goes down 1, until the last term, where it is being raised to the 0 power; and it is not seen written.
in the first term of the expansion (second term of the binomial), raised to the 0 power, which is not seen written.  From there second term's exponent goes up 1, until the last term, where it is being raised to the nth power, which is the exponent on the binomial.
For example, expand (2x+5)4
Substituting in the binomial formula, we get:
(2x+5)4 = (2x)4 + 4(2x)3(5) + (4(3)/2!)(2x)2(5)2 + (4(3)(2)/3!)(2x)(5)3 + (5)4 =16x4 + 160x3 + 600x2 + 1000x+ 625
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