Binomial Expansion using Binomial Theorem & Pascal's Triangle
The Binomial Theorem: predicts the exponents of each term in the product. It is formalized as...
(a+b)^n = (a)^(n-k)(b)^(0+k) for k terms beginning at k=0. Ex: (x + 2)^4 = (x)^4(2)^0 + (x)^3(2)^1 + (x)^2(2)^2 + (x)^1(2)^3 + (x)^0(2)^4 ...now all we need are the final coefficients on each term...we use Pascal's Triangle to help us. |
Pascal's Triangle: predicts the coefficients on each term of the product. It is a self-replicating process starting with 1 and starting and ending each successive row on 1. All interior coefficients are the sum of the two directly above it. To determine which row of coefficients you will use for any given expansion, locate the number of the "power" expansion you are doing (i.e. (a + b)^5, use the row where you see 5 as the second number in.
Ex: (x + 2)^4 = (1) (x)^4(2)^0 + (4) (x)^3(2)^1 + (6) (x)^2(2)^2 + (4) (x)^1(2)^3 + (1) (x)^0(2)^4
= x^4 + 8x^3 + 24x^2 + 32x + 32 , this is your finished product.
Ex: (x + 2)^4 = (1) (x)^4(2)^0 + (4) (x)^3(2)^1 + (6) (x)^2(2)^2 + (4) (x)^1(2)^3 + (1) (x)^0(2)^4
= x^4 + 8x^3 + 24x^2 + 32x + 32 , this is your finished product.
Products & Quotients of Rational ExpressionsLike products of (multiplying) constant rational expressions (fractions), the resulting numerator is a product of the polynomial numerators of the factors (expressions multiplied), and the resulting denominator is a product of the polynomial denominators of the factors.
Like quotients of (dividing) constant rational expressions (fractions), the product is a result of multiplying the dividend (expression being divided) by the reciprocal ("flipped" version of the rational = 1 divided by original value) of the divisor (value by which the dividend is being divided). Before completing the multiplication of polynomial factors, it is recommended that you |
first factor the polynomials in the numerators and denominators completely to reveal any factors common to both the numerator and denominator that may be eliminated ("one-for-one") by division (remember division of a factor by itself = 1, not 0). This allows you to simplify the quotients completely before multiplying to get the final product; avoiding any unnecessary multiplication of factors that would only need to be eliminated by division in the final product.
Sums & Differences of Rational ExpressionsLike constant rational expressions (fractions), in order to add or subtract these expressions, they must have the same denominator.
In the case of polynomial denominators, it is often difficult to figure out what factor is needed to write the rational expressions in terms of a common denominator just from looking at them. So, first factor the denominators. After the've been factored it will be clear which factor(s), if any, they have in common, and those that are present in one denominator, but not the other ("uncommon" factors). Using the multiplicative identity property (a value does not change if multiplied by one), multiply each rational |
expression by a quotient of the "uncommon" factor(s) and itself(themselves). Now, over a common denominator combine the resulting products in the numerators by any applicable addition or subtraction. Remember that in the case of subtracting by a polynomial of multiple terms, you are subtracting EVERY term (not just the first term). In other words, when combining like terms from one numerator to the other, distribute subtraction to each term.
For additional help, check out other great Khan Academy Videos on Rational Expressions & Functions.
For additional help, check out other great Khan Academy Videos on Rational Expressions & Functions.
|
Solving Polynomials by Predicting (and evaluating) Rational RootsWatch the video tutorial on using the "rational root theorem" to find potential "roots" or solutions to polynomial functions. Write a notes summary on the rational roots theorem that highlights the most important ideas and skills in your own words. Write down any questions you still have to bring with you to class.
|