How do you multiply a monomial by a binomial?

Answer 1

just like you would multiply regular numbers. if not than use a calculator

A longer answer:

First, monomials and binomials are both polynomials, where a polynomial is a sum of variables of descending powers, such as: 2x^3 - 4x^2 + 9x^1 + 3x^0 [normally written as: 2x^3 - 4x^2 + 9x + 3 since x^1 = x and x^0 = 1].

A MONomial has exactly one term, e.g.: 2x^3, or 3, while a BInomial has exactly two terms, e.g.: -4x^2 + 9x or x - 5.

To multiply a binomial by a monomial (or vice versa), just multiply each term of the binomial by the monomial. Thus multiply -4x^2 + 9x by 2x^3 to get:

(-4x^2) * (2x^3) + (9x) * (2x^3)

Simplifying, for each term (separated by a plus or minus sign) you collect and multiply the constants and then the variables. (To mutlply the variables group them by like bases and add the exponents):

(-4)(2) (x^2) (x^3) + (9)(2)(x) =

(-8)(x^5) + (18)(x) = -8x^5 +18x

When more than one variable is involved, often the bases (the variable that is raised to some power) are more complex and need to be sorted out carefully. You do this by rearrange the variables in the base (by the rule of multiplicative commutivity).. So if you were to get (in a larger polymonial product) a term like: x*y^2*x^3*y, you would rearrange the variables (usually alphabetically): x* x^3* y^2*y and combine like bases by adding exponents: x^4*y3. This helps a lot as this way you can see that:

x* x^3* y^2*y and

y^2* x^2*y* x^2

are really the same base: x^4*y3 so that you can combine:a monster like

(-7*1)(x* x^3* y^2*y) + (6*2) y^2* x^2*y* x^2) to get:

(-7)( x^4*y3) + (12)(x^4*y3) = 5(x^4*y3)

An example:

(3x^2 + 9y) * (4y^2) =

(3x^2)(4y^2) + (9y)(4y^2) =

12( x^2)( y^2) + 36y^3 (in simplest form)

This principle applied to multiplying two polynomials of any length. Just multiply each term of the first polynomial by each term of the second polynomial, collect terms and simplify as appropriate. It is important to keep track of your steps -- neatness counts!

As you do more (and larger) polynomial multiplications you will notice that this is actually how we do base-ten multiplication: 127 x 32 is really a trinomial multiplied by a binomial:

(1*10^2 + 2*10^1 + 7*10^0) times (3*10^1 + 2*10^0)

Here the base is "10" not "x". If you mutliply out these polynomials you willl get: the horrendous product:

(1*10^2) * (3*10^1) + (2*10^1) * (3*10^1) + (7*10^0)*(3*10^1) +

(1*10^2)( 2*10^0) + (2*10^1) (2*10^1) + (7*10^0) * (2*10^0) =

Simplifying this mess (and multiplying out the powers of ten) you get:

3810 + 254 = 4064.

If you multiply out 127 * 32 the usual way (with 127 on top), you will notice that you will get the two partial products: 254 and 381, but with the 381 "shifted over" one place to the left. This is the same as writing down "3810".

Polynomial multiplication (and division, adn additon and subtraction) are real a more general form of base-ten arithmetic. In base-ten arithemetic as a short cut we use "place value" instead of powers of ten, so we can write "12" instead of "1*10^1 + 2*10^0". the catch is that while we can write "12" as either

"1*10^1 + 2*10^0" or as 2*10^0 + 1*10^1",

we cannot write "12" as either "12" or as "21". In fact "21" is really:

"2*10^1 + 1*10^0", or, "1*10^0 + 2*10^1"

Note the difference!.

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