GED Mathematical Reasoning: Multiplying Polynomials

 

  • When multiplying a term by a term, multiply the coefficients together. Then, from there, multiply the variable parts together – taking it one variable at a time. Multiplying variables involves keeping the base the same and adding the exponents.

 

Example 1

Multiply: (3ab^2)(-4a^3b^5)

 

Multiplying a monomial by a monomial essentially means we are multiplying a single term by another single term. This concept provides the foundation for multiplying polynomials – no matter many terms they contain.

To multiply a monomial by a monomial, begin by multiplying the coefficients together.

For example one, we will begin by multiplying 3 times -4, which is equal to -12. Therefore the coefficient of the final answer will be -12.

From there, we’ll multiply the variable parts together – taking it one variable at a time. Multiplying variables involves keeping the base the same and adding the exponents.

So to multiply “a” – which has an exponent of one – by “a” cubed, we will keep the base of “a” and add 1 plus 3 which is equal to 4. So “a” times “a” cubed is equal to “a” to the fourth power.

Now let’s proceed to the variable “b”. “b” squared times “b” to the fifth power equals “b” to the seventh power, since 2 plus 5 equals 7.

The final answer, then, is:  -12, “a” to the fourth power, “b” to the seventh power.

(3ab^2)(-4a^3b^5)
= (3 \cdot -4)(a \cdot a^3)(b^2 \cdot b^5)

(3)(-4) = -12
(a)(a^3) = (a^1)(a^3) = a^{1 + 3} = a^4

(b^2)(b^5) = b^{2 + 5} = b^7

Final Answer: -12a^{4}b^7

It may seem strange to be adding exponents when the operation in the original problem is multiplication. And this might takes some getting used to.

Here is the reasoning behind it, which may help.

(a)(a^3) can be expanded and written out as:  (a) \cdot (a \cdot a \cdot a)

As you can see, there are four a’s so it makes sense that the product would be “a” to the fourth power.

Likewise, (b^2)(b^5) can be expanded and written out to be: (b \cdot b) \cdot (b \cdot b \cdot b \cdot b \cdot b)

Because there are seven “b’s” the product is “b” to the seventh power.

 

Example 2

Multiply: 7x(6xy + y^2)

 

To multiply a monomial by polynomial, use the distributive property.

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To begin, we will multiply 7x by 6xy. 7 times 6 is equal to 42.

x \cdot x is the same as x^1 \cdot x^1 . We will keep the base of “x” and add the exponents of one plus one, which is equal to 2. So x \cdot x =  x^2

Since the multiplier “7x” doesn’t contain a “y” there is nothing to multiply. We’ll just carry the “y” variable part through to the product.

So, 7x times 6xy is equal to:  42x^2y

7x \cdot 6xy
 = (7 \cdot 6)(x \cdot x)(y)

7 \cdot 6 = 42
x \cdot x = x^1 \cdot x^1 = x^{1 + 1} = x^2
y

So: 7x \cdot 6xy = 42x^2y

By the distributive property we will now multiply 7x times the second term in parenthesis:  “y” squared.

The coefficient of “y” squared is one. Multiplying the coefficients 7 times 1 together gives us 7.

Moving on to the variable parts, these terms have no variables in common. In this case, no other multiplying can be done – we will simply keep the variable parts as they are and write them next to each other in the product.

In other words, the product of 7x and “y” squared is: 7xy^2

7x \cdot y^2 = 7x \cdot 1y^2
 = (7 \cdot 1)(x)(y^2)

7 \cdot 1 = 7
x
y^2

So: 7x \cdot y^2 = 7xy^2

Putting this all together gives us:

7x(6xy + y^2) = 42x^2y + 7xy^2

As a final check, let’s ask ourselves: can this expression be simplified any further by combining like terms.

The answer is no – these terms cannot be added since they are not “like terms.” They both contain the variables “x” and “y” and they both contain exponents of two, but the combination isn’t the same. In the first term, the “x” is squared.  In the second term, the “y” is squared. Since the variable parts don’t match perfectly, these two terms are not ‘like terms’ and cannot be combined.

Before we move on to example 3, I would like to recommend that you use the “stack” method for your work when multiplying polynomials.

The stack method involves multiplying each pair of terms one underneath the other. To state the final result, you can then orient each resulting term horizontally.

For instance, your work for Example 2 might look like this:

7x \cdot 6xy = 42xy^2
7x \cdot y^2 = 7xy^2

Final Answer: 42x^2y +7xy^2

 

Example 3

Multiply: (2x + 3y)(5x - 9y)

 

To multiply a binomial times a binomial, multiply each term in the first set of parenthesis by each term in the second set of parenthesis.

Essentially, we are employing the distributive property twice. First, we will distribute the term “2x” to the terms in the second set of parenthesis. Then, we will distribute the term “3y” to the terms in the second set of parenthesis.

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This is a good example to show how helpful the “stack” method for organizing our work can be.

First, we will multiply 2x times 5x. The result is 10 “x” squared.

Next, we will multiply 2x times -9y. Notice that the subtraction sign in front of the 9y makes that term negative. The result of 2x times -9y is -18xy.

Now we will switch to the second term in the first polynomial and multiply 3y times 5x. The result is positive 15 xy.

Finally, we will multiply 3y times -9y, which is equal to -27 “y” squared. Utilizing the “stack” method for your work would yield something that looks like this:

2x \cdot 5x = 10x^2
2x \cdot -9y = -18xy
3y \cdot 5x = 15xy
3y \cdot -9y = -27y^2

From here, we can orient each result horizontally to state the answer:

(2x + 3y)(5x - 9y) = 10x^2 - 18xy + 15xy - 27y^2

Let’s review this answer to see if it can be simplified further by combining like terms. Since -18 “xy” and 15 “xy” have the same variable part, they can be combined to be:  -3 “xy”. This makes our final result:  10 “x” squared MINUS 3 “xy” MINUS 27 “y” squared.

-18xy + 15xy = (-18 + 15)xy = -3xy

Final Answer: 10x^2 - 3xy - 27y^2

Before we conclude this chapter, I want to mention that we can multiply any two polynomials – no matter the size – with the strategy used in example 3. That is:  each term in the first polynomial is multiplied by each term in the second polynomial using the distributive property repeatedly.

But when it comes to the special case of multiplying a binomial times a binomial, as we did in Example 3, you may hear this strategy referred to as the “FOIL” method.

FOIL stands for: first, outer, inner, last.

Looking back at example 3, notice that we began by multiplying the FIRST two terms of each binomial.

Next we multiplied the OUTER two terms.

Then we multiplied the INNER two terms.

Finally we multiplied the LAST two terms of each binomial.

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