GED Mathematical Reasoning: Dividing Polynomials

 

Dividing a polynomial by a monomial

Step 1:  Write each term in the numerator over the denominator.  If the numerator has more than one term, this will break the expression up into two or more fractions.

Step 2: Factor each numerator and denominator into prime factors

Step 3: One fraction at a time, cancel what is common to the numerator and denominator

Step 4: Simplify what is left

 

Example 1

Simplify:  \frac{9a^3}{3a}

 

In example one, we are being asked to divide a monomial by a monomial. Do you see how the division operation is denoted by a fraction bar? That is common when it comes to polynomial division.

Notice, though, that the direction for the example doesn’t say to “divide” – it says to “simplify.”  That is because dividing polynomials really just means that we need to simplify the fraction using the concept of division. Dividing polynomials involves dividing the numerator and denominator of the fraction by a common expression. To simplify the expression in example one, we will be dividing the numerator and denominator by a common term.

It’s not always easy to wrap our brains around the idea of dividing with variables, however, so it will help if we first “factor” the numerator and denominator into prime factors. Factoring is the process of writing a number or expression as a product. Put another way, factoring “undoes” multiplication. A factor is what gets multiplied to equal a number or term.

Factoring a term into its “prime factors” means that each factor is prime. Something that’s prime is divisible only by itself and one. For example, factoring the term: 9a^3 into prime factors gives the result:  3 times 3 times a times a times a.

 9a^3 = 3 \cdot 3\cdot a\cdot a \cdot a

Factoring the term: 3a into prime factors gives the result:  3 times a.

From here, we can simplify the expression in example one by “canceling out” what is common to the numerator and denominator. The numerator and denominator have a 3 and an “a” in common, which can be canceled leaving us with just: 3a^2

\frac{9a^3}{3a} = \frac{3 \cdot 3 \cdot a \cdot a \cdot a}{3 \cdot a}

\frac{9a^3}{3a} = \frac{3 \cdot a \cdot a}{1} = 3a^2

 

Example 2

Simplify:  \frac{4x - 8}{2}

 

When dividing any size polynomial by a monomial, we will use the technique we just discussed in example one, with an additional step at the beginning of the process.

We can rewrite the expression in example two as the difference of two fractions: 4x over 2 minus 8 over 2.

\frac{4x - 8}{2} = \frac{4x}{2} - \frac{8}{2}

We are allowed to do this because we are simply breaking the terms in the numerator up over the common denominator. From here, we can implement the same strategy we used in example one to simplify each resulting fraction. So factor the numerator and denominator of each fraction into prime factors.

The denominator 2 is already a prime number.

As for the numerators… 4x factors into:  2 times 2 times x. 8 factors into: 2 times 2 times 2

 \frac{4x}{2}-\frac{8}{2} = \frac{2 \cdot 2 \cdot x}{2} - \frac{2 \cdot 2 \cdot 2}{2}

Now examine each fraction separately. What do the numerator and denominator of each fraction have in common? The numerator and denominator of the first fraction have one 2 in common. The numerator and denominator of the second fraction also have one 2 in common. When we cancel the two’s, which means we are dividing by that common factor of 2, we get the result of: 2x minus 4.

\frac{4x}{2} - \frac{8}{2} = 2 \cdot x - 2 \cdot 2
\frac{4x}{2} - \frac{8}{2} = 2x - 4

 

Example 3

Simplify: \frac{25y^2 + 15y^3}{10y}

 

Step 1:  Write each term in the numerator over the denominator.

This gives us: 25y^2 over 10y plus 15y^3 over 10y.

\frac{25y^2 + 15y^3}{10y} = \frac{25y^2}{10y} + \frac{15y^3}{10y} 

Step 2: We’ll factor each numerator and denominator into prime factors. 25y^2 factors into 5 times 5 times y times y. 15y^3 factors into 3 times 5 times y times y times y. And  factors into 2 times 5 times y.

\frac{25y^2}{10y} + \frac{15y^3}{10y} = \frac{5 \cdot 5 \cdot y \cdot y}{2 \cdot 5 \cdot y} + \frac{3 \cdot 5 \cdot y \cdot y \cdot y}{2 \cdot 5 \cdot y}

Step 3: calls for us to cancel what is common to the numerator and denominator, one fraction at a time. The first fraction has one 5 and one “y” in common. The second fraction also has one 5 and one “y” in common. When we cancel the five’s and “y’s” in both fractions, we are really dividing by a common factor of 5y.

\frac{25y^2}{10y} + \frac{15y^3}{10y} = \frac{5 \cdot 5 \cdot y \cdot y}{2 \cdot 5 \cdot y} + \frac{3 \cdot 5 \cdot y \cdot y \cdot y}{2 \cdot 5 \cdot y} = \frac{5 \cdot y}{2} + \frac{3 \cdot y \cdot y}{2}

Finally, let’s simplify what’s left.

We can combine y times y to equal “y” squared in the numerator of the second fraction. And then we can put the two fractions together, and write the final answer as one fraction over the common denominator of 2.

 \frac{5 \cdot y}{2} + \frac{3 \cdot y \cdot y}{2}

 \frac{5y}{2} + \frac{3y^2}{2}

\frac{5y + 3y^2}{2}

 

Example 4

Simplify: \frac{a(b - 2) - 7(b - 2)}{b - 2}

 

Example 4 is an example that looks a little different than the first few examples. In fact, it’s a unique case. Notice that the denominator is not a monomial. It’s the binomial expression: “b minus 2.” Also notice that the numerator contains the difference of two expressions, which both contain that binomial expression “b minus 2.”

Here’s what happens when we use the same process for simplifying that we used in the previous examples. We’ll first write each expression in the numerator over the denominator of “b minus 2.”

\frac{a(b - 2) - 7(b - 2)}{b - 2} = \frac{a(b - 2)}{b - 2} - \frac{7(b - 2)}{b - 2}

From here, notice that the numerator and denominator of the first fraction have a “b minus 2” in common.  Also, the numerator and denominator of the second fraction have a “b minus 2” in common.

We can cancel what is common to the numerator and denominator, one fraction at a time. And basically what we’re doing here is dividing the numerator and denominator by a common factor of “b minus 2.”

\frac{a(b - 2) - 7(b - 2)}{b - 2} = \frac{a(b-2)}{b-2} - \frac{7(b-2)}{b-2} = a - 7

What we’re left with is a final answer of: a minus 7.

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