12.4. Division of Algebraic Expressions Continued

Consider (7x2 + 14x) ÷ (x + 2)

We shall factorise (7x2 + 14x) first to check and match factors with the denominator:

7x2 + 14x = (7 × x × x) + (2 × 7 × x)

= 7 × x × (x + 2)

Now (7x2 + 14x) ÷ (x + 2) = 7x.

In this example, we start with the polynomial 7x² + 14x. Before performing the division, we factor the polynomial. This means we rewrite it in a form that reveals its components. We notice that both terms have a common factor of 7x. Thus, we can express it as 7x(x + 2). Now, we perform the division by the binomial (x + 2). Since (x + 2) is a factor of the polynomial, we can cancel it, simplifying our expression down to just 7x. This shows us that dividing a polynomial by one of its factors gives a simpler polynomial as a result.

Imagine you have a box of chocolates that comes in two sizes: small and large. In this code, you can represent the total number of chocolates in terms of their size. If you know that a certain number of chocolates can be combined into larger boxes, you can easily find out how many boxes you are left with after distributing them into the larger sections. Just like here, when you divide, you're essentially reorganizing items into simpler groups.

Example 15: Divide 44(x4 – 5x3 – 24x2) by 11x (x – 8)

Solution: Factorising 44(x4 – 5x3 – 24x2), we get

44(x4 – 5x3 – 24x2) = 2 × 2 × 11 × x2(x2 – 5x – 24) = 2 × 2 × 11 × x2(x – 8)(x + 3) (after factoring further)

Now dividing:

= 4x²(x + 3).

Here, we approach a slightly more complex polynomial division. First, we start with the polynomial 44(x⁴ – 5x³ – 24x²). Our goal is to factor the expression to identify parts that we can cancel out during division. We factor by taking out the common factors (2 × 2 × 11 × x²) and then factoring the quadratic part using methods appropriate for quadratics. After factoring, we have 4x²(x + 3) remaining after canceling the common terms with the divisor 11x(x – 8). This example emphasizes how breaking down polynomials before division can simplify the process and yield clearer results.

Think of dividing a fruit basket shared among friends. You might have various types of fruits mixed together, and you need to separate them into types (apples, oranges, bananas) before dividing. By organizing (factoring) your basket first, it becomes easier to see how many pieces of each fruit each person can receive. Similarly, in polynomial division, recognizing and factoring out common terms simplifies the division process.

Example 16: Divide z(5z2 – 80) by 5z(z + 4)

Solution: Dividend = z(5z2 – 80)

=z[(5 × z2) – (5 × 16)]

=5z(z + 4)(z – 4) using the identity (a² – b²) = (a + b)(a – b).

Thus, z(5z² – 80) ÷ 5z(z + 4) = (z – 4).

In this example, we encounter a polynomial expression that is suitable for factoring based on known identities, like the difference of squares. We identify a pattern in z(5z² - 80) which allows us to express it as z times a factored form (common algebraic identity). Then we can proceed to cancel out the common factors with the divisor 5z(z + 4). This ultimately reduces our result to just (z - 4). This technique demonstrates how recognizing identities can significantly streamline polynomial division.

Imagine you have a recipe for a large cake (our polynomial) that includes layers of different flavors. If you know the total size that's meant for, say, 5 guests, and you want to see how many servings you could get out of it based on the individual flavors, factorizing helps you split (or divide) that cake effectively. By using what you know about each flavor (identity), you can quickly find you're left with smaller manageable portions.

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