Algebra I, Article 79. How to Divide One Polynomial by Another

In article 79, we are learning how to divide one polynomial by another.  

To master this lesson, you must read SLOWLY and make sure you understand every step in the lesson.

First, the teacher shows how two polynomials are multiplied so that we can see how the product is formed.  The product of multiplication becomes the dividend in division, so in division, we are seeking to move backwards to find that polynomial which, when multiplied by the divisor produced the dividend.

Second, the teacher explains to us that before we can divide, we must make sure that the leading letter is arranged in the same order in both the dividend and in the divisor.  The “leading letter” is that letter whose exponents decrease from left to right, as the letter a does here:  2a3 − 3a2b + ab2.

Third, the key to understanding this division is explained in observation 3d:  

“If we divide 2a3by a, the result, 2a2, will be the term of the quotient by which a – b was first multiplied. If we now multiply a − b by 2a2, and subtract the product from the dividend, there will remain -a2b + ab2, which is the product of a − b by the other term of the quotient. Dividing -a2b by a, we find this unknown term. Multiplying a − b by it, and subtracting the product, nothing remains.”

Fourth, we have this summarized into a rule in the textbook.

  • Step 1. Arrange the dividend and divisor with reference to the leading letter, and place the divisor on the right of the dividend. 
  • Step 2. Divide the first term of the dividend by the first term of the divisor, for the first term of the quotient. Multiply the divisor by this term, and subtract the product from the dividend. 
  • Step 3. Divide the first term of the remainder by the first term of the divisor, for the second term of the quotient. Multiply the divisor by this term, and subtract the product from the last remainder. 
  • Step 4. Proceed in the same manner, and if you obtain 0 for a remainder, the division is said to be exact.

Now, let us walk through the first example in the textbook.

1. Divide 6a2 − 13ax + 6x2 by 2a − 3x.

Step 1. Arrange the dividend and divisor with reference to the leading letter, and place the divisor on the right of the dividend. 

6a2 − 13ax + 6x2 | 2a − 3x.

Step 2a. Divide the first term of the dividend (6a2) by the first term of the divisor (2a), for the first term of the quotient. 

6a2 ÷ 2a = 3a, because 2a × 3a = 6a2.

Step 2b. Multiply the divisor (2a − 3x) by this term (3a)

(2a – 3x) × 3a = 6a2 − 9ax

Step 2c. …and subtract the product (6a2 − 9ax) from the dividend (6a2 − 13ax + 6x2). 

(6a2 – 13ax + 6x2) − (6a2 − 9ax) = 6a2 – 13ax + 6x2 − 6a2 + 9ax  = -4ax + 6x2

(Note that the signs of the subtrahend changed when removed from parentheses.)

Step 3a. Divide the first term of the remainder (-4ax) by the first term of the divisor (2a), for the second term of the quotient. 

-4ax ÷ 2a = -2x, because 2a × -2x = -4ax.

Step 3b. Multiply the divisor (2a − 3x) by this term (-2x), 

(2a − 3x) × -2x = -4ax + 6x2

Step 3c.  and subtract the product (-4ax + 6x2) from the last remainder (-4ax + 6x2).

(-4ax + 6x2) − (-4ax + 6x2) = 0
-4ax + 6x2 + 4ax − 6x2 = 0

Step 4. Proceed in the same manner, and if you obtain 0 for a remainder, the division is said to be exact.

Here’s what it looks like on paper, with the steps of the rule labeled in red on the left:

I hope this makes this process clear.  If you need help, please reply.

God bless,
Mr. William C. Michael, Headmaster
Classical Liberal Arts Academy

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