الرياضيات (بالإنجليزية) الصف الثالث الإعدادي، الأسبوع 6: Multiplying and dividing rational algebraic fractions, with their domains

Operations on rational algebraic fractions

الأسبوع 6 · العمليات على الكسور الجبرية النسبية: الجمع والطرح والضرب والقسمة والمعكوسان وخواص العمليات · 3 دروس

Multiplying and dividing rational algebraic fractions, with their domains

Mathematics Prep 3 Term 1, Unit 2, Lesson 5

GoalMultiply and divide rational fractions, and find the domain of the result, including the extra restriction in a division.

New words

reciprocal · the fraction turned upside down: the reciprocal of p ÷ q is q ÷ pcancel across · cancel a factor of one numerator against a factor of the other denominatordivisor · the fraction we divide by, the second oneextra restriction · in a division, the zeros of the divisor’s numerator are removed as well

Explanation

To multiply, factorise everything, cancel any factor that is on the top of one fraction and on the bottom of either, and multiply what is left. The domain is found from both ORIGINAL denominators. To divide p ÷ q by r ÷ s, multiply the first fraction by the reciprocal of the second: (p ÷ q) × (s ÷ r). Invert only the second fraction. A division needs one more restriction: we cannot divide by zero, so r ≠ 0 as well. The domain of a quotient therefore removes the zeros of q, of s AND of r. This last restriction is easy to forget because r does not appear in a denominator until we invert.

Operation	Rule	Domain removes zeros of
multiply	(p ÷ q) × (r ÷ s) = pr ÷ qs	q, s
divide	(p ÷ q) ÷ (r ÷ s) = ps ÷ qr	q, s, r

Worked examples

Multiply (x + 2) ÷ (x² − 4x + 3) × (x² − 1) ÷ (x² + 3x + 2). Factorise: x² − 4x + 3 = (x − 1)(x − 3), x² − 1 = (x − 1)(x + 1), x² + 3x + 2 = (x + 1)(x + 2). Domain: R − {1, 3, −1, −2}. The product is (x + 2)(x − 1)(x + 1) ÷ ((x − 1)(x − 3)(x + 1)(x + 2)). Everything cancels except (x − 3): the result is 1 ÷ (x − 3), with domain R − {−2, −1, 1, 3}.

Divide (x + 1) ÷ (x − 2) by (x − 3) ÷ (x + 4). Multiply by the reciprocal of the second: (x + 1) ÷ (x − 2) × (x + 4) ÷ (x − 3) = (x + 1)(x + 4) ÷ ((x − 2)(x − 3)). Domain: zeros of the denominators x − 2 and x + 4, and of the divisor’s numerator x − 3: R − {2, −4, 3}. Notice that −4 is not visible in the final fraction, but it is still excluded.

Divide (x² − 4) ÷ (x + 3) by (x − 2) ÷ (x² − 9). Multiply by (x² − 9) ÷ (x − 2): (x − 2)(x + 2)(x − 3)(x + 3) ÷ ((x + 3)(x − 2)) = (x + 2)(x − 3). Domain: the denominators x + 3 and x² − 9 give −3 and 3; the divisor’s numerator x − 2 gives 2; R − {−3, 3, 2}. Check at x = 1: (−3 ÷ 4) ÷ (−1 ÷ −8) = −3/4 ÷ 1/8 = −6 and (3)(−2) = −6 ✓.

Class activity

Flip the right one. Divide 1/2 ÷ 3/4 using the rule (invert the second only). Then try inverting the first instead and compare. Which is right? Check by multiplying your answer by 3/4.

Practice

Multiply x ÷ (x + 1) × (x² − 1) ÷ x².

Divide (x² − 9) ÷ (x + 2) by (x − 3) ÷ (x + 2).

Multiply 6 ÷ (x² − 1) × (x − 1) ÷ 3.

Divide (x² + x) ÷ (x − 4) by (x + 1) ÷ (x² − 16) and give the domain.

In the last question, is x = −1 allowed although x(x + 4) has a value there? Explain.