math-2017:A.APR.6math-2017:A.APR.7
math-2017:A.APR.6 Rewrite simple rational expressions in different forms; write a(x)/b(x) in the form q(x) + r(x)/b(x), where a(x), b(x), q(x), and r(x) are polynomials with the degree of r(x) less than the degree of b(x), using inspection, long division, or, for the more complicated examples, a computer algebra system.
math-2017:A.APR.7 Understand that rational expressions form a system analogous to the rational numbers, closed under addition, subtraction, multiplication, and division by a nonzero rational expression; add, subtract, multiply, and divide rational expressions.
Learning goal: Rewrite simple rational expressions in different forms, writing a(x)/b(x) as q(x) + r(x)/b(x) where the degree of r(x) is less than the degree of b(x), by inspection, by long division or, for complicated cases, with a computer algebra system; and understand that rational expressions form a system analogous to the rational numbers, closed under addition, subtraction, multiplication and division by a nonzero rational expression, so that these four operations can be carried out.
Scope note. A.APR.7 is marked (+) in Ohio's Learning Standards for mathematics, which Ohio uses to indicate a standard "for students who are planning on taking advanced courses". It is taught here in full, as extension content for students on that path.
What a rational expression is. A rational expression is a quotient of two polynomials, a(x)/b(x), with b(x) not the zero polynomial. Examples: (x + 5)/(x + 2), 1/x, (2x² + 3x − 1)/(x − 1). A rational expression is undefined at any value of x that makes the denominator zero, so (x + 5)/(x + 2) has no value at x = −2. These excluded values travel with the expression through every rewrite.
The analogy with rational numbers. A rational number is a quotient of integers with nonzero denominator, and an improper fraction such as 23/5 can be rewritten as a mixed form, 4 + 3/5, with the remainder 3 smaller than the divisor 5. A rational expression can be rewritten the same way: a(x)/b(x) = q(x) + r(x)/b(x), where q(x) is the quotient and r(x) is the remainder, and "smaller" means the degree of r(x) is less than the degree of b(x). The two forms are equal wherever the expression is defined. The rewritten form shows things the original hides: for large x the fraction r(x)/b(x) becomes small, so the expression behaves like q(x).
Rewriting by inspection. When the numerator is close to a multiple of the denominator, rewrite it directly. (x + 5)/(x + 2): write the numerator as (x + 2) + 3, so the expression is (x + 2)/(x + 2) + 3/(x + 2) = 1 + 3/(x + 2), valid for x ≠ −2. Here q(x) = 1 and r(x) = 3, of degree 0, less than the degree 1 of the denominator.
Rewriting by long division: a fully worked model. Rewrite (2x² + 3x − 1)/(x − 1). Divide the leading term 2x² by the leading term x to get 2x, the first term of the quotient. Multiply back: 2x(x − 1) = 2x² − 2x. Subtract from the numerator: (2x² + 3x − 1) − (2x² − 2x) = 5x − 1. Divide 5x by x to get 5, the next quotient term. Multiply back: 5(x − 1) = 5x − 5. Subtract: (5x − 1) − (5x − 5) = 4. The remainder 4 has degree 0, less than the degree 1 of x − 1, so the division stops. Therefore (2x² + 3x − 1)/(x − 1) = 2x + 5 + 4/(x − 1) for x ≠ 1. Check by multiplying back: (x − 1)(2x + 5) + 4 = 2x² + 5x − 2x − 5 + 4 = 2x² + 3x − 1. The diagram lists these steps in order. Note the link with the Remainder Theorem: the numerator at x = 1 is 2 + 3 − 1 = 4, the remainder.
A divisor of degree 2. Rewrite (x³ + 1)/(x² + 1). Divide x³ by x² to get x; multiply back, x(x² + 1) = x³ + x; subtract: (x³ + 1) − (x³ + x) = 1 − x. The remainder 1 − x has degree 1, less than the degree 2 of the divisor, so stop: (x³ + 1)/(x² + 1) = x + (1 − x)/(x² + 1). This denominator is never zero for real x, so no value is excluded. For more complicated cases, such as a degree-6 numerator over a degree-3 denominator with awkward coefficients, a computer algebra system performs the same algorithm without arithmetic slips; the method is unchanged, only the bookkeeping is delegated.
Closure: the four operations. Rational numbers are closed under addition, subtraction, multiplication and division by a nonzero rational number: the result is again a quotient of integers. Rational expressions behave the same way, and the reasons are the same. To add or subtract, use a common denominator: 1/x + 2/(x + 1) = (x + 1)/(x(x + 1)) + 2x/(x(x + 1)) = (3x + 1)/(x(x + 1)), for x ≠ 0, −1. The numerator 3x + 1 and the denominator x(x + 1) are polynomials, so the sum is a rational expression. For 3/(x − 2) − 1/(x + 2), the common denominator is (x − 2)(x + 2), giving (3(x + 2) − (x − 2))/((x − 2)(x + 2)) = (3x + 6 − x + 2)/((x − 2)(x + 2)) = (2x + 8)/((x − 2)(x + 2)), for x ≠ ±2. The subtraction sign must apply to the whole of x − 2, which is the step students skip. To multiply, multiply numerators and denominators, then cancel common factors: (x² − 9)/(x + 1) · (x + 1)/(x − 3) = ((x − 3)(x + 3)(x + 1))/((x + 1)(x − 3)) = x + 3, but only for x ≠ −1 and x ≠ 3, because the original expression is undefined at those values even though the simplified one is not. To divide, multiply by the reciprocal of the divisor, which must be a nonzero rational expression: (x/(x + 1)) ÷ (x²/(x² − 1)) = x/(x + 1) · (x − 1)(x + 1)/x² = (x − 1)/x, for x ≠ 0, ±1. In each case, since polynomials are closed under addition, subtraction and multiplication, the numerator and denominator that result are polynomials, and the denominator is nonzero as a polynomial; that is why the system is closed. Division by the zero expression is excluded for the same reason 5 ÷ 0 is excluded among rational numbers.
The misconception to remove. Students cancel terms instead of factors: writing (x + 5)/(x + 2) as 5/2 by "cancelling the x". Only a factor common to the whole numerator and the whole denominator may be cancelled. Test it: at x = 1, (1 + 5)/(1 + 2) = 2, not 5/2. The correct rewrite is 1 + 3/(x + 2). A second error is dropping excluded values after simplifying; the simplified expression is equal to the original only where the original was defined.
Worked rewrite with a quadratic numerator: Write (x² + 4x + 7)/(x + 1) in the form q(x) + r(x)/(x + 1). Long division: x² ÷ x = x; x(x + 1) = x² + x; subtract to get 3x + 7. Then 3x ÷ x = 3; 3(x + 1) = 3x + 3; subtract to get 4. So the expression is x + 3 + 4/(x + 1), for x ≠ −1. Check: (x + 1)(x + 3) + 4 = x² + 4x + 3 + 4 = x² + 4x + 7. The Remainder Theorem agrees: the numerator at x = −1 is 1 − 4 + 7 = 4.
Worked addition: Add (x + 1)/(x − 1) and (x − 1)/(x + 1). The common denominator is (x − 1)(x + 1) = x² − 1. The sum is ((x + 1)² + (x − 1)²)/(x² − 1) = ((x² + 2x + 1) + (x² − 2x + 1))/(x² − 1) = (2x² + 2)/(x² − 1), for x ≠ ±1. The numerator and denominator are polynomials, so the sum is a rational expression, as closure requires. Check at x = 2: 3/1 + 1/3 = 10/3, and (8 + 2)/(4 − 1) = 10/3.
Worked division: Compute ((x² − 4)/(2x)) ÷ ((x + 2)/(4x²)). Multiply by the reciprocal: (x² − 4)/(2x) · (4x²)/(x + 2) = ((x − 2)(x + 2) · 4x²)/(2x(x + 2)). Cancel the factor (x + 2) and reduce 4x²/(2x) to 2x: the result is 2x(x − 2), for x ≠ 0 and x ≠ −2. The divisor (x + 2)/(4x²) is nonzero as an expression, so the division is allowed. Check at x = 3: (5/6) ÷ (5/36) = (5/6)(36/5) = 6, and 2·3·1 = 6.
Worked long division with a negative quotient term: Write (3x² − 2x + 5)/(x + 2) as q(x) + r(x)/(x + 2). Long division: 3x² ÷ x = 3x; 3x(x + 2) = 3x² + 6x; subtract to get −8x + 5; −8x ÷ x = −8; −8(x + 2) = −8x − 16; subtract to get 21. Result: 3x − 8 + 21/(x + 2), x ≠ −2. The remainder 21 equals the numerator at x = −2: 12 + 4 + 5 = 21.
Representation task: Make a two-column comparison headed "Rational numbers" and "Rational expressions". In the first row show 23/5 = 4 + 3/5 beside (2x² + 3x − 1)/(x − 1) = 2x + 5 + 4/(x − 1), and explain in one sentence what "the remainder is smaller than the divisor" means in each column. In the next four rows give one worked example of addition, subtraction, multiplication and division in each column, with the excluded values stated for every expression. In a final row, state the one thing you may not divide by in each system.
Independent performance: Rewrite (x³ − 2x)/(x − 3) and (4x² + 1)/(2x − 1) in the form q(x) + r(x)/b(x) by long division, checking each by multiplying back. Then, with the expressions A = (x + 3)/(x − 1) and B = 2/(x + 1), compute A + B, A − B, A·B and A ÷ B, simplifying each to a single fraction in factored form and stating every excluded value. Success criteria: each division shows the multiply-back-and-subtract steps and the check; each remainder has degree less than its divisor; each of the four results is a quotient of polynomials with excluded values listed; and one sentence explains why the four results had to be rational expressions.
An adult may read the prompts aloud. Answer independently with words, objects, drawings or a demonstration before opening the answer guide. Award 1 point for a correct result or claim and 1 point for a matching explanation or demonstration; where a question asks only for a result (a corrected sentence, a spelling, a count), a fully correct result earns both points, and the answer guide shows what a complete response contains. Record prompting separately. The total (two points per question) is local practice feedback, not an Ohio proficiency level.
Rewrite (x + 7)/(x + 3) in the form q(x) + r(x)/b(x) by inspection, and state the excluded value.
Write the numerator as (x + 3) + 4. Then (x + 7)/(x + 3) = (x + 3)/(x + 3) + 4/(x + 3) = 1 + 4/(x + 3), with q(x) = 1 and r(x) = 4, whose degree 0 is less than the degree 1 of x + 3. The expression is undefined at x = −3.
Use long division to write (x² + x + 1)/(x + 1) as q(x) + r(x)/(x + 1), and check by multiplying back.
x² ÷ x = x; x(x + 1) = x² + x; subtracting leaves 1. The remainder 1 has degree 0 < 1, so the result is x + 1/(x + 1), for x ≠ −1. Check: (x + 1)·x + 1 = x² + x + 1.
Add 2/(x + 3) and 5/(x − 3), and explain why the result is still a rational expression.
Common denominator (x + 3)(x − 3): (2(x − 3) + 5(x + 3))/((x + 3)(x − 3)) = (2x − 6 + 5x + 15)/((x + 3)(x − 3)) = (7x + 9)/((x + 3)(x − 3)), for x ≠ ±3. The numerator 7x + 9 and the denominator (x + 3)(x − 3) are polynomials because polynomials are closed under multiplication and addition, so the sum is a quotient of polynomials, a rational expression.
Simplify (x² − 1)/(x² + 2x + 1) and state the values of x for which the original and the simplified expression are not both defined.
Factor: (x − 1)(x + 1)/((x + 1)(x + 1)). Cancel the common factor (x + 1): (x − 1)/(x + 1). The original is undefined at x = −1 (denominator zero), and so is the simplified form. Only whole factors are cancelled; cancelling the x² terms to get −1/(2x + 1) would be wrong, as x = 2 shows: the original is 3/9 = 1/3, not −1/5.
Compute ((x² − 9)/(x + 1)) · ((x + 1)/(x − 3)) and list every excluded value.
Multiply: ((x − 3)(x + 3)(x + 1))/((x + 1)(x − 3)) = x + 3. Excluded values come from every denominator in the original: x = −1 and x = 3. The simplified expression x + 3 equals the product only for x ≠ −1, 3, because the original is undefined there.
Explain in what sense rational expressions form a system analogous to the rational numbers, and name the one operation that needs a restriction.
Both are quotients: rational numbers are integers over nonzero integers, rational expressions are polynomials over nonzero polynomials. Both are closed under addition, subtraction, multiplication and division, meaning the result is again in the system, because the numerators and denominators produced are integers (or polynomials) and the denominators stay nonzero. The restriction is division: the divisor must be nonzero, since dividing by 0 (or by the zero polynomial) is undefined.
Divide (x/(x + 1)) by (x²/(x² − 1)), simplify, and check with x = 2.
Multiply by the reciprocal: x/(x + 1) · (x² − 1)/x² = x(x − 1)(x + 1)/((x + 1)x²) = (x − 1)/x, for x ≠ 0, 1, −1. Check at x = 2: (2/3) ÷ (4/3) = (2/3)(3/4) = 1/2, and (2 − 1)/2 = 1/2.