Simplifying Rational Expressions

Simplifying Rational Expressions Lesson 11-1

Rational Expression • Any fraction that has a variable in the numerator or the denominator. An extended value of a rational expression is when the denominator is undefined or equals zero. In this case, x can not be -6. Therefore, x = -6 is a extended value.

Problem 1 • What is the simplified version of ? State any extended values. Find what the numerator and the denominator have in common. We want to cancel terms. = x – 1 can be cancelled since it appears in both the numerator and the denominator and it is being MULTIPLIED by a number. Solution:

Got it? 1 , a ≠ 0 , d ≠ -2 , n ≠ ; none

Problem 2 • What is the simplified form of ? State any excluded values. = = Look at the second part…what values of x would make the denominator zero? Those are the excluded values. x 2 and x -3

Got it? 2 Simplify and give the excluded values. , x ≠ -2, x ≠ 4 , x ≠ 1 , z ≠ -2, z ≠ - , c ≠ -3, x ≠ -2

Opposite Numerators and Denominators The numerator and denominator of are opposites. To simplify the expression, we can factor out -1 from (3 – x) and can rewrite the denominator as -1(-3 + x) or -1(x – 3). We can now simplify .

Problem 3 Simplify and state any excluded values. = Set aside since that is fully simplified. What’s left is .

Problem 3 can be simplified even further. Because they are opposites, we can factor a -1 in the numerator. (HINT: factor -1 from the part of the fraction that makes the variable negative) = = -1 Now add the first part.

Problem 3 Finding excluded values: Take a look at the original problem . What number can x NOT be? 7x – 14 = 0 7x = 14 x = 2 x can not be 2.

Got it? 3 -1, x ≠ -2.5 -y – 4 , y ≠ 4 - , d ≠ - , d ≠ - , z ≠ - 1, z ≠ 1

Problem 4 A square has a side length of 6x + 2. A rectangle with a width of 3x + 1 has the same area as the square. What is the length of the rectangle? (6x + 2)(6x + 2) = L (3x + 1) L = L = L = 4(3x + 1) L = 12x + 4

Complete #30 as a class.

Multiplying and Dividing Rational Expressions Lesson 11-2

Key Concept x = ÷ =

Problem 1: What is the product? a. ∙ , where a 0

Problem 1: What is the product? b. ∙ where x 0, x -3

Got it? Find the product. • ∙ • ∙ , y ≠ 0 , x ≠ 2, x ≠ 3

Problem 2: Using Factoring What is the product of ∙ ? Factor what you can. ∙ = = 2

Got it? Find the product Find the product of ∙ = 3x(x + 1)

Problem 3: Multiplying by a Polynomial What is the product of and (m2 + m – 6)?

Got it? 3 Multiply these expressions ∙ (6x2 – 13x + 6) (x – 7)(3x – 2)

Dividing Rational Expressions Step 1: Change the division sign to a multiplication sign. Step 2: Take the reciprocal of the second part. Step 3: Multiply and simplify.

Problem 4: Dividing Expressions • ÷ • •

Problem 4: Got it? ÷

Problem 5: Dividing Expressions • ÷ (x2 – 3x – 4) • •

Problem 5: Got it? ÷ (z - 1)

Problem 6: Complex Expressions • ÷ •

Problem 6: Got it?

Dividing Polynomials Lesson 11-3

Problem 1: Dividing by a Monomial What is (9x3 – 6x2 + 15x) ÷ 3x2? = Cross off 3x in numerator and denominator. Distribute the x to each term. - + = 3x – 2 +

Problem 1: Got it? What is (4d3 + 10d2 + 3d) ÷ 2d2? 2d + 5 +

Problem 2: Dividing by a Binomial Compute 22 ÷ 5 using long division. You will have a remainder. Now try, (3d2 – 4d + 13) ÷ (d + 3) on the board.

Problem 2: Got it? (2m2 – m + 3) ÷ (m + 1) Answer: 2m – 3

Problem 3: Dividing with a Zero Coefficient The width w of a rectangle is 3z – 1. The area A of the rectangle is 18z3 – 8z + 2. What is an expression for the length? Step 1: We need to represent all of the degrees of z. 18z3 – 8z + 2 = 18z3+ 0z2 – 8z + 2 Use18z3 + 0z2 – 8z + 2 when dividing. Complete on the board.

Problem 3: Got it? Divide: (h3 – 4h + 12) ÷ (h + 3) Answer: h2 – 3h + 5 -

Problem 4: Reordering Terms before Dividing What is (-10x – 1 + 4x2) ÷ (-3 + 2x) Step 1: We need to reorder them from greatest to smallest degree. (-10x – 1 + 4x2) = (4x2 – 10x – 1) (-3 + 2x) = (2x – 3) Now Divide. Complete on the board.

Problem 4: Got it? Divide: (21a + 2 + 18a2) ÷ (5 + 6a) Answer: 3a + 1 -

Adding and Subtracting Rational Expressions Lesson 11-4

Review of adding and subtracting fractions: Remember? Like Denominators: add the numerator and keep the denominator the same. Difference Denominators: find the least common multiple and change the denominators so they look the same. Then add the numerators.

Problem 1 and 2: Adding and Subtracting with Like Denominators • + + = • - = = = = =

Problem 1 and 2 Got it? Add or subtract. • + = • - =

Problem 3: Adding with Unlike Denominators What is the sum + ? Step 1: Find the lowest common multiple of the denominators. 6x = 2 ∙ 3 ∙ x 2x2 = 2 ∙ x ∙ x LCM = 2 ∙ 3 ∙ x ∙ x = 6x2 Step 2: Rewrite each rational expression using the LCD and add. + = + =

Problem 3: Got it? What is the sum + ? .

Problem 4: Subtracting with Unlike Denominators What is the difference - ? Step 1: Find the LCD. Since there are no common factors the lowest common denominator is (d – 1)(d + 2). Step 2: Rewrite each rational expression using the LCD and add. - = =

Problem 4: Got it? What is the sum - ? .

Solving Rational Equations Lesson 11-5

Rational Equation Rational = fraction Equation = has an equal sign Example: Multiply both sides by the LCD, 9. 6x = 12 x = 2

Problem 1 • What is the solution of ? The LCD is 12x. 5x – 6 = 4 5x = 10 x = 2

Got it? 1 x = -3 x = 5.29 or

Simplifying Rational Expressions