
LESSON 5 : RATIONAL EXPRESSIONS PART 1
Rational expressions are quotients whose numerator and denominator are both polynomials [monomials, binomial, trinomial, etc…]
Examples:
Review of Basic Factoring methods:
1. Common Factoring:
Factor 6x^{3} – 15x
Solution: 6x^{3} – 15x = 3x(2x^{2} – 5) **
Find the HIGHEST COMMON FACTOR for each term  “3x”
** Divide “3x” into each term to get the second factor  “2x^{2} – 5”
** Check by expanding
2. Difference of Squares: Formula  a^{2} – b^{2 }= (a – b)(a + b)
Factor 49x^{2} – 64y^{2}
Solution: 49x^{2} – 64y^{2} = (7x – 8y)(7x + 8y)
Factor x^{2} – 9y^{2}
Solution: x^{2} – 9y^{2} = (x – 3y)(x + 3y)
3. Simple Trinomials: Form x^{2} + bx + c [Coefficient of x^{2 }is 1]
a) Factor x^{2} + 5x + 6
Solution: Recall x^{2} + 5x + 6 = (x + __ )(x + __ ) ** We need two numbers that multiply to +6 and add to +5
** Check all pairs of factors of 6: {1, 6} adds to 7
{2, 3} adds to 5
Hence
x^{2} + 5x + 6 = (x + 3 )(x + 2) ** Check by expanding
b) Factor x^{2} – 2x – 35
Solution: Recall x^{2}  2x  35 = (x + __ )(x + __ ) **
We need two numbers that multiply to 35 and add to 2
** Check all pairs of factors of 35: {1,
35} no combination adds to “2”
{5,
7} 7 + 5 = 2
Hence
x^{2} – 2x  35 = (x + 5 )(x – 7) ** Check by expanding
4. Hard Trinomials: Form ax^{2} + bx + c [Coefficient of x^{2 } does not equal 1]
a) Factor 6m^{2} – 5m – 4
Solution:
We use the method of decomposition (although there are other methods)
We decompose the middle term “5m” into two parts using the two clues:
Multiply
to (6)(4) = 24 and
Add to “5”
Strategy: List all
pairs of factors of “24” and find the pair that adds to “5” ** {1, 24} cannot obtain “5” with these two factors;
1+24=25; 124= 23
** {2, 12} cannot
obtain “5” with these two factors
** {3, 8} 8 + 3 = 5 Choose these two factors
Hence 6m^{2}
– 5m – 4 = 6m^{2} – 8m + 3m –4 ** “5m” broken into two parts “8m + 3m”
=
2m(3m – 4) + 1(3m – 4) ** Group by twos and common factor
= (3m – 4)(2m + 1) ** check by expanding
b) Factor 2x^{2} – 9x + 4
Solution:
We decompose the middle term “9x” into two parts using the two clues:
Multiply
to (2)(4) = +8 and
Add to “9”
Strategy: List all
pairs of factors of “+8” and find the pair that adds to “9” ** {2, 4} cannot obtain “9” with these two factors;
2+4=6; 24= 2
** {1, 8} 8 + 1 =
9 Choose these two factors
Hence 2x^{2}
– 9x + 4 = 2x^{2} –8x – 1x + 4 ** “9x” broken into two parts “8x – 1x”
=
2x(x – 4) – 1(x – 4) ** Group by twos and common factor
=
(x – 4)(2x – 1) ** check by expanding
Example 1:
Solutions: