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Operations with rational expressions
Expressions with integers (polynomials)
NOTABLE PRODUCTS:
\latex{(a + b)^2 = a^2 + 2ab + b^2}.
\latex{(a - b)^2 = a^2 - 2ab + b^2}.
\latex{(a + b + c)^2 = a^2 + b^2 + c^2 + 2ab + 2ac + 2bc}.
\latex{(a + b)^3 = a^3 + 3a^2b + 3ab^2 + b^3}.
\latex{(a - b)^3 = a^3 - 3a^2b + 3ab^2 - b^3}.
\latex{(a + b)\times(a - b) = a^2 - b^2}.
\latex{(a - b)^2 = a^2 - 2ab + b^2}.
\latex{(a + b + c)^2 = a^2 + b^2 + c^2 + 2ab + 2ac + 2bc}.
\latex{(a + b)^3 = a^3 + 3a^2b + 3ab^2 + b^3}.
\latex{(a - b)^3 = a^3 - 3a^2b + 3ab^2 - b^3}.
\latex{(a + b)\times(a - b) = a^2 - b^2}.
METHODS OF FACTORIZATION
- factoring out;
- factoring out using grouping;
- using notable identities;
- using the factored form of a quadratic equation.
Example 1
Expand the parentheses and group the terms:
- \latex{(2a - 3)\times(4a + 1) - (3a - 5)};
- \latex{(4b + 3)\times(4b - 3) - (3b - 2)\times(3b + 2)}.
Solution (a)
After expanding the parentheses and grouping the terms:
\latex{(2a - 3)\times(4a + 1) - (3a - 5) = 8a^2 + 2a - 12a - 3 - 3a + 5 = \\ = 8a^2 - 13a + 2.}
Solution (b)
Apply the identity for the product of sum and difference:
\latex{(4b + 3)\times(4b - 3) - (3b - 2)\times(3b + 2) = \\ = 16b^2 - 9 - 9b^2 + 4 = 7b^2 - 5.}
Example 2
Compute the following and order the terms of the resulting polynomials in a decreasing order of powers of \latex{x}:
- \latex{(3x + 2)^2 - (2x - 3)^2};
- \latex{(2x - 1)^3 - (x + 4)^3};
- \latex{(2x + y - 3)^2}.
Solution
- \latex{(3x + 2)^2 - (2x - 3)^2 = 9x^2 + 12x + 4 - (4x^2 - 12x + 9) = \\ = 9x^2 + 12x + 4 - 4x^2 + 12x - 9 = 5x^2 + 24x - 5;}
- \latex{(2x - 1)^3 - (x + 4)^3 = 8x^3 - 12x^2 + 6x - 1 - (x^3 + 12x^2 + 48x + 64) = \\= 8x^3 - 12x^2 + 6x - 1 - x^3 - 12x^2 - 48x - 64 = 7x^3 - 24x^2-42x - 65;}
- \latex{(2x + y - 3)^2 = 4x^2 + y^2 + 9 + 4xy - 12x - 6y = \\= 4x^2 + 4xy - 12x + y^2 - 6y + 9.}
Example 3
Which two-term expression’s square is it:
- \latex{25x^2 + 10x + 1};
- \latex{x^6 - 14x^3 + 49};
- \latex{x4 - 8x^2y + 16y^2}?
Solution
- \latex{25x^2 + 10x + 1 = (5x + 1)^2 = (- 5x - 1)^2};
- \latex{x^6 - 14x^3 + 49 = (x^3 - 7)^2 = (7 - x^3)^2};
- \latex{x^4 - 8x^2y + 16y^2 = (x^2 - 4y)^2 = (4y - x^2)^2}.
Example 4
Factorize the following expressions:
- \latex{6x^2 - 4x^3 + 8x^2};
- \latex{2x^2 + 4x + xy + 2y};
- \latex{9x^2 - 25y^2};
- \latex{3x^2 - 12x + 12};
- \latex{x^2 + 2x - 15}.
Solution (a)
Factor out \latex{2x^2} from every term:
\latex{6x^4 - 4x^3 + 8x^2 = 2x^2\times(3x^2 - 2x + 4)}.
The discriminant of the quadratic equation \latex{3x^2 - 2x + 4 = 0} is \latex{D = 4 - 48 = - 44}, thus the equation does not have a solution, therefore the quadratic polynomial \latex{3x^2 - 2x + 4} cannot be factorized.
Solution (b)
Factor out \latex{2x} from the first two terms and \latex{ y } from the last two, and after a second factoring:
\latex{2x^2 + 4x + xy + 2y = 2x \times(x + 2) +y \times(x + 2) =\\ = (x + 2)\times(2x + y).}
Solution (c)
Notice the notable product:
\latex{9x^2 - 25y^2 = (3x + 5y)\times(3x - 5y).}
Solution (d)
After factoring \latex{3} out the parentheses contain the square of a two-term expression:
\latex{3x^2 - 12x + 12x + 12 = 3\times(x^2 - 4x + 4) = 3\times(x - 2)^2.}
Solution (e)
Method I
Complete the square and apply some notable product identities:
\latex{x^2 + 2x - 15 = x^2 + 2x + 1 - 1 - 15 = (x + 1)^2 - 16 = \\= (x + 1)^2 - 4^2 = (x + 1 + 4)\times(x + 1 - 4) = (x + 5)\times(x - 3).}
Method II
Use the factored form of a quadratic equation.
Solve the equation \latex{x^2 + 2x - 15 = 0}:
Solve the equation \latex{x^2 + 2x - 15 = 0}:
\latex{x_{1,2}= \frac{- 2 \pm \sqrt {4 + 60}}{2} = \begin{cases} x_1 = 3 \\ x_2 = - 5. \end{cases}}
According to the factored form,
\latex{x^2 + 2x - 15 = (x - 3)\times(x + 5).}
Algebraic fractions
Example 5
Simplify the following fractions:
- \latex{\frac{a^2+4a+4}{3a+6}};
- \latex{\frac{2b^3 - 12b^2 + 18b}{b^3 - 9b}}.
Solution (a)
The first step is to find the domain: \latex{3a + 6 \neq 0}, that is, \latex{a\neq - 2}.
Factorize both the numerator and the denominator, and simplify with the common factors:
Factorize both the numerator and the denominator, and simplify with the common factors:
\latex{\frac{a^2+4a+4}{3a+6}=\frac{(a+2)^2}{3\times(a+2)}=\frac{a+2}{3}.}
Solution (b)
The denominator cannot be zero: \latex{b^3 - 9b\neq 0}.
Using factorization:
Using factorization:
\latex{b^3 - 9b = b\times(b^2 - 9) = b\times(b + 3)\times(b - 3)}.
Thus \latex{b\neq 0}, \latex{b\neq-3}, \latex{b\neq 3}.
Factoring out helps us factorize the numerator as well:
\latex{\frac{2b^3-12b^2\times318b}{b^3-9b}=\frac{2b\times(b^2-6b+9)}{b\times(b-3)\times(b+3)}=\\=\frac{2b\times(b - 3)^2}{b\times(b-3)\times(b+3)}=\frac{2\times(b-3)}{b+3}.}
Example 6
Simplify the following expression as much as possible:
\latex{\frac{2a+1}{a-3}-\frac{a-1}{a+3}-\frac{11a+9}{a^2-9}}.
Solution
The fractions are valid if their denominators are not equal to zero: \latex{a \neq3} and \latex{a \neq - 3}.
Since \latex{a^2 - 9=(a - 3)\times(a + 3)}, this product can be the common denominator:
Since \latex{a^2 - 9=(a - 3)\times(a + 3)}, this product can be the common denominator:
\latex{\frac{2a+1}{a-3}-\frac{a-1}{a+3}-\frac{11a+9}{a^2-9}=\\=\frac{(2a+1)\times(a+3)}{(a-3)\times(a+3)}-\frac{(a-1)\times(a-3)}{(a-3)\times(a+3)}-\frac{11a+9}{(a-3)\times(a+3)}=\\=\frac{2a^2+6a+a+3-a^2+3a-3-11a-9}{(a-3)\times(a+3)}=\frac{a^2-9}{(a-3)\times(a+3)}=1}.
Example 7
Compute the expression for \latex{\frac{3b-1}{b^2-b}-\frac{4}{b+1}+\frac{2b-6}{b^2-1}} for \latex{b=\frac{1}{23}.}
Solution
Factorize the denominators:
\latex{b^2 - b = b\times(b - 1)};
\latex{b^2 - 1 = (b - 1)\times(b + 1)}.
\latex{b^2 - 1 = (b - 1)\times(b + 1)}.
It can be seen using the factored forms that the fractions are valid if \latex{b \neq 0}; \latex{b \neq 1}; \latex{b \neq - 1}. The value \latex{b} given can be substituted, but first simplify the expression.
The common denominator is \latex{b\times(b - 1)\times(b + 1)}.
\latex{\frac{3b-1}{b^2-b}-\frac{4}{b+1}+\frac{2b-6}{b^2-1}=\\=\frac{(3b-1)\times(b+1)}{b\times(b-1)\times(b+1)}-\frac{4\times b\times(b-1)}{b\times(b-1)\times(b+1)}+\frac{(2b-6)}{b\times(b-1)\times(b+1)}=\\=\frac{3b^2+3b-b-1-4b^2+4b+2b^2-6b}{b\times(b-1)\times(b+1)}=\\=\frac{b^2-1}{b\times(b-1)\times(b+1)}=\frac{1}{b}.}
If \latex{b=\frac{1}{23}}, then substitution gives the value \latex{23}.
Exercises
{{exercise_number}}. Factorize the following expressions:
- \latex{8a^2 - 6a};
- \latex{25b^4 - b^2};
- \latex{28c^2 + 84c + 63}.
{{exercise_number}}. Let \latex{a, b, c, d} be four consecutive natural numbers in an increasing order. Prove that \latex{a + b^2 + c^3} is divisible by \latex{d^2}.
{{exercise_number}}. Compute the following expressions for \latex{x = 2004}:
- \latex{\frac{2x^2-16x+32}{4x-16}};
- \latex{\frac{x^2+6x+5}{x^2+10x+25}+{\frac{x+9}{x+5}}}.
{{exercise_number}}. Simplify the following expressions:
- \latex{\frac{2x - 1}{2x-6}+\frac{x - 1}{x+3}-\frac{2x^2+1}{x^2-9}};
- \latex{\frac{3}{b^2+b}-\frac{5}{b^2-b}+\frac{8}{b^3-b}};
- \latex{\left\lgroup\frac{8}{x^2-4}-\frac{2}{3x-6}+\frac{2}{x+2}\right\rgroup \times \frac{x-2}{x-4}}.
