What should be added to \(\frac{1}{{(x - 2)(x - 4)}}\) to get \(\frac{2x - 5}{{(x^2-5x+6)(x - 4)}}\) ?
Finding the Expression to Add to a Rational Fraction
The question asks us to find an expression that, when added to the given rational expression \(\frac{1}{{(x - 2)(x - 4)}}\), results in the expression \(\frac{2x - 5}{{(x^2-5x+6)(x - 4)}}\).
Let the unknown expression be denoted by \(Y\). According to the problem statement, we have the following equation:
\(\frac{1}{{(x - 2)(x - 4)}} + Y = \frac{2x - 5}{{(x^2-5x+6)(x - 4)}}\)
To find the expression \(Y\), we need to subtract the first term from the expression on the right side of the equation:
\(Y = \frac{2x - 5}{{(x^2-5x+6)(x - 4)}} - \frac{1}{{(x - 2)(x - 4)}}\)
Factoring the Denominator
Let's first simplify the denominator \(x^2-5x+6\) in the first term. We look for two numbers that multiply to +6 and add up to -5. These numbers are -2 and -3. So, we can factor the quadratic expression as:
\(x^2 - 5x + 6 = (x - 2)(x - 3)\)
Now, substitute this factored form back into the equation for \(Y\):
\(Y = \frac{2x - 5}{{(x - 2)(x - 3)(x - 4)}} - \frac{1}{{(x - 2)(x - 4)}}\)
Subtracting Rational Expressions
To subtract these two rational expressions, we need a common denominator. The denominators are \({(x - 2)(x - 3)(x - 4)}\) and \({(x - 2)(x - 4)}\). The least common denominator (LCD) is \({(x - 2)(x - 3)(x - 4)}\).
The first fraction already has the LCD. For the second fraction, \(\frac{1}{{(x - 2)(x - 4)}}\), we need to multiply the numerator and the denominator by \((x - 3)\) to get the LCD:
\(\frac{1}{{(x - 2)(x - 4)}} \times \frac{(x - 3)}{(x - 3)} = \frac{1 \cdot (x - 3)}{{(x - 2)(x - 4)(x - 3)}} = \frac{x - 3}{{(x - 2)(x - 3)(x - 4)}}\)
Now, we can perform the subtraction with the common denominator:
\(Y = \frac{2x - 5}{{(x - 2)(x - 3)(x - 4)}} - \frac{x - 3}{{(x - 2)(x - 3)(x - 4)}}\)
\(Y = \frac{(2x - 5) - (x - 3)}{{(x - 2)(x - 3)(x - 4)}}\)
Simplifying the Numerator
Next, simplify the numerator:
\((2x - 5) - (x - 3) = 2x - 5 - x + 3\)
\(= (2x - x) + (-5 + 3)\)
\(= x - 2\)
So, the expression for \(Y\) becomes:
\(Y = \frac{x - 2}{{(x - 2)(x - 3)(x - 4)}}\)
Final Simplification
Assuming \(x \neq 2\), we can cancel the common factor \((x - 2)\) from the numerator and the denominator:
\(Y = \frac{^{\cancel{(x - 2)}}1}{_{\cancel{(x - 2)}}(x - 3)(x - 4)}\)
\(Y = \frac{1}{{(x - 3)(x - 4)}}\)
Expanding the Denominator
Finally, let's expand the denominator \({(x - 3)(x - 4)}\):
\((x - 3)(x - 4) = x \cdot x - 4x - 3x + (-3)(-4)\)
\(= x^2 - 7x + 12\)
So, the expression that should be added is \(\frac{1}{x^2 - 7x + 12}\).
Comparing with Options
Let's compare our result with the given options:
- Option 1: \(\frac{1}{{({x^2} - 7x + 12)}}\)
- Option 2: \(\frac{1}{{({x^2} + 7x + 12)}}\)
- Option 3: \(\frac{1}{{({x^2} - 7x - 12)}}\)
- Option 4: \(\frac{1}{{({x^2} + 7x - 12)}}\)
Our calculated expression \(\frac{1}{{x^2 - 7x + 12}}\) matches Option 1.
| Step | Description | Expression / Result |
|---|---|---|
| 1 | Set up the equation | \(\frac{1}{{(x - 2)(x - 4)}} + Y = \frac{2x - 5}{{(x^2-5x+6)(x - 4)}}\) |
| 2 | Isolate Y | \(Y = \frac{2x - 5}{{(x^2-5x+6)(x - 4)}} - \frac{1}{{(x - 2)(x - 4)}}\) |
| 3 | Factor denominator \(x^2-5x+6\) | \((x - 2)(x - 3)\) |
| 4 | Rewrite Y with factored denominator | \(Y = \frac{2x - 5}{{(x - 2)(x - 3)(x - 4)}} - \frac{1}{{(x - 2)(x - 4)}}\) |
| 5 | Find Common Denominator (LCD) | \({(x - 2)(x - 3)(x - 4)}\) |
| 6 | Rewrite 2nd term with LCD | \(\frac{x - 3}{{(x - 2)(x - 3)(x - 4)}}\) |
| 7 | Subtract fractions | \(Y = \frac{(2x - 5) - (x - 3)}{{(x - 2)(x - 3)(x - 4)}}\) |
| 8 | Simplify numerator | \(x - 2\) |
| 9 | Result after numerator simplification | \(Y = \frac{x - 2}{{(x - 2)(x - 3)(x - 4)}}\) |
| 10 | Cancel common factor \((x-2)\) | \(Y = \frac{1}{{(x - 3)(x - 4)}}\) |
| 11 | Expand denominator \((x - 3)(x - 4)\) | \(x^2 - 7x + 12\) |
| 12 | Final Expression for Y | \(\frac{1}{{x^2 - 7x + 12}}\) |
Revision Table: Key Concepts
| Concept | Description |
|---|---|
| Subtracting Rational Expressions | To subtract fractions with different denominators, find the least common denominator (LCD), rewrite each fraction with the LCD, and then subtract the numerators, keeping the common denominator. |
| Factoring Quadratics | Factoring a quadratic expression like \(ax^2 + bx + c\) involves finding two binomials \((px + q)(rx + s)\) that multiply to give the original expression. For \(x^2 + bx + c\) with a leading coefficient of 1, find two numbers that multiply to \(c\) and add to \(b\). |
| Simplifying Rational Expressions | Simplify a rational expression by canceling common factors that appear in both the numerator and the denominator. Note that this cancellation is valid only where the factors are non-zero. |
Additional Information: Operations with Rational Expressions
Rational expressions are fractions where the numerator and denominator are polynomials. Performing arithmetic operations (addition, subtraction, multiplication, division) with rational expressions is similar to working with numerical fractions.
- Addition/Subtraction: Requires a common denominator. Find the LCD, rewrite expressions, then add or subtract numerators.
- Multiplication: Multiply the numerators together and multiply the denominators together. Simplify the resulting expression by canceling common factors before or after multiplying.
- Division: To divide by a rational expression, multiply by its reciprocal.
Factoring polynomials is a crucial skill when working with rational expressions, as it helps in finding common denominators and simplifying expressions by identifying common factors.
Always remember to consider the values of variables for which the denominators are zero, as these values are excluded from the domain of the rational expression.
Similar Questions
If (x 2- 1) is a factor of ax 4+ bx 3+ cx 2+ dx + e, then which one of the following is correct?
If α, β and γ are the zeros of the polynomial f(x) = ax 3+ bx 2+ cx + d, then α 2+ β 2+ γ 2is equal to
Let f(x) and g(x) be two polynomials (with real coefficients) having degree 3 and 4 respectively. What is the degree of f(x) g(x)
If α and β are the two zeros of the polynomial 25x 2– 15x + 2, then what is a quadratic polynomial whose zeros are (2α) -1 and (2β) -1 ?
The expression \(\frac{{\left( {{x^3} - 1} \right)\left( {{x^2} - 9x + 14} \right)}}{{\left( {{x^2} + x + 1} \right)\left( {{x^2} - 8x + 7} \right)}}\) simplifies to
The HCF and the LCM of two polynomials are 3x + 1 and 30x 3 + 7x 2 - 10x - 3 respectively. If one polynomial is 6x 2 + 5x + 1, then what is the other polynomial?
If (6x + 4y) / (6x - 4y) = 8/6 then what is the value of x 2 / y 2?
What is the sum of the linear factors (in x and y) of the expression
2x 2 + xy - 3y 2 ?
What is (x - a) (x - b) (x - c) equal to?
For what value of k can the expression x 3+ kx 2– 7x + 6 be resolved into three linear factors?
Important Questions from Polynomials
If y 2= y + 7, then what is the value of y 3?
Factorize x 2- y 2- 9z 2+ 6yz
If one of the zeros of the polynomial x 3+ ax 2+ bx + c is - 1, then the product of other two zeros is equal to :
If a(a + b + c) 2 = 1792; b(a + b + c) 2 = 1536; c(a + b + c) 2 = 768, then what will be the value of b?
If x = 3 so, what is the value of x 2 + 2x + 5 ?
