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Sagot :
To handle the simplification of the given rational function [tex]\(\frac{4x^2 + 3x + 5}{7x^2 + 2x + 1}\)[/tex], let's proceed with the following steps:
1. Check for Factorization:
- First, check if the numerator [tex]\(4x^2 + 3x + 5\)[/tex] or the denominator [tex]\(7x^2 + 2x + 1\)[/tex] can be factored. Factoring might simplify the rational function.
2. Factor the Polynomials:
- We attempt to factor the numerator polynomial [tex]\(4x^2 + 3x + 5\)[/tex]. To factor it, we look for two numbers that multiply to the product of the leading coefficient (4) and the constant term (5), which is 20, and simultaneously add up to the middle coefficient (3). However, there are no such pairs of integers. Thus, [tex]\(4x^2 + 3x + 5\)[/tex] is not factorable using integer coefficients.
- Next, we attempt to factor the denominator polynomial [tex]\(7x^2 + 2x + 1\)[/tex]. Similarly, we look for two numbers that multiply to the product of the leading coefficient (7) and the constant term (1), which is 7, and add up to the middle coefficient (2). There are no such pairs of integers either. Thus, [tex]\(7x^2 + 2x + 1\)[/tex] is not factorable using integer coefficients.
3. Simplify the Rational Function:
- Since neither the numerator nor the denominator can be factored further, we can't simplify the rational function by canceling common factors.
4. Conclusion:
- Therefore, the rational function [tex]\(\frac{4x^2 + 3x + 5}{7x^2 + 2x + 1}\)[/tex] remains in its original form.
5. Domain Consideration:
- The rational function is undefined where the denominator equals zero. To find these points, solve the quadratic equation [tex]\(7x^2 + 2x + 1 = 0\)[/tex]. Using the quadratic formula [tex]\(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)[/tex], where [tex]\(a = 7\)[/tex], [tex]\(b = 2\)[/tex], and [tex]\(c = 1\)[/tex]:
[tex]\[ x = \frac{-2 \pm \sqrt{2^2 - 4 \cdot 7 \cdot 1}}{2 \cdot 7} = \frac{-2 \pm \sqrt{4 - 28}}{14} = \frac{-2 \pm \sqrt{-24}}{14} = \frac{-2 \pm 2i\sqrt{6}}{14} = \frac{-1 \pm i\sqrt{6}}{7} \][/tex]
- The roots [tex]\(\frac{-1 + i\sqrt{6}}{7}\)[/tex] and [tex]\(\frac{-1 - i\sqrt{6}}{7}\)[/tex] are complex and therefore don't affect the real-valued domain of the function.
In conclusion, the rational function is [tex]\(\frac{4x^2 + 3x + 5}{7x^2 + 2x + 1}\)[/tex] and it can't be simplified further. It is defined for all real [tex]\(x\)[/tex] since the roots of the denominator are complex.
1. Check for Factorization:
- First, check if the numerator [tex]\(4x^2 + 3x + 5\)[/tex] or the denominator [tex]\(7x^2 + 2x + 1\)[/tex] can be factored. Factoring might simplify the rational function.
2. Factor the Polynomials:
- We attempt to factor the numerator polynomial [tex]\(4x^2 + 3x + 5\)[/tex]. To factor it, we look for two numbers that multiply to the product of the leading coefficient (4) and the constant term (5), which is 20, and simultaneously add up to the middle coefficient (3). However, there are no such pairs of integers. Thus, [tex]\(4x^2 + 3x + 5\)[/tex] is not factorable using integer coefficients.
- Next, we attempt to factor the denominator polynomial [tex]\(7x^2 + 2x + 1\)[/tex]. Similarly, we look for two numbers that multiply to the product of the leading coefficient (7) and the constant term (1), which is 7, and add up to the middle coefficient (2). There are no such pairs of integers either. Thus, [tex]\(7x^2 + 2x + 1\)[/tex] is not factorable using integer coefficients.
3. Simplify the Rational Function:
- Since neither the numerator nor the denominator can be factored further, we can't simplify the rational function by canceling common factors.
4. Conclusion:
- Therefore, the rational function [tex]\(\frac{4x^2 + 3x + 5}{7x^2 + 2x + 1}\)[/tex] remains in its original form.
5. Domain Consideration:
- The rational function is undefined where the denominator equals zero. To find these points, solve the quadratic equation [tex]\(7x^2 + 2x + 1 = 0\)[/tex]. Using the quadratic formula [tex]\(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)[/tex], where [tex]\(a = 7\)[/tex], [tex]\(b = 2\)[/tex], and [tex]\(c = 1\)[/tex]:
[tex]\[ x = \frac{-2 \pm \sqrt{2^2 - 4 \cdot 7 \cdot 1}}{2 \cdot 7} = \frac{-2 \pm \sqrt{4 - 28}}{14} = \frac{-2 \pm \sqrt{-24}}{14} = \frac{-2 \pm 2i\sqrt{6}}{14} = \frac{-1 \pm i\sqrt{6}}{7} \][/tex]
- The roots [tex]\(\frac{-1 + i\sqrt{6}}{7}\)[/tex] and [tex]\(\frac{-1 - i\sqrt{6}}{7}\)[/tex] are complex and therefore don't affect the real-valued domain of the function.
In conclusion, the rational function is [tex]\(\frac{4x^2 + 3x + 5}{7x^2 + 2x + 1}\)[/tex] and it can't be simplified further. It is defined for all real [tex]\(x\)[/tex] since the roots of the denominator are complex.
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