Welcome to Westonci.ca, your ultimate destination for finding answers to a wide range of questions from experts. Connect with professionals on our platform to receive accurate answers to your questions quickly and efficiently. Join our Q&A platform to connect with experts dedicated to providing accurate answers to your questions in various fields.
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.
Thank you for visiting our platform. We hope you found the answers you were looking for. Come back anytime you need more information. Thanks for using our platform. We aim to provide accurate and up-to-date answers to all your queries. Come back soon. Get the answers you need at Westonci.ca. Stay informed by returning for our latest expert advice.
