Knowing that [tex]x^2+5 x+6[/tex] is a factor of [tex]x^4+r x^2+s[/tex], find the values of coefficients without using long or synthetic division. Describe the steps of your solution using mathematical terminology stating any theorems learned in this unit.
Answer (2)
Factor the quadratic x 2 + 5 x + 6 into ( x + 2 ) ( x + 3 ) .
Apply the factor theorem by substituting x = − 2 and x = − 3 into x 4 + r x 2 + s = 0 , resulting in two equations.
Solve the system of equations 16 + 4 r + s = 0 and 81 + 9 r + s = 0 to find r and s .
The solution is r = − 13 , s = 36 .
Explanation
Problem Analysis We are given that x 2 + 5 x + 6 is a factor of x 4 + r x 2 + s , and we need to find the values of r and s . We are instructed not to use long or synthetic division. The key idea here is to use the factor theorem.
Factoring the Quadratic First, we factor the quadratic x 2 + 5 x + 6 . We are looking for two numbers that multiply to 6 and add to 5. These numbers are 2 and 3. Therefore, x 2 + 5 x + 6 = ( x + 2 ) ( x + 3 ) .
Applying the Factor Theorem Since x 2 + 5 x + 6 is a factor of x 4 + r x 2 + s , it means that ( x + 2 ) and ( x + 3 ) are also factors of x 4 + r x 2 + s . By the factor theorem, if ( x − a ) is a factor of a polynomial P ( x ) , then P ( a ) = 0 . Thus, x = − 2 and x = − 3 are roots of the equation x 4 + r x 2 + s = 0 .
Substituting x = -2 Substituting x = − 2 into x 4 + r x 2 + s = 0 , we get ( − 2 ) 4 + r ( − 2 ) 2 + s = 0 , which simplifies to 16 + 4 r + s = 0 .
Substituting x = -3 Substituting x = − 3 into x 4 + r x 2 + s = 0 , we get ( − 3 ) 4 + r ( − 3 ) 2 + s = 0 , which simplifies to 81 + 9 r + s = 0 .
Solving the System of Equations Now we have a system of two linear equations with two variables:
16 + 4 r + s = 0
81 + 9 r + s = 0
We can solve this system by subtracting the first equation from the second equation to eliminate s : ( 81 + 9 r + s ) − ( 16 + 4 r + s ) = 0 − 0 , which simplifies to 65 + 5 r = 0 .
Finding r Solving for r , we get 5 r = − 65 , so r = − 13 .
Finding s Substituting r = − 13 into the first equation 16 + 4 r + s = 0 , we get 16 + 4 ( − 13 ) + s = 0 , which simplifies to 16 − 52 + s = 0 , so − 36 + s = 0 . Therefore, s = 36 .
Final Answer Thus, the values of the coefficients are r = − 13 and s = 36 .
Examples
Understanding polynomial factorization is crucial in many engineering applications, such as signal processing and control systems. For example, when designing a filter to remove noise from a signal, engineers often need to find the roots of a polynomial. Knowing that a certain quadratic expression is a factor of a higher-degree polynomial simplifies the process of finding these roots, which are essential for designing effective filters. Similarly, in control systems, the stability of a system can be determined by analyzing the roots of a characteristic polynomial. If a factor of this polynomial is known, the analysis becomes much easier, allowing engineers to ensure the system operates safely and efficiently.
To determine the coefficients r and s for the polynomial x 4 + r x 2 + s given that x 2 + 5 x + 6 is a factor, we apply the factor theorem and derive two equations from the roots -2 and -3. Solving the resulting system yields r = − 13 and s = 36 . Thus, the values are r = − 13 and s = 36 .
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