We are given that ( x + 1 ) is a factor of p ( x ) = 5 x 4 + 7 x 3 − 2 x 2 − 3 x + c .
We substitute x = − 1 into p ( x ) and set it equal to zero: p ( − 1 ) = 5 ( − 1 ) 4 + 7 ( − 1 ) 3 − 2 ( − 1 ) 2 − 3 ( − 1 ) + c = 0 .
Simplifying the expression, we get 5 − 7 − 2 + 3 + c = 0 , which simplifies to − 1 + c = 0 .
Solving for c , we find that c = 1 , so the final answer is 1 .
Explanation
Understanding the Problem We are given the polynomial p ( x ) = 5 x 4 + 7 x 3 − 2 x 2 − 3 x + c and we need to find the value of c such that ( x + 1 ) is a factor of p ( x ) . This means that when we substitute x = − 1 into the polynomial, the result should be zero, i.e., p ( − 1 ) = 0 .
Substituting x = -1 To find the value of c , we substitute x = − 1 into the polynomial p ( x ) :
p ( − 1 ) = 5 ( − 1 ) 4 + 7 ( − 1 ) 3 − 2 ( − 1 ) 2 − 3 ( − 1 ) + c p ( − 1 ) = 5 ( 1 ) + 7 ( − 1 ) − 2 ( 1 ) − 3 ( − 1 ) + c p ( − 1 ) = 5 − 7 − 2 + 3 + c p ( − 1 ) = − 1 + c
Solving for c Since ( x + 1 ) is a factor of p ( x ) , we must have p ( − 1 ) = 0 . Therefore, − 1 + c = 0 Solving for c , we get: c = 1
Final Answer Thus, the value of c that makes ( x + 1 ) a factor of the polynomial p ( x ) is 1 .
Examples
Polynomial factorization is used in many areas of engineering and physics. For example, in control systems, the characteristic equation of a system is a polynomial, and the roots of this polynomial determine the stability of the system. If we know that ( x + 1 ) is a factor, we can reduce the degree of the polynomial and find the other roots more easily. Also, in signal processing, polynomial factorization is used to design filters and analyze signals.
The value of c so that ( x + 1 ) is a factor of the polynomial p ( x ) is 1 . This is found by substituting x = − 1 into the polynomial and setting the result equal to zero. The calculation simplifies to c = 1 .
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