Factor the polynomial by grouping: 5 x 3 − x 2 − 15 x + 3 = ( 5 x − 1 ) ( x 2 − 3 ) .
Set each factor to zero: 5 x − 1 = 0 and x 2 − 3 = 0 .
Solve for x in each equation: x = 5 1 and x = ± 3 .
The zeros of p ( x ) are 5 1 , 3 , − 3 .
Explanation
Problem Analysis We are given the polynomial p ( x ) = 5 x 3 − x 2 − 15 x + 3 and we want to find its zeros. This means we want to find all values of x such that p ( x ) = 0 . We will attempt to factor the polynomial by grouping.
Grouping and Factoring We group the terms as ( 5 x 3 − x 2 ) + ( − 15 x + 3 ) . Then, we factor out x 2 from the first group and − 3 from the second group to get x 2 ( 5 x − 1 ) − 3 ( 5 x − 1 ) .
Factoring out Common Factor Now we factor out the common factor ( 5 x − 1 ) to obtain ( 5 x − 1 ) ( x 2 − 3 ) .
Setting Polynomial to Zero We set the factored polynomial equal to zero: ( 5 x − 1 ) ( x 2 − 3 ) = 0 .
Solving for x To solve for x , we set each factor equal to zero: 5 x − 1 = 0 and x 2 − 3 = 0 .
Solving the First Factor Solving 5 x − 1 = 0 for x , we get 5 x = 1 , so x = 5 1 .
Solving the Second Factor Solving x 2 − 3 = 0 for x , we get x 2 = 3 , so x = ± 3 .
Final Answer Therefore, the zeros of the polynomial are 5 1 , 3 , − 3 .
Examples
Finding the zeros of a polynomial can help us understand the behavior of systems modeled by polynomials. For example, in engineering, the stability of a system can be determined by finding the roots of a characteristic polynomial. If the real parts of the roots are negative, the system is stable. This also has applications in physics, such as determining the equilibrium points of a system.