Use the factor theorem and trial and error to find a linear factor of these polynomials.
a. [tex]P(x)=x^3+2 x^2+7 x+6[/tex]
b. [tex]P(x)=x^3+2 x^2-x-2[/tex]
c. [tex]P(x)=x^3+x^2+x+6[/tex]
d. [tex]P(x)=x^3-2 x-4[/tex]

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Answered by Anonymous | 2025-07-04

Use the factor theorem, which states that if P ( a ) = 0 , then ( x − a ) is a factor of P ( x ) .
Test integer values (factors of the constant term) to find a root 'a' such that P ( a ) = 0 for each polynomial.
a. For P ( x ) = x 3 + 2 x 2 + 7 x + 6 , P ( − 1 ) = 0 , so ( x + 1 ) is a factor.
b. For P ( x ) = x 3 + 2 x 2 − x − 2 , P ( 1 ) = 0 and P ( − 2 ) = 0 , so ( x − 1 ) and ( x + 2 ) are factors.
c. For P ( x ) = x 3 + x 2 + x + 6 , P ( − 2 ) = 0 , so ( x + 2 ) is a factor.
d. For P ( x ) = x 3 − 2 x − 4 , P ( 2 ) = 0 , so ( x − 2 ) is a factor.
The linear factors are: a. x + 1 ​ , b. x − 1 ​ or x + 2 ​ , c. x + 2 ​ , d. x − 2 ​ .

Explanation

Understanding the Problem and Strategy We are given four cubic polynomials and asked to find a linear factor for each using the factor theorem and trial and error. The factor theorem states that if P ( a ) = 0 , then ( x − a ) is a factor of P ( x ) . We will test integer values (factors of the constant term) to find a root 'a' such that P ( a ) = 0 for each polynomial.

Finding a Linear Factor for Polynomial a a. P ( x ) = x 3 + 2 x 2 + 7 x + 6 . We test factors of 6: ± 1 , ± 2 , ± 3 , ± 6 . Let's try x = − 1 : P ( − 1 ) = ( − 1 ) 3 + 2 ( − 1 ) 2 + 7 ( − 1 ) + 6 = − 1 + 2 − 7 + 6 = 0 . Since P ( − 1 ) = 0 , ( x − ( − 1 )) = ( x + 1 ) is a factor.

Finding a Linear Factor for Polynomial b b. P ( x ) = x 3 + 2 x 2 − x − 2 . We test factors of -2: ± 1 , ± 2 . Let's try x = 1 : P ( 1 ) = ( 1 ) 3 + 2 ( 1 ) 2 − 1 − 2 = 1 + 2 − 1 − 2 = 0 . Since P ( 1 ) = 0 , ( x − 1 ) is a factor. Let's try x = − 2 : P ( − 2 ) = ( − 2 ) 3 + 2 ( − 2 ) 2 − ( − 2 ) − 2 = − 8 + 8 + 2 − 2 = 0 . Since P ( − 2 ) = 0 , ( x − ( − 2 )) = ( x + 2 ) is a factor.

Finding a Linear Factor for Polynomial c c. P ( x ) = x 3 + x 2 + x + 6 . We test factors of 6: ± 1 , ± 2 , ± 3 , ± 6 . Let's try x = − 2 : P ( − 2 ) = ( − 2 ) 3 + ( − 2 ) 2 + ( − 2 ) + 6 = − 8 + 4 − 2 + 6 = 0 . Since P ( − 2 ) = 0 , ( x − ( − 2 )) = ( x + 2 ) is a factor.

Finding a Linear Factor for Polynomial d d. P ( x ) = x 3 − 2 x − 4 . We test factors of -4: ± 1 , ± 2 , ± 4 . Let's try x = 2 : P ( 2 ) = ( 2 ) 3 − 2 ( 2 ) − 4 = 8 − 4 − 4 = 0 . Since P ( 2 ) = 0 , ( x − 2 ) is a factor.

Final Answer Therefore, the linear factors are: a. x + 1 b. x − 1 or x + 2 c. x + 2 d. x − 2


Examples
The factor theorem is useful in various engineering applications, such as control systems design, where engineers need to analyze the stability of a system. The characteristic equation of a system can be represented as a polynomial, and finding its roots (which are the values that make the polynomial equal to zero) helps determine the system's stability. If the roots have negative real parts, the system is stable; otherwise, it's unstable. The factor theorem helps in factoring the characteristic polynomial to find these roots, aiding in the design and analysis of stable control systems. For example, if a system's characteristic equation is s 3 + 5 s 2 + 8 s + 4 = 0 , we can use the factor theorem to find that ( s + 1 ) and ( s + 2 ) are factors, leading to roots that indicate stability.

Answered by GinnyAnswer | 2025-07-04