One factor of [tex]f(x)=5 x^3+5 x^2-170 x+280[/tex] is [tex](x+7)[/tex]. What are all the roots of the function? Use the Remainder Theorem.
A. [tex]x=-4, x=-2[/tex], or [tex]x=7[/tex]
B. [tex]x=-7, x=2[/tex], or [tex]x=4[/tex]
C. [tex]x=-7, x=5[/tex], or [tex]x=280[/tex]
D. [tex]x=-280, x=-5[/tex], or [tex]x=7[/tex]
Answer (1)
Divide the given cubic function f ( x ) by the factor ( x + 7 ) to obtain the quadratic factor 5 x 2 − 30 x + 40 .
Simplify the quadratic factor by dividing by 5, resulting in x 2 − 6 x + 8 = 0 .
Factor the quadratic equation as ( x − 2 ) ( x − 4 ) = 0 , which gives the roots x = 2 and x = 4 .
Combine the root from the given factor ( x + 7 ) , which is x = − 7 , with the roots from the quadratic factor to get the complete set of roots: x = − 7 , x = 2 , x = 4 .
Explanation
Understanding the Problem We are given the cubic function f ( x ) = 5 x 3 + 5 x 2 − 170 x + 280 and told that ( x + 7 ) is a factor. Our goal is to find all the roots of f ( x ) , meaning all values of x for which f ( x ) = 0 . We will use the fact that ( x + 7 ) is a factor to simplify the problem.
Using the Factor Theorem Since ( x + 7 ) is a factor of f ( x ) , we can divide f ( x ) by ( x + 7 ) to find the remaining quadratic factor. This is based on the Factor Theorem, which states that if ( x − a ) is a factor of f ( x ) , then f ( a ) = 0 .
Polynomial Division We perform polynomial division of f ( x ) by ( x + 7 ) .
\multicolumn 2 r 5 x 2 − 30 x + 40 \cline 2 − 5 x + 7 5 x 3 + 5 x 2 − 170 x + 280 \multicolumn 2 r 5 x 3 + 35 x 2 \cline 2 − 3 \multicolumn 2 r 0 − 30 x 2 − 170 x \multicolumn 2 r − 30 x 2 − 210 x \cline 3 − 4 \multicolumn 2 r 0 40 x + 280 \multicolumn 2 r 40 x + 280 \cline 4 − 5 \multicolumn 2 r 0 0
The quotient is 5 x 2 − 30 x + 40 .
Finding the Quadratic Factor Now we have f ( x ) = ( x + 7 ) ( 5 x 2 − 30 x + 40 ) . To find the remaining roots, we set the quadratic factor equal to zero and solve for x : 5 x 2 − 30 x + 40 = 0 . We can simplify this equation by dividing by 5: x 2 − 6 x + 8 = 0 .
Solving the Quadratic Equation We can solve the quadratic equation x 2 − 6 x + 8 = 0 by factoring. We look for two numbers that multiply to 8 and add to -6. These numbers are -2 and -4. So, we can factor the quadratic as ( x − 2 ) ( x − 4 ) = 0 .
Finding the Roots Setting each factor to zero, we get x − 2 = 0 or x − 4 = 0 . Thus, the roots are x = 2 and x = 4 .
Final Answer Therefore, the roots of the function f ( x ) = 5 x 3 + 5 x 2 − 170 x + 280 are x = − 7 , x = 2 , and x = 4 .
Examples
Understanding the roots of a polynomial can help engineers design stable structures. For example, if f ( x ) represents the stress on a bridge as a function of temperature x , finding the roots tells us at what temperatures the stress is zero, which is crucial for safety. Similarly, in electrical engineering, roots of characteristic equations determine the stability of circuits. Knowing the roots helps in predicting and preventing failures in real-world applications.
