one root of $f(x)=x^{3}+10x^{2}-25x - 250$ is $x=-10$. what are all the roots of the function? use the…

one root of $f(x)=x^{3}+10x^{2}-25x - 250$ is $x=-10$. what are all the roots of the function? use the remainder theorem.\n$x=-25$ or $x = 10$\n$x=-25,x = 1$, or $x = 10$\n$x=-10$ or $x = 5$\n$x=-10,x=-5$, or $x = 5$

one root of $f(x)=x^{3}+10x^{2}-25x - 250$ is $x=-10$. what are all the roots of the function? use the remainder theorem.\n$x=-25$ or $x = 10$\n$x=-25,x = 1$, or $x = 10$\n$x=-10$ or $x = 5$\n$x=-10,x=-5$, or $x = 5$

Answer

Answer:

D. $x = - 10,x=-5$, or $x = 5$

Explanation:

Step1: Use the Remainder Theorem

Since $x=-10$ is a root of $f(x)=x^{3}+10x^{2}-25x - 250$, then $(x + 10)$ is a factor of $f(x)$.

Step2: Perform polynomial long - division

Divide $x^{3}+10x^{2}-25x - 250$ by $x + 10$. [ \begin{align*} \frac{x^{3}+10x^{2}-25x - 250}{x + 10}&=\frac{x^{2}(x + 10)-25(x + 10)}{x + 10}\ &=x^{2}-25 \end{align*} ]

Step3: Factor the resulting quadratic

We know that $x^{2}-25=(x + 5)(x - 5)$ (using the difference - of - squares formula $a^{2}-b^{2}=(a + b)(a - b)$ where $a=x$ and $b = 5$).

Step4: Find all roots

Set $f(x)=(x + 10)(x + 5)(x - 5)=0$. Then $x=-10$ or $x=-5$ or $x = 5$.