Question

one factor of $f(x) = 5x^3 + 5x^2 - 170x + 280$ is $(x + 7)$. what are all the roots of the function? use the remainder theorem. \
$\bigcirc$ $x = -4$, $x = -2$, or $x = 7$ \
$\bigcirc$ $x = -7$, $x = 2$, or $x = 4$ \
$\bigcirc$ $x = -7$, $x = 5$, or $x = 280$ \
$\bigcirc$ $x = -280$, $x = -5$, or $x = 7$

Explanation:

Step 1: Recall the Factor Theorem

If \((x + a)\) is a factor of \(f(x)\), then \(f(-a)=0\). Given \((x + 7)\) is a factor, so \(x=-7\) is a root. We can use polynomial division or synthetic division to factor \(f(x)\) further. Let's use synthetic division with root \(-7\) for \(f(x)=5x^{3}+5x^{2}-170x + 280\).

The coefficients are \(5,5,-170,280\).
Bring down the \(5\).
Multiply \(5\times(-7)=-35\), add to next coefficient: \(5+(-35)=-30\).
Multiply \(-30\times(-7)=210\), add to next coefficient: \(-170 + 210 = 40\).
Multiply \(40\times(-7)=-280\), add to last coefficient: \(280+(-280)=0\). So the quotient polynomial is \(5x^{2}-30x + 40\).

Step 2: Factor the Quotient Polynomial

Factor out a \(5\) from \(5x^{2}-30x + 40\): \(5(x^{2}-6x + 8)\).
Factor \(x^{2}-6x + 8\): find two numbers that multiply to \(8\) and add to \(-6\), which are \(-2\) and \(-4\). So \(x^{2}-6x + 8=(x - 2)(x - 4)\).
Thus, \(f(x)=5(x + 7)(x - 2)(x - 4)\).

Step 3: Find the Roots

Set \(f(x)=0\): \(5(x + 7)(x - 2)(x - 4)=0\).
Since \(5
eq0\), we solve \((x + 7)=0\), \((x - 2)=0\), \((x - 4)=0\).
So \(x=-7\), \(x = 2\), \(x = 4\).

Answer:

B. \(x = -7\), \(x = 2\), or \(x = 4\)