Question

the value of a polynomial at $x = 1$ is the remainder when the polynomial is divided by ____.\
$\bigcirc$ $-1$\
$\bigcirc$ $x - 1$\
$\bigcirc$ $1$\
$\bigcirc$ $x + 1$\
\

1. which of these are not potential rational roots of $8x^3 + 15x^2 - 7x - 5$?\

$\square$ $\pm 8$\
$\square$ $\pm \frac{1}{2}$\
$\square$ $\pm \frac{1}{4}$\
$\square$ $\pm 5$\
$\square$ $\pm \frac{5}{8}$

Explanation:

First Question (Multiple Choice)

By the Remainder Theorem, if a polynomial \( f(x) \) is divided by \( x - a \), the remainder is \( f(a) \). Here, we want the remainder when the polynomial is divided by a divisor such that the remainder is \( f(1) \). So, setting \( a = 1 \), the divisor is \( x - 1 \) because when we divide \( f(x) \) by \( x - 1 \), the remainder is \( f(1) \) (from the Remainder Theorem). The other options: dividing by a constant (-1 or 1) is not how the Remainder Theorem works, and \( x + 1=x-(-1) \) would give the remainder \( f(-1) \), not \( f(1) \).

Step 1: Recall the Rational Root Theorem

The Rational Root Theorem states that any potential rational root \( \frac{p}{q} \) of a polynomial \( a_nx^n + \dots + a_1x + a_0 \) (with integer coefficients) has \( p \) as a factor of the constant term \( a_0 \) and \( q \) as a factor of the leading coefficient \( a_n \).

For the polynomial \( 8x^3 + 15x^2 - 7x - 5 \), the leading coefficient \( a_n = 8 \) and the constant term \( a_0=- 5 \).

Step 2: Find possible \( p \) and \( q \)

• Factors of \( a_0=-5 \) (so \( p \)): \( \pm1, \pm5 \)
• Factors of \( a_n = 8 \) (so \( q \)): \( \pm1, \pm2, \pm4, \pm8 \)

Step 3: Determine potential rational roots \( \frac{p}{q} \)

Possible rational roots are \( \frac{\pm1}{\pm1},\frac{\pm1}{\pm2},\frac{\pm1}{\pm4},\frac{\pm1}{\pm8},\frac{\pm5}{\pm1},\frac{\pm5}{\pm2},\frac{\pm5}{\pm4},\frac{\pm5}{\pm8} \), which simplifies to \( \pm1, \pm\frac{1}{2}, \pm\frac{1}{4}, \pm\frac{1}{8}, \pm5, \pm\frac{5}{2}, \pm\frac{5}{4}, \pm\frac{5}{8} \)

Step 4: Identify non - potential roots

Looking at the options:

• \( \pm8 \): For \( \frac{p}{q}=\pm8 \), we would need \( p = \pm8 \) and \( q=\pm1 \), but \( p \) must be a factor of \( - 5 \), and 8 is not a factor of \( - 5 \) (since factors of \( - 5 \) are \( \pm1,\pm5 \)). So \( \pm8 \) are not potential rational roots.
• \( \pm\frac{1}{2} \): \( p=\pm1 \) (factor of - 5), \( q = \pm2 \) (factor of 8), so this is a potential root.
• \( \pm\frac{1}{4} \): \( p=\pm1 \) (factor of - 5), \( q=\pm4 \) (factor of 8), so this is a potential root.
• \( \pm5 \): \( p = \pm5 \) (factor of - 5), \( q=\pm1 \) (factor of 8), so this is a potential root.
• \( \pm\frac{5}{8} \): \( p=\pm5 \) (factor of - 5), \( q=\pm8 \) (factor of 8), so this is a potential root.

So \( \pm8 \) is not a potential rational root because 8 is not a factor of the constant term (-5), so \( \frac{\pm8}{\pm1} \) (since \( q = 1 \) here) has \( p=\pm8 \), but 8 is not a factor of - 5.

Answer:

B. \( x - 1 \)

Second Question (Which are NOT potential rational roots)