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We begin by first looking for rational zeros. We can apply the rational zero theorem because the polynomial has integer coefficients.

[tex]\[ n(x) = 3x^3 - x^2 - 39x + 13 \][/tex]

Possible rational zeros:

[tex]\[
\begin{aligned}
\frac{\text{Factors of } 13}{\text{Factors of } 3} & = \frac{\pm 1, \pm 13}{\pm 1, \pm 3} \\
& = \pm 1, \pm \frac{1}{3}, \pm 13, \pm \frac{13}{3}
\end{aligned}
\][/tex]

Part 1 of 6

Next, use synthetic division and the remainder theorem to determine if any of the numbers in the list is a zero of [tex]\( n \)[/tex].

Test [tex]\( x = 1 \)[/tex]:

The remainder is [tex]\(\square\)[/tex].

Sagot :

GinnyAnsweridn
To determine if [tex]\( x = 1 \)[/tex] is a zero of the polynomial [tex]\( n(x)=3x^3-x^2-39x+13 \)[/tex], we need to use the Remainder Theorem, which states that if a polynomial [tex]\( p(x) \)[/tex] is divided by [tex]\( x-a \)[/tex], the remainder of this division is [tex]\( p(a) \)[/tex]. Thus, we evaluate [tex]\( n(1) \)[/tex].

We use synthetic division for this:

1. Begin with the coefficients of the polynomial: [tex]\( 3, -1, -39, 13 \)[/tex].

[tex]\[ \begin{array}{r|rrrr} 1 & 3 & -1 & -39 & 13 \\ \hline & 3 & 2 & -37 & -24 \\ \end{array} \][/tex]

- Bring down the leading coefficient (3):
- [tex]\( 3 \)[/tex]

- Multiply 3 by 1 (the value being tested):
- [tex]\( 3 \cdot 1 = 3 \)[/tex]

- Add this product to the next coefficient:
- [tex]\( -1 + 3 = 2 \)[/tex]

- Multiply 2 by 1:
- [tex]\( 2 \cdot 1 = 2 \)[/tex]

- Add this product to the next coefficient:
- [tex]\( -39 + 2 = -37 \)[/tex]

- Multiply -37 by 1:
- [tex]\( -37 \cdot 1 = -37 \)[/tex]

- Add this product to the next coefficient:
- [tex]\( 13 + (-37) = -24 \)[/tex]

The remainder when [tex]\( x = 1 \)[/tex] is tested is:

[tex]\[ \boxed{-24} \][/tex]

Since the remainder is not zero, [tex]\( x = 1 \)[/tex] is not a zero of the polynomial [tex]\( n(x) \)[/tex].
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