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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]\[
\text{Factors of } 13: \pm 1, \pm 13
\][/tex]
[tex]\[
\text{Factors of } 3: \pm 1, \pm 3
\][/tex]
[tex]\[
\text{Possible rational zeros:} \pm 1, \pm \frac{1}{3}, \pm 13, \pm \frac{13}{3}
\][/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]\( n(x) \)[/tex].

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

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

Is [tex]\( \square \)[/tex] a zero of [tex]\( n(x) \)[/tex]?


Sagot :

Sure, let's go through the solution to see if [tex]\( x = 12 \)[/tex] is a zero of the polynomial [tex]\( n(x) = 3x^3 - x^2 - 39x + 13 \)[/tex] by using synthetic division and the remainder theorem.

### Synthetic Division Steps:

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

Given [tex]\( x = 12 \)[/tex]:

1. Write down the coefficients:
The coefficients of [tex]\( n(x) \)[/tex] are [tex]\( [3, -1, -39, 13] \)[/tex].

2. Perform synthetic division:
- Start with the leading coefficient [tex]\( 3 \)[/tex].
- Multiply by [tex]\( x = 12 \)[/tex] and add the result to the next coefficient.

Let's perform the steps:

1. Bring down the first coefficient (3):
[tex]\[ 3 \][/tex]

2. Multiply by [tex]\( 12 \)[/tex] and add to the second coefficient [tex]\( -1 \)[/tex]:
[tex]\[ 3 \cdot 12 = 36 \][/tex]
[tex]\[ 36 + (-1) = 35 \][/tex]

3. Multiply the result by [tex]\( 12 \)[/tex] and add to the third coefficient [tex]\( -39 \)[/tex]:
[tex]\[ 35 \cdot 12 = 420 \][/tex]
[tex]\[ 420 + (-39) = 381 \][/tex]

4. Multiply the result by [tex]\( 12 \)[/tex] and add to the constant term [tex]\( 13 \)[/tex]:
[tex]\[ 381 \cdot 12 = 4572 \][/tex]
[tex]\[ 4572 + 13 = 4585 \][/tex]

### Remainder Result:
Based on the synthetic division, the remainder is [tex]\( 4585 \)[/tex].

The Remainder Theorem states that if you substitute a number [tex]\( x = c \)[/tex] into a polynomial [tex]\( n(x) \)[/tex] and the result is 0, then [tex]\( x = c \)[/tex] is a zero of the polynomial.

In this case, substituting [tex]\( x = 12 \)[/tex] into the polynomial yielded the remainder [tex]\( 4585 \)[/tex], which is not zero.

### Conclusion:
- The remainder is [tex]\( 4585 \)[/tex].
- Since the remainder is not zero, [tex]\( x = 12 \)[/tex] is not a zero of the polynomial [tex]\( n(x) \)[/tex].