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Consider this polynomial, where [tex]\(a\)[/tex] is an unknown real number:
[tex]\[ p(x) = x^4 + 5x^3 + ax^2 - 3x + 11 \][/tex]

The remainder of the quotient of [tex]\( p(x) \)[/tex] and [tex]\( (x+1) \)[/tex] is 17. Braulio uses synthetic division to find the value of [tex]\(a\)[/tex], and Zahra uses the remainder theorem to find the value of [tex]\(a\)[/tex].

Braulio's Work:
[tex]\[
\begin{array}{l}
a + 14 = 17 \\
a = 3 \\
\end{array}
\][/tex]

Zahra's Work:
[tex]\[
\begin{aligned}
p(-1) &= (-1)^4 + 5(-1)^3 + a(-1)^2 - 3(-1) + 11 \\
&= 1 - 5 + a + 3 + 11 \\
&= a + 10
\end{aligned}
\][/tex]

Given [tex]\(p(-1) = 17\)[/tex]:
[tex]\[
a + 10 = 17 \\
a = 7
\][/tex]

Braulio's Synthetic Division:
[tex]\[
\begin{array}{cccccc}
1 & 5 & a & -3 & 11 \\
\downarrow & & -1 & 4-a & 7-a \\
\hline
1 & 4 & a-1 & 4-a & 18
\end{array}
\][/tex]

Verify using the Remainder Theorem:
[tex]\[
\begin{aligned}
p(-1) &= 1 - 5 + 7 + 3 + 11 \\
&= 17
\end{aligned}
\][/tex]

Who correctly found the value of [tex]\(a\)[/tex]?
- Braulio correctly found the value of [tex]\(a\)[/tex] because he used synthetic division.
- Zahra correctly found the value of [tex]\(a\)[/tex] because she used the remainder theorem.


Sagot :

To find the value of [tex]\(a\)[/tex] in the polynomial [tex]\( p(x) = x^4 + 5x^3 + ax^2 - 3x + 11 \)[/tex] when the remainder of the division by [tex]\((x+1)\)[/tex] is 17, we can use two different methods: synthetic division and the remainder theorem. Let's see how Braulio and Zahra approached this problem.

### Braulio's Method: Synthetic Division

1. Set up synthetic division for the polynomial [tex]\( p(x) \)[/tex] with the divisor [tex]\((x + 1)\)[/tex], which has the root [tex]\(-1\)[/tex].

The coefficients of [tex]\( p(x) \)[/tex] are [tex]\([1, 5, a, -3, 11]\)[/tex].

2. Perform synthetic division using [tex]\(-1\)[/tex]:

[tex]\[ \begin{array}{r|rrrrr} & 1 & 5 & a & -3 & 11 \\ -1 & & & & & \\ \hline & 1 & 4 & a-4 & -7 & 17 \\ \end{array} \][/tex]

Here’s the step-by-step calculation:

[tex]\[ \begin{array}{r|rrrrr} & 1 & 5 & a & -3 & 11 \\ -1 & & -1 & -4 & -(a-4) & -(a-14) \\ \hline & 1 & 4 & a-4 & -7 & a-14 \\ \end{array} \][/tex]

- The first coefficient is 1; it goes directly down.
- Multiply -1 (root) with 1 (first coefficient) and add to 5 (second coefficient): [tex]\(5 - 1 = 4\)[/tex].
- Multiply -1 with 4 (obtained value) and add to [tex]\(a\)[/tex]: [tex]\(a - 4\)[/tex].
- Multiply -1 with [tex]\(a-4\)[/tex] and add to -3: [tex]\(-3 - (a-4) = -3 - a + 4 = 1 - a\)[/tex].
- Multiply -1 with [tex]\(1 - a\)[/tex] and add to 11: [tex]\(11 - (1 - a) = 11 - 1 + a = 10 + a\)[/tex].

The last value is the remainder when divided by [tex]\((x+1)\)[/tex].

We are given that this remainder is 17:

[tex]\[ a - 14 = 17 \implies a = 17 + 14 = 31 \][/tex]

Performing a quick check:

- [tex]\(1 \rightarrow \text{drops down as } 1\)[/tex]
- [tex]\(5 + (-1 \times 1) = 5 - 1 = 4\)[/tex]
- [tex]\(a + (-1 \times 4) = a - 4\)[/tex]
- [tex]\(-3 + (-1 \times (a-4)) = -3 - a + 4 = 1 - a\)[/tex]
- [tex]\(11 + (-1 \times (1 - a)) = 11 - 1 + a = 10 + a\)[/tex]

3. Therefore, Braulio's value for [tex]\(a\)[/tex] is [tex]\(31\)[/tex].

### Zahra's Method: Remainder Theorem

1. According to the remainder theorem, if a polynomial [tex]\( f(x) \)[/tex] is divided by [tex]\( x - c \)[/tex], the remainder is [tex]\( f(c) \)[/tex].

Here, we need to find [tex]\( p(-1) \)[/tex]:

[tex]\[ p(-1) = (-1)^4 + 5(-1)^3 + a(-1)^2 - 3(-1) + 11 \][/tex]

2. Substitute [tex]\(-1\)[/tex] into [tex]\( p(x) \)[/tex]:

[tex]\[ \begin{aligned} p(-1) &= 1 + 5(-1) + a(1) - 3(-1) + 11 \\ &= 1 - 5 + a + 3 + 11 \\ &= 1 - 5 + 3 + 11 + a \\ &= 1 - 5 + 14 + a \\ &= 10 + a \end{aligned} \][/tex]

3. We are given [tex]\( p(-1) = 17 \)[/tex], so set up the equation:

[tex]\[ 10 + a = 17 \implies a = 17 - 10 = 7 \][/tex]

### Conclusion:

- Braulio, using synthetic division, correctly found the value of [tex]\(a\)[/tex] to be [tex]\(31\)[/tex].
- Zahra, using the remainder theorem, correctly found the value of [tex]\(a\)[/tex] to be [tex]\(7\)[/tex].