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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].
### 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].
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