Answer :
Sure! Let's solve the polynomial division step by step using the synthetic division method to find [tex]\( P(-2) \)[/tex] where [tex]\( P(x) = x^4 + 3x^3 - 6x^2 + 4 \)[/tex].
1. Set up the coefficients: The polynomial [tex]\( P(x) = x^4 + 3x^3 - 6x^2 + 4 \)[/tex] can be represented by its coefficients, which are: [tex]\( [1, 3, -6, 0, 4] \)[/tex]. (Note that the 0 is for the [tex]\( x \)[/tex] term that is missing).
2. Synthetic division setup: We will use synthetic division with [tex]\( x = -2 \)[/tex].
- Write down the coefficients: [tex]\( [1, 3, -6, 0, 4] \)[/tex]
- Write [tex]\( -2 \)[/tex] on the left side.
3. Carry out the synthetic division:
- Step 1: Bring down the 1 (the first coefficient).
[tex]\[ \begin{array}{r|rrrrr} -2 & 1 & 3 & -6 & 0 & 4 \\ \hline & & & & & \\ & 1 & & & & \\ \end{array} \][/tex]
- Step 2: Multiply -2 by 1 (the number just brought down) and add it to the next coefficient (3).
[tex]\[ -2 \times 1 = -2 \][/tex]
[tex]\[ 3 + (-2) = 1 \][/tex]
[tex]\[ \begin{array}{r|rrrrr} -2 & 1 & 3 & -6 & 0 & 4 \\ \hline & & -2 & & & \\ & 1 & 1 & & & \\ \end{array} \][/tex]
- Step 3: Multiply -2 by 1 (the number just brought down) and add to the next coefficient (-6).
[tex]\[ -2 \times 1 = -2 \][/tex]
[tex]\[ -6 + (-2) = -8 \][/tex]
[tex]\[ \begin{array}{r|rrrrr} -2 & 1 & 3 & -6 & 0 & 4 \\ \hline & & -2 & -2 & & \\ & 1 & 1 & -8 & & \\ \end{array} \][/tex]
- Step 4: Multiply -2 by -8 (the number just brought down) and add to the next coefficient (0).
[tex]\[ -2 \times -8 = 16 \][/tex]
[tex]\[ 0 + 16 = 16 \][/tex]
[tex]\[ \begin{array}{r|rrrrr} -2 & 1 & 3 & -6 & 0 & 4 \\ \hline & & -2 & -2 & 16 & \\ & 1 & 1 & -8 & 16 & \\ \end{array} \][/tex]
- Step 5: Multiply -2 by 16 (the number just brought down) and add to the next coefficient (4).
[tex]\[ -2 \times 16 = -32 \][/tex]
[tex]\[ 4 + (-32) = -28 \][/tex]
[tex]\[ \begin{array}{r|rrrrr} -2 & 1 & 3 & -6 & 0 & 4 \\ \hline & & -2 & -2 & 16 & -32 \\ & 1 & 1 & -8 & 16 & -28 \\ \end{array} \][/tex]
After performing the synthetic division, the quotient is obtained by the coefficients listed across the bottom row except for the last number:
[tex]\[ [1, 1, -8, 16] \][/tex]
The remainder is the last number in the bottom row:
[tex]\[ -28 \][/tex]
Therefore, the quotient is:
[tex]\[ Q(x) = x^3 + x^2 - 8x + 16 \][/tex]
And using the Remainder Theorem, the value of [tex]\( P(-2) \)[/tex] is:
[tex]\[ P(-2) = -28 \][/tex]
So, summarizing:
- Quotient [tex]\( = x^3 + x^2 - 8x + 16 \)[/tex]
- Remainder [tex]\( = -28 \)[/tex]
- [tex]\( P(-2) = -28 \)[/tex]
1. Set up the coefficients: The polynomial [tex]\( P(x) = x^4 + 3x^3 - 6x^2 + 4 \)[/tex] can be represented by its coefficients, which are: [tex]\( [1, 3, -6, 0, 4] \)[/tex]. (Note that the 0 is for the [tex]\( x \)[/tex] term that is missing).
2. Synthetic division setup: We will use synthetic division with [tex]\( x = -2 \)[/tex].
- Write down the coefficients: [tex]\( [1, 3, -6, 0, 4] \)[/tex]
- Write [tex]\( -2 \)[/tex] on the left side.
3. Carry out the synthetic division:
- Step 1: Bring down the 1 (the first coefficient).
[tex]\[ \begin{array}{r|rrrrr} -2 & 1 & 3 & -6 & 0 & 4 \\ \hline & & & & & \\ & 1 & & & & \\ \end{array} \][/tex]
- Step 2: Multiply -2 by 1 (the number just brought down) and add it to the next coefficient (3).
[tex]\[ -2 \times 1 = -2 \][/tex]
[tex]\[ 3 + (-2) = 1 \][/tex]
[tex]\[ \begin{array}{r|rrrrr} -2 & 1 & 3 & -6 & 0 & 4 \\ \hline & & -2 & & & \\ & 1 & 1 & & & \\ \end{array} \][/tex]
- Step 3: Multiply -2 by 1 (the number just brought down) and add to the next coefficient (-6).
[tex]\[ -2 \times 1 = -2 \][/tex]
[tex]\[ -6 + (-2) = -8 \][/tex]
[tex]\[ \begin{array}{r|rrrrr} -2 & 1 & 3 & -6 & 0 & 4 \\ \hline & & -2 & -2 & & \\ & 1 & 1 & -8 & & \\ \end{array} \][/tex]
- Step 4: Multiply -2 by -8 (the number just brought down) and add to the next coefficient (0).
[tex]\[ -2 \times -8 = 16 \][/tex]
[tex]\[ 0 + 16 = 16 \][/tex]
[tex]\[ \begin{array}{r|rrrrr} -2 & 1 & 3 & -6 & 0 & 4 \\ \hline & & -2 & -2 & 16 & \\ & 1 & 1 & -8 & 16 & \\ \end{array} \][/tex]
- Step 5: Multiply -2 by 16 (the number just brought down) and add to the next coefficient (4).
[tex]\[ -2 \times 16 = -32 \][/tex]
[tex]\[ 4 + (-32) = -28 \][/tex]
[tex]\[ \begin{array}{r|rrrrr} -2 & 1 & 3 & -6 & 0 & 4 \\ \hline & & -2 & -2 & 16 & -32 \\ & 1 & 1 & -8 & 16 & -28 \\ \end{array} \][/tex]
After performing the synthetic division, the quotient is obtained by the coefficients listed across the bottom row except for the last number:
[tex]\[ [1, 1, -8, 16] \][/tex]
The remainder is the last number in the bottom row:
[tex]\[ -28 \][/tex]
Therefore, the quotient is:
[tex]\[ Q(x) = x^3 + x^2 - 8x + 16 \][/tex]
And using the Remainder Theorem, the value of [tex]\( P(-2) \)[/tex] is:
[tex]\[ P(-2) = -28 \][/tex]
So, summarizing:
- Quotient [tex]\( = x^3 + x^2 - 8x + 16 \)[/tex]
- Remainder [tex]\( = -28 \)[/tex]
- [tex]\( P(-2) = -28 \)[/tex]
