Answer :
To find [tex]\( p(-3) \)[/tex] using synthetic substitution for the polynomial [tex]\( p(x) = x^3 + 6x^2 + 7x - 25 \)[/tex], we will follow these steps:
1. Identify the coefficients of the polynomial:
The polynomial is [tex]\( p(x) = x^3 + 6x^2 + 7x - 25 \)[/tex].
The coefficients are: 1 (for [tex]\( x^3 \)[/tex]), 6 (for [tex]\( x^2 \)[/tex]), 7 (for [tex]\( x \)[/tex]), and -25 (constant term).
2. Set up the synthetic division:
We are evaluating [tex]\( p(x) \)[/tex] at [tex]\( x = -3 \)[/tex]. Arrange the coefficients in a row:
[tex]\[ \begin{array}{r|rrrr} -3 & 1 & 6 & 7 & -25 \\ \end{array} \][/tex]
3. Perform the synthetic substitution step-by-step:
- Step 1: Bring down the first coefficient (1) to the bottom row:
[tex]\[ \begin{array}{r|rrrr} -3 & 1 & 6 & 7 & -25 \\ & & & & \\ & 1 & & & \\ \end{array} \][/tex]
- Step 2: Multiply the first coefficient by [tex]\( -3 \)[/tex] and write the result under the second coefficient:
[tex]\[ 1 \times (-3) = -3 \][/tex]
[tex]\[ \begin{array}{r|rrrr} -3 & 1 & 6 & 7 & -25 \\ & & -3& & \\ & 1 & & & \\ \end{array} \][/tex]
- Step 3: Add the second coefficient (6) and the result from the multiplication (-3):
[tex]\[ 6 + (-3) = 3 \][/tex]
[tex]\[ \begin{array}{r|rrrr} -3 & 1 & 6 & 7 & -25 \\ & & -3& & \\ & 1 & 3 & & \\ \end{array} \][/tex]
- Step 4: Multiply the result (3) by [tex]\( -3 \)[/tex] and write it under the third coefficient:
[tex]\[ 3 \times (-3) = -9 \][/tex]
[tex]\[ \begin{array}{r|rrrr} -3 & 1 & 6 & 7 & -25 \\ & & -3& -9& \\ & 1 & 3 & & \\ \end{array} \][/tex]
- Step 5: Add the third coefficient (7) and the result from the multiplication (-9):
[tex]\[ 7 + (-9) = -2 \][/tex]
[tex]\[ \begin{array}{r|rrrr} -3 & 1 & 6 & 7 & -25 \\ & & -3& -9& \\ & 1 & 3 & -2& \\ \end{array} \][/tex]
- Step 6: Multiply the result (-2) by [tex]\( -3 \)[/tex] and write it under the fourth coefficient:
[tex]\[ -2 \times (-3) = 6 \][/tex]
[tex]\[ \begin{array}{r|rrrr} -3 & 1 & 6 & 7 & -25 \\ & & -3& -9& 6 \\ & 1 & 3 & -2 & \\ \end{array} \][/tex]
- Step 7: Add the fourth coefficient (-25) and the result from the multiplication (6):
[tex]\[ -25 + 6 = -19 \][/tex]
[tex]\[ \begin{array}{r|rrrr} -3 & 1 & 6 & 7 & -25 \\ & & -3& -9& 6 \\ & 1 & 3 & -2 & -19 \\ \end{array} \][/tex]
So, we see that the result of [tex]\( p(-3) \)[/tex] is -19. Thus:
[tex]\[ p(-3) = -19 \][/tex]
1. Identify the coefficients of the polynomial:
The polynomial is [tex]\( p(x) = x^3 + 6x^2 + 7x - 25 \)[/tex].
The coefficients are: 1 (for [tex]\( x^3 \)[/tex]), 6 (for [tex]\( x^2 \)[/tex]), 7 (for [tex]\( x \)[/tex]), and -25 (constant term).
2. Set up the synthetic division:
We are evaluating [tex]\( p(x) \)[/tex] at [tex]\( x = -3 \)[/tex]. Arrange the coefficients in a row:
[tex]\[ \begin{array}{r|rrrr} -3 & 1 & 6 & 7 & -25 \\ \end{array} \][/tex]
3. Perform the synthetic substitution step-by-step:
- Step 1: Bring down the first coefficient (1) to the bottom row:
[tex]\[ \begin{array}{r|rrrr} -3 & 1 & 6 & 7 & -25 \\ & & & & \\ & 1 & & & \\ \end{array} \][/tex]
- Step 2: Multiply the first coefficient by [tex]\( -3 \)[/tex] and write the result under the second coefficient:
[tex]\[ 1 \times (-3) = -3 \][/tex]
[tex]\[ \begin{array}{r|rrrr} -3 & 1 & 6 & 7 & -25 \\ & & -3& & \\ & 1 & & & \\ \end{array} \][/tex]
- Step 3: Add the second coefficient (6) and the result from the multiplication (-3):
[tex]\[ 6 + (-3) = 3 \][/tex]
[tex]\[ \begin{array}{r|rrrr} -3 & 1 & 6 & 7 & -25 \\ & & -3& & \\ & 1 & 3 & & \\ \end{array} \][/tex]
- Step 4: Multiply the result (3) by [tex]\( -3 \)[/tex] and write it under the third coefficient:
[tex]\[ 3 \times (-3) = -9 \][/tex]
[tex]\[ \begin{array}{r|rrrr} -3 & 1 & 6 & 7 & -25 \\ & & -3& -9& \\ & 1 & 3 & & \\ \end{array} \][/tex]
- Step 5: Add the third coefficient (7) and the result from the multiplication (-9):
[tex]\[ 7 + (-9) = -2 \][/tex]
[tex]\[ \begin{array}{r|rrrr} -3 & 1 & 6 & 7 & -25 \\ & & -3& -9& \\ & 1 & 3 & -2& \\ \end{array} \][/tex]
- Step 6: Multiply the result (-2) by [tex]\( -3 \)[/tex] and write it under the fourth coefficient:
[tex]\[ -2 \times (-3) = 6 \][/tex]
[tex]\[ \begin{array}{r|rrrr} -3 & 1 & 6 & 7 & -25 \\ & & -3& -9& 6 \\ & 1 & 3 & -2 & \\ \end{array} \][/tex]
- Step 7: Add the fourth coefficient (-25) and the result from the multiplication (6):
[tex]\[ -25 + 6 = -19 \][/tex]
[tex]\[ \begin{array}{r|rrrr} -3 & 1 & 6 & 7 & -25 \\ & & -3& -9& 6 \\ & 1 & 3 & -2 & -19 \\ \end{array} \][/tex]
So, we see that the result of [tex]\( p(-3) \)[/tex] is -19. Thus:
[tex]\[ p(-3) = -19 \][/tex]
