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Sagot :
Sure, let's use synthetic division to verify that [tex]\( x = -4 \)[/tex] is a solution to the polynomial equation [tex]\( x^3 + 8x^2 + 19x + 12 = 0 \)[/tex].
### Step-by-Step Solution:
1. Write down the coefficients of the polynomial:
[tex]\[ 1, 8, 19, 12 \][/tex]
2. Set up the synthetic division with the root [tex]\( x = -4 \)[/tex]. Write [tex]\( -4 \)[/tex] to the left and the coefficients in a row:
[tex]\[ \begin{array}{r|rrrr} -4 & 1 & 8 & 19 & 12 \\ \end{array} \][/tex]
3. Bring down the leading coefficient (1) directly below the line:
[tex]\[ \begin{array}{r|rrrr} -4 & 1 & 8 & 19 & 12 \\ & & & & \\ & 1 & & & \\ \end{array} \][/tex]
4. Multiply the root [tex]\( -4 \)[/tex] with the value you just brought down (1) and write it under the next coefficient (8):
[tex]\[ \begin{array}{r|rrrr} -4 & 1 & 8 & 19 & 12 \\ & & -4 & & \\ & 1 & & & \\ \end{array} \][/tex]
5. Add these two values (8 and -4) and write the result below the line:
[tex]\[ \begin{array}{r|rrrr} -4 & 1 & 8 & 19 & 12 \\ & & -4 & 3 & \\ & 1 & 4 & & \\ \end{array} \][/tex]
6. Multiply the root [tex]\( -4 \)[/tex] with the new value (4) and write the result under the next coefficient (19):
[tex]\[ \begin{array}{r|rrrr} -4 & 1 & 8 & 19 & 12 \\ & & -4 & -16 & \\ & 1 & 4 & & \\ \end{array} \][/tex]
7. Add these two values (19 and -16) and write the result:
[tex]\[ \begin{array}{r|rrrr} -4 & 1 & 8 & 19 & 12 \\ & & -4 & -16 & 3 \\ & 1 & 4 & 3 & \\ \end{array} \][/tex]
8. Multiply the root [tex]\( -4 \)[/tex] by the new result (3) and write it under the last coefficient (12):
[tex]\[ \begin{array}{r|rrrr} -4 & 1 & 8 & 19 & 12 \\ & & -4 & -16 & -12 \\ & 1 & 4 & 3 & \\ \end{array} \][/tex]
9. Add the final pair of values (12 and -12) to get the final value:
[tex]\[ \begin{array}{r|rrrr} -4 & 1 & 8 & 19 & 12 \\ & & -4 & -16 & -12 \\ & 1 & 4 & 3 & 0 \\ \end{array} \][/tex]
The result at the bottom row shows the coefficients of the quotient when the polynomial is divided by [tex]\( x + 4 \)[/tex]. The last value is 0, which confirms that [tex]\( x = -4 \)[/tex] is a root of the polynomial [tex]\( x^3 + 8x^2 + 19x + 12 = 0 \)[/tex].
Thus, [tex]\( x = -4 \)[/tex] is indeed a solution of the equation.
### Step-by-Step Solution:
1. Write down the coefficients of the polynomial:
[tex]\[ 1, 8, 19, 12 \][/tex]
2. Set up the synthetic division with the root [tex]\( x = -4 \)[/tex]. Write [tex]\( -4 \)[/tex] to the left and the coefficients in a row:
[tex]\[ \begin{array}{r|rrrr} -4 & 1 & 8 & 19 & 12 \\ \end{array} \][/tex]
3. Bring down the leading coefficient (1) directly below the line:
[tex]\[ \begin{array}{r|rrrr} -4 & 1 & 8 & 19 & 12 \\ & & & & \\ & 1 & & & \\ \end{array} \][/tex]
4. Multiply the root [tex]\( -4 \)[/tex] with the value you just brought down (1) and write it under the next coefficient (8):
[tex]\[ \begin{array}{r|rrrr} -4 & 1 & 8 & 19 & 12 \\ & & -4 & & \\ & 1 & & & \\ \end{array} \][/tex]
5. Add these two values (8 and -4) and write the result below the line:
[tex]\[ \begin{array}{r|rrrr} -4 & 1 & 8 & 19 & 12 \\ & & -4 & 3 & \\ & 1 & 4 & & \\ \end{array} \][/tex]
6. Multiply the root [tex]\( -4 \)[/tex] with the new value (4) and write the result under the next coefficient (19):
[tex]\[ \begin{array}{r|rrrr} -4 & 1 & 8 & 19 & 12 \\ & & -4 & -16 & \\ & 1 & 4 & & \\ \end{array} \][/tex]
7. Add these two values (19 and -16) and write the result:
[tex]\[ \begin{array}{r|rrrr} -4 & 1 & 8 & 19 & 12 \\ & & -4 & -16 & 3 \\ & 1 & 4 & 3 & \\ \end{array} \][/tex]
8. Multiply the root [tex]\( -4 \)[/tex] by the new result (3) and write it under the last coefficient (12):
[tex]\[ \begin{array}{r|rrrr} -4 & 1 & 8 & 19 & 12 \\ & & -4 & -16 & -12 \\ & 1 & 4 & 3 & \\ \end{array} \][/tex]
9. Add the final pair of values (12 and -12) to get the final value:
[tex]\[ \begin{array}{r|rrrr} -4 & 1 & 8 & 19 & 12 \\ & & -4 & -16 & -12 \\ & 1 & 4 & 3 & 0 \\ \end{array} \][/tex]
The result at the bottom row shows the coefficients of the quotient when the polynomial is divided by [tex]\( x + 4 \)[/tex]. The last value is 0, which confirms that [tex]\( x = -4 \)[/tex] is a root of the polynomial [tex]\( x^3 + 8x^2 + 19x + 12 = 0 \)[/tex].
Thus, [tex]\( x = -4 \)[/tex] is indeed a solution of the equation.
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