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Sagot :
Let's find the value of [tex]\( k \)[/tex] such that [tex]\( x + k \)[/tex] is a factor of the polynomial [tex]\( x^2 + kx + k - 2 \)[/tex].
### Step-by-Step Solution:
1. Factor Theorem Application:
According to the Factor Theorem, [tex]\( x + k \)[/tex] is a factor of the polynomial [tex]\( x^2 + kx + k - 2 \)[/tex] if and only if substituting [tex]\( x = -k \)[/tex] into the polynomial yields zero.
2. Substitution and Setting the Polynomial to Zero:
Substitute [tex]\( x = -k \)[/tex] into the polynomial:
[tex]\[ (-k)^2 + k(-k) + k - 2 = 0 \][/tex]
3. Simplify the Substitution:
Simplify the expression:
[tex]\[ k^2 - k^2 + k - 2 = 0 \][/tex]
[tex]\[ 0 + k - 2 = 0 \][/tex]
[tex]\[ k - 2 = 0 \][/tex]
4. Solve for [tex]\( k \)[/tex]:
Solving the equation [tex]\( k - 2 = 0 \)[/tex], we find:
[tex]\[ k = 2 \][/tex]
Therefore, the value of [tex]\( k \)[/tex] for which [tex]\( x + k \)[/tex] is a factor of [tex]\( x^2 + k x + k - 2 \)[/tex] is [tex]\( k = 2 \)[/tex].
### Step-by-Step Solution:
1. Factor Theorem Application:
According to the Factor Theorem, [tex]\( x + k \)[/tex] is a factor of the polynomial [tex]\( x^2 + kx + k - 2 \)[/tex] if and only if substituting [tex]\( x = -k \)[/tex] into the polynomial yields zero.
2. Substitution and Setting the Polynomial to Zero:
Substitute [tex]\( x = -k \)[/tex] into the polynomial:
[tex]\[ (-k)^2 + k(-k) + k - 2 = 0 \][/tex]
3. Simplify the Substitution:
Simplify the expression:
[tex]\[ k^2 - k^2 + k - 2 = 0 \][/tex]
[tex]\[ 0 + k - 2 = 0 \][/tex]
[tex]\[ k - 2 = 0 \][/tex]
4. Solve for [tex]\( k \)[/tex]:
Solving the equation [tex]\( k - 2 = 0 \)[/tex], we find:
[tex]\[ k = 2 \][/tex]
Therefore, the value of [tex]\( k \)[/tex] for which [tex]\( x + k \)[/tex] is a factor of [tex]\( x^2 + k x + k - 2 \)[/tex] is [tex]\( k = 2 \)[/tex].
Answer:
2
Step-by-step explanation:
To determine the value of \( k \) for which \( x + k \) is a factor of \( x^2 + kx + k - 2 \), we proceed as follows:
Since \( x + k \) is a factor, the polynomial \( x^2 + kx + k - 2 \) must be divisible by \( x + k \). By the factor theorem, if \( x + k \) is a factor, then \( x = -k \) is a root of the polynomial.
Substitute \( x = -k \) into \( x^2 + kx + k - 2 \):
\[
(-k)^2 + k(-k) + k - 2 = k^2 - k^2 + k - 2 = k - 2
\]
For \( x + k \) to be a factor, \( k - 2 \) must equal \( 0 \) (since the remainder when \( x^2 + kx + k - 2 \) is divided by \( x + k \) should be zero):
\[
k - 2 = 0
\]
\[
k = 2
\]
Therefore, the value of \( k \) for which \( x + k \) is a factor of \( x^2 + kx + k - 2 \) is \( \boxed{2} \).
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