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Sagot :
To find the value of [tex]\( k \)[/tex] in the polynomial [tex]\( p(x) = 6x^2 + 5x + k \)[/tex], given that one of the roots of the polynomial is [tex]\( x = -1 \)[/tex], we can use the fact that if [tex]\( x = -1 \)[/tex] is a root, then substituting [tex]\( x = -1 \)[/tex] into the polynomial should yield zero. Let’s do this step-by-step:
1. Substitute [tex]\( x = -1 \)[/tex] into the polynomial:
[tex]\[ p(-1) = 6(-1)^2 + 5(-1) + k \][/tex]
2. Calculate each term:
- First term:
[tex]\[ 6(-1)^2 = 6 \cdot 1 = 6 \][/tex]
- Second term:
[tex]\[ 5(-1) = -5 \][/tex]
3. Combine the terms:
[tex]\[ p(-1) = 6 + (-5) + k = 1 + k \][/tex]
4. Set [tex]\( p(-1) \)[/tex] equal to zero because [tex]\( -1 \)[/tex] is a root:
[tex]\[ 1 + k = 0 \][/tex]
5. Solve for [tex]\( k \)[/tex]:
[tex]\[ k = -1 \][/tex]
Thus, the value of [tex]\( k \)[/tex] is [tex]\(-1\)[/tex].
1. Substitute [tex]\( x = -1 \)[/tex] into the polynomial:
[tex]\[ p(-1) = 6(-1)^2 + 5(-1) + k \][/tex]
2. Calculate each term:
- First term:
[tex]\[ 6(-1)^2 = 6 \cdot 1 = 6 \][/tex]
- Second term:
[tex]\[ 5(-1) = -5 \][/tex]
3. Combine the terms:
[tex]\[ p(-1) = 6 + (-5) + k = 1 + k \][/tex]
4. Set [tex]\( p(-1) \)[/tex] equal to zero because [tex]\( -1 \)[/tex] is a root:
[tex]\[ 1 + k = 0 \][/tex]
5. Solve for [tex]\( k \)[/tex]:
[tex]\[ k = -1 \][/tex]
Thus, the value of [tex]\( k \)[/tex] is [tex]\(-1\)[/tex].
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