Step-by-step explanation:
We can use synthetic division to test if (x-2) is a factor of the polynomial x^3 + kx^2 -7x -10.
First, we set up the synthetic division table:
2 | 1 k -7 -10
|___ 2k 2k^2-10k -24k+38
We need (x-2) to be a factor of the polynomial if and only if the remainder or last term is equal to zero. From the table, the remainder is -24k + 38.
Thus:
-24k + 38 = 0
Solving for k:
-24k = -38
k = 38/24
k = 19/12
Therefore, for k = 19/12, (x-2) is a factor of the polynomial x^3 + kx^2 - 7x - 10.
Answer:
-4/7
Correct option is C)
Let p(x)=x
3
+kx
2
−2x+k+4
As x+k is a factor
∴p(−k)=0
p(−k)=−k
+k
+2k+k+4⇒3k+4
As 3k+4=0
∴k=
−4
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Answers & Comments
Step-by-step explanation:
We can use synthetic division to test if (x-2) is a factor of the polynomial x^3 + kx^2 -7x -10.
First, we set up the synthetic division table:
2 | 1 k -7 -10
|___ 2k 2k^2-10k -24k+38
We need (x-2) to be a factor of the polynomial if and only if the remainder or last term is equal to zero. From the table, the remainder is -24k + 38.
Thus:
-24k + 38 = 0
Solving for k:
-24k = -38
k = 38/24
k = 19/12
Therefore, for k = 19/12, (x-2) is a factor of the polynomial x^3 + kx^2 - 7x - 10.
Verified answer
Answer:
-4/7
Step-by-step explanation:
Correct option is C)
Let p(x)=x
3
+kx
2
−2x+k+4
As x+k is a factor
∴p(−k)=0
p(−k)=−k
3
+k
3
+2k+k+4⇒3k+4
As 3k+4=0
∴k=
3
−4
Was this answer helpful?
mark as brainliest