Subject Test Math 2

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Subject Test Math 2

Question 3 of 28

If $x - 2$ is a factor of $x^{3} + k x^{2} + 12 x - 8$ , then $k =$

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$negative 6$

Correct Answer:

Yes

$negative 3$

Correct Answer:

No

$2$

Correct Answer:

No

$3$

Correct Answer:

No

$6$

Correct Answer:

No

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The correct answer is A.

By the Factor Theorem, $x - 2$ is a factor of $x^{3} + k x^{2} + 12 x - 8$ only when $2$ is a root of $x^{3} + k x^{2} + 12 x - 8,$ that is, ${(2)}^{3} + {k (2)}^{2} + 12 (2) - 8 = 0$ , which simplifies to $4 k + 24 = 0$ . Therefore, $k = - 6$ .

Alternatively, one can perform the division of $x^{3} + k x^{2} + 12 x - 8$ by $x - 2$ and then find a value for $k$ so that the remainder of the division is $0$ .

$x^{3} + k x^{2} + 12 x - 8 = x^{2} (x - 2) + (2 + k) x^{2} + 12 x - 8$

$text end text text end text equals x squared left parenthesis x minus 2 right parenthesis plus left parenthesis 2 plus k right parenthesis x left parenthesis x minus 2 right parenthesis plus left parenthesis 2 left parenthesis 2 plus k right parenthesis plus 12 right parenthesis x minus 8$

$= x^{2} (x - 2) + (2 + k) x (x - 2) + (2 (2 + k) + 12) (x - 2) + (2 (2 (2 + k) + 12) - 8)$

$= (x^{2} + (2 + k) x + (2 (2 + k) + 12)) (x - 2) + (8 + 4 k + 24 - 8)$

$= (x^{2} + (2 + k) x + (2 (2 + k) + 12)) (x - 2) + (4 k + 24)$

Since the remainder is $4 k + 24$ , the value of $k$ must satisfy $4 k + 24 = 0$ . Therefore, $k = - 6$ .

Question Difficulty:

easy

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Subject Test Math 2

Question 3 of 28

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View Correct Answer