Question

find the value of k if 2x+1 is a factor of (3k+2)x^3+(k-2)​

Answer 1

Step-by-step explanation:

2x+1 is a factor

2x+1=0

x=-1/2

(3k+2)x^3+(k-2)

(3k+2)(-1/2)^3+(k-2)=0

(3k-2)(-1/8)+(k-2)=0

-3k/8+k+1/4-2=0

5k/8-7/8=0

5k/8=7/8

k =7/5

Answer 2

Answer:

We'll use the factor theorem here.

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$\sf{2x + 1 = 0}$

$\sf{x = \frac{ - 1}{2} }$

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Putting in the value of x in p(x)

$\sf{(3k + 2) \frac{ - 1}{2}^{3} + (k - 2) = 0 }$

(We put the value 0 because we know that 2x+1 is a factor of p(x) and it's remainder will obviously be 0 considering the given information)

$\sf{(3k + 2) \frac{ - 1}{8} + (k - 2) = 0}$

$\sf{ \frac{ - 3k}{8} - \frac{2}{8} + k - 2= 0 }$

$\sf{ \frac{ - 3k + 8k}{8} + ( \frac{ - 2 - 16}{8}) = 0 }$

$\sf{ \frac{ - 5k}{8} = \frac{9}{4} }$

$\sf{k = \frac{9}{4} \times \frac{ - 8}{5} }$

$\sf{ k = \frac{ - 18}{5} }$

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