Question

Find the value of k for which (3x+2)is a factor of x³+kx²–7x+k

Answer 1

Heya!

Here is yr answer......


Given,

(3x+2) is a factor of x³+kx²-7x+k

3x+2 =0

3x = -2

x = -2/3


Now,

x³+kx²-7x+k = 0

(-2/3)³ + k(-2/3)² - 7(-2/3) + k = 0

-8/27 + k(4/9) +14/3 + k = 0

-8/27 + 14/3 + 4k/9+ k = 0

-8+126/27 + 4k+9k/9 = 0

118/27 + 13k/9 = 0

13/9k = -118/27

k = -118/27 × 9/13

k = -118/39

k = 3



Hope it hlpz..

Answer 2

Hello dear friend !!
Answer is here ☺☺✌
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Given (3x + 2) is a factor of χ² +κχ² - 7χ + κ

we have –

$= > 3x + 2 = 0 \\ = > 3x = 0 - 2 \\ = > x = \frac{ - 2}{3}$



Since , ( 3x + 2 ) is a factor of χ³ + κχ² - 7χ + κ.

Hence , it will be satisfy the above Polynomial .

$= > {x}^{3} + k {x}^{2} - 7x + k = 0 \\ \\ = > ( \frac{ - 2}{3} {)}^{3} + k( \frac{ - 2}{3} {)}^{2} - 7( \frac{ - 2}{3} ) + k = 0 \\ \\ = > \frac{ - 8}{27} + k( \frac{4}{9} ) + \frac{14}{3} + k = 0 \\ \\ = > \frac{ - 8}{27} + \frac{14}{3} + \frac{4k}{9} + k = 0 \\ \\ = > \frac{ - 8 + 126}{27} + \frac{4k + 9k}{9} = 0 \\ \\ = > \frac{118}{27} + \frac{13k}{9} = 0 \\ \\ = > \frac{13k}{9} = 0 - \frac{118}{27} = \frac{ - 118}{27} \\ \\ = > k = \frac{ - 118}{27} \times \frac{9}{13} \\ \\ = > k = \frac{ - 118}{39} = - 3 \\ \\ = > k = - 3$

THEREFORE , k = -3 ...

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Hope it's helps you.
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