Question

Find the value of k, if kx`3+9x² +4x –10 is divided by( x + 3) leaves a remainder - 22.
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Answer 1

Answer:

Given p(x) = kx^3 + 9x^2 - 4x - 10.

Given g(x) = x - 3.

By the remainder theorem, we get

= > x - 3 = 0

= > x = 3.

plug x = 3 in p(x), we get

= > k(3)^3 + 9(3)^2 - 4(3) - 10 = -22

= > 27k + 81 - 12 - 10 = -22

= > 27k + 59 = -22

= > 27k = -81

= > k = -3.

Therefore the value of k = -3.

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Step-by-step explanation:

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Answer 2

Given:-

Find the value of k, if kx³ + 9x² + 4x – 10 is divided by (x + 3) leaves a remainder - 22.

Solution:-

Given f(x) = Kx² + 9x² + 4x - 10 and the zero of (x + 3) is - 3.

f(x) = - 22

f(-3) = K(-3)³ + 9(-3)² + 4(-3) - 10 = - 22

${\Large\ ⇒}\sf\ \: K(-27) \: + \: 9(9) \: + \: 4(-3) \: - \: 10 \: = \: 22$

${\Large\ ⇒}\sf\ \: K \: + \: 81 \: - \: 12 \: - \: 10 \: = \: - 22$

${\Large\ ⇒}\sf\ \: - 27 K \: + \: 81 \: - \: \cancel{22} \: = \: - \: 22$

${\Large\ ⇒}\sf\ \: - 27 K \: \: \cancel{=} \: \: 81$

${\Large\ ⇒} \: \sf\ \: + \: 27 K \: \: \cancel{=} \: \: 81$

${\Large\ ⇒}\sf\ \: K = \cancel \frac{81}{27}$

${\Large\ ⇒} \: \: {\Large{\boxed{\red{\sf{K = 3.}}}}}$

$\small$

∴Hence, K = 3.