Question

Please solve this .
Find the value of K if the division of
$kx { }^{3} + 9x {}^{2} + 4x - 10$
by x+ 3 leaves a reminder -22.

Most satisfying answer will mark as brainliest.
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Answer 1

Answer :-

Value of k is 3.

Explanation :-

Let f(x) = kx³ + 9x² + 4x - 10

Given

When f(x) is divided by (x + 3) leaves the remainder - 22

Finding the zero of (x + 3)

x + 3 = 0

x = - 3

By Remainder theorem, f(-3) is the remainder

⇒ f(-3) = - 22

⇒ k(-3)³ + 9(-3)² + 4(-3) - 10 = - 22

⇒ k(-27) + 9(9) - 12 - 10 = - 22

⇒ - 27k + 81 - 22 = - 22

⇒ - 27k + 59 = - 22

⇒ 59 + 22 = 27k

⇒ 81 = 27k

⇒ 81/27 = k

⇒ 3 = k

⇒ k = 3

∴ the value of k is 3.

Answer 2

ANSWER:-

Given:

kx³ + 9x² +4x-10 by x+3 is a Factor.

Remainder: -22

To find:

The value of k.

Solution:

Let p(x)= kx³ +9x² +4x-10

According to the question:

p(x) leaves the remainder is -22 when divided by (x+3).

=) x+3 =0

=) x= -3

Therefore,

➨p(-3)=k(-3)³ + 9(-3)² + 4(-3)- 10= -22

➨p(-3)=k(-27) + 9(9) + (-12) - 10= -22

➨p(-3)= -27k + 81 - 12 - 10= -22

➨p(-3)= -27k +81 -22= -22

➨p(-3)= -27k + 59= -22

➨p(-3)= -27k = -22 - 59

➨p(-3)= -27k= -81

➨p(-3)= k= -81/-27

Therefore,

➨k= 3

Thus,

The value of k is 3.

Hope it helps ☺️