Question

Find the value of K if the division of Kx^3+9x^2+4x-10 leaves a remainder -22

Answer 1

The value of K if the division of Kx^3+9x^2+4x-10 by (x-3) leaves a remainder -22 is -35/9

Step-by-step explanation:

The given polynomial is

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

Dividing $p(x)$ by $x-3$ leaves remainder -22

Therefore, by remainder theorem, the value of the polynomial at x=3 will be -22

Therefore,

$p(3)=-22$

$k(3)^3+9(3)^2+4\times 3-10=-22$

$\implies 27k+81+12-10=-22$

$\implies 27k=-105$

$\implies k=\frac{-105}{27}$

$\implies k-\frac{35}{9}$

Hope this helps.

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Answer 2

Step-by-step explanation:

Given that,

remainder is -22

k=?

(X+3)=0

X=-3

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

k(-27)+9(9)-12-10=-22

-27k+81-22=-22

-27k+81=-22+22

-27k+81=0

-27k=-81

27k=81

k=81/27

k=3