Question

If kx³+9x²+4x-10 is dividing by x+3 it leaves the remainder 5, find the value of k​

Answer 1

• x + 3 = 0

=> x = - 3

» Put value of x in kx³ + 9x² + 4x - 10 = 5

=> k(-3)³ + 9(-3)² + 4(-3) - 10 = 5

=> - 27k + 81 - 12 - 10 = 5

=> - 27k + 59 = 5

=> - 27k = 5 - 59

=> - 27k = - 54

=> 27k = 54

_______________________________

$\textbf{k = 2}$

___________ [$\bold{ANSWER}$]

Answer 2

Answer:

k = 2.

Step-by-step explanation:

Given :

$\large p(x)=kx^3+9x^2+4x-10 \ and \ g(x)=x+3$

P ( x ) is divisible by x + 3 means x + 3 is factor of p ( x )

So zeroes of g ( x )

x  + 3 = 0

x = - 3

Now putting x = - 3 in  p ( x )

$\large p(x)=kx^3+9x^2+4x-10\\\\\\\large p(-3)=k(-3)^3+9(-3)^2+4(-3)-10\\\\\\\large p(-3)=-27k+81-12-10\\\\\\\large p(-3)=-27k+81-12\\\\\\\large \text{ p(-3)=-27k+59 It is given that remainder is equal to 5 }\\\\\\\large p(-3)=-27k+59=5\\\\\\\large -27k=5-59\\\\\\\large -27k=-54\\\\\\\large k=2$

Thus we get answer k = 2.