If g(x)= -2(x^2-6x-12)+3(k-x) and g(x) is evenly divisible by x, then what's the value of k?

Question

If g(x)=-2(x²-6x-12)+3(k-x) and g(x) is evenly divisible by x, then what's the value of k?

Answer 1

Set   g(x)=0.

 

 

Recall that if we have a root, then the factor that is associated by that root is evenly divisible to the polynomial.  In this case, x=0 is a root because the factor is x.

 

-2(x² - 6x - 12) + 3(k - x) = 0

 

-2x² + 12x + 24 + 3k - 3x = 0

 

24 + 3k = 0

 

 

Solving for k, we get

 

3k = -24

 

 k = -8

 

 

We can check this by plugging this value of k into g(x) and dividing g(x) by x.  We should get a remainder of 0.

 

g(x) = -2x² + 12x + 24 - 24 - 3x

 

 

g(x) / x =

 

(-2x² + 9x) / x =

 

-2x + 9

 

 

We get a quotient of (-2x + 9) with no remainder.