Question

Find the value of k for which x^2+(k-1)x+k^2-16 is exactly divisible by (x-3) but not divisible by x+4

Answer 1

Answer:

Step-by-step explanation:

F(x) =x² + (k-1)x +k² - 16

F(3) = 9 +3k-3 + k²-16

SInce it is exactly divisible by x-3 , Therefore

9 +3k-3 + k²-16 =0

k²+3k = 10

k² + 3k -10 = 0

Factorising we get

k²-2k+5k-10

k(k-2) + 5 (k-2)

(k+5)(k-2)

Therefore k = -5 or k =2

Answer 2

x-3=0
x=3
x^2+(k-1)x+k^2-16
3^2+(k-1)3+k^2-16
9+3k-3+k^2-16
-10+3k+k^2=0