If #f(x)=x^2+3x+k# is divided by #x+k#, remainder is #0#. What is the value of #k#?

2 Answers
salamat
Mar 28, 2017

# k= 0 and 2#

Explanation:

#(x^2 + 3 x + k )/ (x + k)#

we use long division to solve it.
#x^2 + 3 x + k -> x * (x + k)#
#-(x^2 +kx)#
........................
#3 x - kx + k#
#(3 - k)x + k ->(3 - k) * (x + k)#
#-((3 - k)x + (3 - k)k)#
.......................
#k - (3 - k)k = k -3k + k^2 = -2k + k^2#

to be a divisible,
#-2k + k^2 = 0#
#k(-2 + k) = 0#

# k= 0 and 2#

Shwetank Mauria
Mar 28, 2017

#k=0# or #k=2#

Explanation:

We can use Remainder Theorem here. According to this theorem, if a polynomial #f(x)# is divided by #x-a#, the remainder is #f(a)#.

In other words, if #f(x)# is divisible by #x-a#, #f(a)=0#.

Now #f(x)=x^2+3x+k# to be divisible by #x+k# or #x-(-k)#,

we should have #f(-k)=(-k)^2+3xx(-k)+k=0#

or #k^2-3k+k=0#

i.e. #k(k-2)=0#

For this we must have either #k=0# or #k-2=0#

i.e. either #k=0# or #k=2#

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