Answer :
Answer:
Step-by-step exTo determine the value of
k, we start with the polynomial given:
(
,
)
=
2
3
−
2
−
2
+
3
−
P(x,y)=k
2
x
3
−ky
2
−ky
2
+3ky−k
We know that
−
3
x−3 is a factor of
(
,
)
P(x,y). This means
(
3
,
)
=
0
P(3,y)=0.
Let's substitute
=
3
x=3 into
(
,
)
P(x,y):
(
3
,
)
=
2
⋅
3
3
−
2
−
2
+
3
⋅
−
P(3,y)=k
2
⋅3
3
−ky
2
−ky
2
+3k⋅y−k
(
3
,
)
=
2
⋅
27
−
2
2
+
3
−
P(3,y)=k
2
⋅27−2ky
2
+3ky−k
Since
(
3
,
)
=
0
P(3,y)=0, we have:
2
⋅
27
−
2
2
+
3
−
=
0
k
2
⋅27−2ky
2
+3ky−k=0
Now, we will evaluate this equation for a specific value of
y that allows us to simplify and solve for
k.
Let's try
=
1
y=1:
(
3
,
1
)
=
2
⋅
27
−
2
⋅
1
2
+
3
⋅
1
−
P(3,1)=k
2
⋅27−2k⋅1
2
+3k⋅1−k
(
3
,
1
)
=
27
2
−
2
+
3
−
P(3,1)=27k
2
−2k+3k−k
(
3
,
1
)
=
27
2
+
P(3,1)=27k
2
+k
Since
(
3
,
1
)
=
0
P(3,1)=0, substitute and solve for
k:
27
2
+
=
0
27k
2
+k=0
(
27
+
1
)
=
0
k(27k+1)=0
So,
=
0
k=0 or
=
−
1
27
k=−
27
1
.
To find the specific value of
k when
−
3
x−3 is a factor, we need
≠
0
k
=0 (as
=
0
k=0 would make the polynomial trivially zero for all
x).
Therefore, the value of K is-1/27