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put x=-2 in the polynomial
x^3-3x^2-4x+k
(-2)^3 -3(-2)^2 -4(-2)+k
-8-12+8+k=0
-20+8+k=0
-12+k=0
K= 12
x^3-3x^2-4x+k
(-2)^3 -3(-2)^2 -4(-2)+k
-8-12+8+k=0
-20+8+k=0
-12+k=0
K= 12
SouravBangera1:
Thanks
Answered by
2
Hello friend
Let p(x)=x³-3x²-4x+k
It is given that (x+2) divides the polynomial p(x)
So x+2 is a factor of p(x)
So when p(x) is divided by x+2 it yields a remainder 0
By remainder theorem
y+2=0
x=-2
So p(-2)=0
p(-2)=-2³-3(-2)²-4(-2)+k
0= -8-12+8+k
0= -12+k
k=12
Hope it helps!
Let p(x)=x³-3x²-4x+k
It is given that (x+2) divides the polynomial p(x)
So x+2 is a factor of p(x)
So when p(x) is divided by x+2 it yields a remainder 0
By remainder theorem
y+2=0
x=-2
So p(-2)=0
p(-2)=-2³-3(-2)²-4(-2)+k
0= -8-12+8+k
0= -12+k
k=12
Hope it helps!
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