find the value of a and b so that the polynomial x cube - 4x square+ax+b is exactly divisible by x-2 as well as x+1
Answers
Hi friends how are you
This is your answer
Hello Mate!
Factor theorum states that p( x ) = 0
x = 1, 2
p( x ) = x^3 -10x^2+ax+ b
0 = 1 - 10 + a + b
0 = - 9 + a + b ------(1)
0 = 2^3 - 10(2)^2 + a(2) + b
0 = 8 - 40 + 2a + b
0 = - 32 + 2a + b ------(2)
(1) - (2)
-9 + a + b
-( - 32 + 2a - b )
23 - a = 0
a = 23
keeping value of a in eq. 1
0 = - 9 + 23 + b
-14 = b
Hope it helps you
Answer:
here is ur answer bud
given x-2 is the factor of p(x)=x^3-4x^2+ax+b
so, x-2=0
x=2 put in p(x)
therefore (2)^3-4(2)^2+a(2)+b=0
8-16+2a+b=0
-8+2a+b=0–----––(I)
now, x+1 is also a factor of p(x)
therefore, x+1=0
x=-1 put in p(x)
(-1)^3-4(-1)^2+a(-1)+b=0
-1-4-a+b=0
-5-a+b=0–--------(ii)
on comparing (I) and (ii) we get,
-8+2a+b=-5-a+b
2a+a+b-v=8-5
3a=3
a=1 put in (ii)
-5-1+b=0
-6+b=0
b=6
hence a=1 and b=6
hope u understand
thanx
by Prachi