Question

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

Answer 1

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 2

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