find the values of A and B so that (x+1) and (x-2) are factors of x^3+ax^2+2x+b
Answers
Step-by-step explanation:
x+1=0
x=-1
P(x) =x cube +ax square + 2x +b
=(-1)cube + a(-1)square +2(-1)+b
=-1+a-2+b
=a+b-3 - 1
x-2=0
x=2
P(x) =xcube + ax square +2x+ b
=(2) cube +a(2)square +2(2) +b
=8 + 4a +4 +b
=4a+b+12 - 2
Using elimination method
( a+b-3=0) - 1
(4a+b +12=0) - 2
Subtracting 2 from 1
a+ b - 3=0
- 4a+b+12=0
(-) (-) (-)
=-3a +0b - 15=0
a=-15/3
a=-5
In eqn 1
a+b=3
-5 +b=3
b=3+5
b=8
a= - 5
b= 8
Hopefully this will help you.....
given (x + 1) and (x - 2) are the factors of p(x) = x^3 + ax^2 + 2x + b
equating both the factors by 0 we get,
- x = -1
- x = 2
when x = -1, p(-1) = (-1)^3 + a(-1)^2 + 2(-1) + b
⇒ -1 + a - 2 + b = 0
⇒ -3 + a + b = 0
⇒ a + b - 3 = 0 -------(i)
when x = 2, p(2) = (2)^3 + a(2)^2 + 2(2) + b
⇒ 8 + 4a + 4 + b = 0
⇒ 12 + 4a + b = 0
⇒ 4a + b + 12 = 0 --------(ii)
subtracting equation (i) from (ii)
⇒ 4a + b + 12 - (-3 + a + b) = 0
⇒ 4a + b + 12 + 3 - a - b = 0
⇒ 4a - a + 15 = 0
⇒ 3a = -15
⇒ a = -15/3
⇒ a = -5
putting value of a in (i)
⇒ -5 + b = 3
⇒ b = 3 + 5
⇒ b = 8
hence, a = -5 and b = 8