If( X + 2) and( X +3) are the factors of 2x^3+ax^2+bx-18, find the values of a and b
Answers
Answer: Let p(x) = x3 + ax2 + bx +6
(x-2) is a factor of the polynomial x3 + ax2 + b x +6
p(2) = 0
p(2) = 23 + a.22 + b.2 +6 =8+4a+2b+6 =14+ 4a+ 2b = 0
7 +2 a +b = 0
b = - 7 -2a -(i)
x3 + ax2 + bx +6 when divided by (x-3) leaves remainder 3.
p(3) = 3
p(3) = 33 + a.32 + b.3 +6= 27+9a +3b +6 =33+9a+3b = 3
11+3a +b =1 => 3a+b =-10 => b= -10-3a -(ii)
Equating the value of b from (ii) and (i) , we have
(- 7 -2a) = (-10 - 3a)
a = -3
Substituting a = -3 in (i), we get
b = - 7 -2(-3) = -7 + 6 = -1
Thus the values of a and b are -3 and -1 respectively.
Step-by-step explanation:
Answer:
a = 8
b = 18
Step-by-step explanation:
Given,
p(x) = 2x^3+ax^2+bx-18
Factors,
(x + 2) and (x + 3)
By using factor theorem, we get
x = -2, -3
Now by applying the value of x = -2 in p(x) , we get
2×(-2)^3 + a×(-2)^2 + b×(-2) + 18 = 0
=> -16 + 4a -2b + 18 = 0
=> 4a - 2b + 2 = 0
=> 2a - b + 2 = 0
=> 2a = b - 2
=> a = (b - 2) / 2 ...... (i)
Now applying the value of x = -3 in p(x) , we get
2×(-3)^3 + a×(-3)^2 + b×(-3) + 18 = 0
=> -54 + 9a - 3b + 18 = 0
=> 9a - 3b - 36 = 0
=> 3a - b - 12 = 0 ..... (ii)
From equation (i) and (ii) , we get
=> 3{(b-2) /2} - b - 12 = 0
=> (3b - 6) / 2 - b - 12 = 0
=> (3b - 6 - 2b) / 2 = 12
=> b - 6 = 24
=> b = 24 - 6
=> b = 18
Hence, value of b = 18
From equation (i) we get,
a = (b-2) /2 = (18-2) / 2 = 16/2 = 8
Hence, value of a = 8
Thus,
a = 8
b = 18