find the value of a and b so that 2x3+ax2+x+b has x+2 and 2x-1 as factor
Answers
Answer:
f(x)=2x3+ax2−bx+3
At x=2
f(2)=15
f(1)=0
f(x)=2x3+ax2−bx+3
f(1)= 2+a-b+3=0
a-b+5=0 ____A
f(x)= 2x3+ax2−bx+3
f(2)=2(23)+a(22)−2b+3=15
4a-2b=-4
Multiply A by 2 and subtract from above equation
4a-2b=-4
2a-2b+10=0
2a-10=-4
2a= 6
a=3
From A
3-b+5=0
8-b=0
b=8
So a=3 and b=8
Therefore the value of a = 5 and b = -2.
Step-by-step explanation:
Given f(x) = 2x^3 + ax^2 + x + b.
Given that 2x - 1 and x + 2 are the factors of f(x).
=> 2x - 1 = 0
=> x = 1/2
Plug x = 1/2 in f(x), we get
=> 2(1/2)^3 + a(1/2)^2 + (1/2) + b = 0
=> 1/4 + a/4+1/2+b=0
=> 3/4 + a/4 + b = 0
=> a+ 3+ 4b = 0
=> a + 4b = -3 (1)
When x + 2:
=> x + 2 = 0
=> x = -2.
Plug x = -2 in f(x), we get
=> 2(-2)^3 + (a)(-2)^2 + (-2) + b = 0
=> -16 + 4a - 2 + b = 0
=> 4a + b - 18 = 0
=> 4a + b = 18 (2)
On solving (1) * 4 & (2), we get
4a +16b-12
4a + b = 18
15b = -30
b = -30/15.
b = -2
Substitute b = -2in (1), we get
=> a + 4b = -3
=> a + 4(-2) = -3
=> a - 8 = -3
=> a = -3 + 8
a = 5.
Therefore the value of a = 5 and b = -2.