3 2
If 3x+ax+bx+4 has x+1 as factor and leaves remainder 2 when divided by x-1, find the value of a and b.
f(x)=3x^3+ax^2+bx+4
f(1)=2 => 3+a+b+4 = 2 ....(1)
f(-1)=0 => -3+a-b+4=0 ....(2)
can you solve the system defined by (1) and (2) with a and b as unknowns?
Hint: both a and b are negative.
To find the value of 'a' and 'b', we need to use the Remainder Theorem and Synthetic Division.
Given that x + 1 is a factor of the polynomial 3x + ax + bx + 4, we can set x + 1 equal to zero and solve for x.
x + 1 = 0
x = -1
Since x + 1 is a factor, when we substitute x = -1 into the polynomial, the result should be zero.
Substituting x = -1 into 3x + ax + bx + 4:
3(-1) + a(-1) + b(-1) + 4 = 0
-3 - a - b + 4 = 0
-a - b + 1 = 0
Now, we know that there is a remainder of 2 when the polynomial is divided by x - 1. So we can set up another equation using the Remainder Theorem.
According to the Remainder Theorem, if there is a remainder of 2 when a polynomial P(x) is divided by x - 1, then P(1) should be equal to 2.
Substituting x = 1 into 3x + ax + bx + 4:
3(1) + a(1) + b(1) + 4 = 2
3 + a + b + 4 = 2
a + b + 7 = 2
a + b = -5
Now, we have two equations:
Equation 1: -a - b + 1 = 0
Equation 2: a + b = -5
We can solve this system of equations by substitution or elimination.
Subtracting Equation 1 from Equation 2:
(a + b) - (-a - b) = -5 - 1
2a = -6
a = -3
Substituting the value of a into Equation 2:
(-3) + b = -5
b = -5 + 3
b = -2
Therefore, the value of a is -3 and the value of b is -2.