If the polynomial ax3+2x2+5x+b is divided by (x2−1) and gives a remainder of (6x+5), then a+3b equals....
Explanation
Given that the polynomial ax3+2x2+5x+b divided by (x2−1) has a remainder of (6x+5).
What we can do is find the roots of the equation x2−1.
x2−1=0
(x+1)(x−1)=0
x=−1 or x=1
Substitute for x = -1
P(x)=ax3+2x2+5x+b
P(−1)=a(−1)3+2(−1)2+5(−1)+b
P(−1)=−a+b−3
Substitute for x = 1
P(x)=ax3+2x2+5x+b
P(1)=a(1)3+2(1)2+5(1)+b
P(1)=a+b+7
Substitute roots into the remainder
Then substitute the roots of x2−1 into the remainder (6x+5).
For x=−1
P(x)=6x+5
P(−1)=6(−1)+5
−a+b−3=−6+5
−a+b=2...(1)
For x=1
P(x)=6x+5
P(1)=6(1)+5
a+b+7=6+5
a+b=4...(2)
Eliminate the equations
Next, eliminate the first and second equations to get the values of a and b.
−a+b=2
a+b=4−
−2a=−2
a=1
Substitute the value of a into the second equation to get the value of b.
a+b=4
1+b=4
b=3
Therefore, the value of a+3b=1+3(3)=10.