If x³+ax²+bx+6 has (x-2)as a factor and leaves 3 remainder when divided by (x-3), find the value of a and b
Answers
Answer:
The value of a and b are -3 and -1 respectively.
Step-by-step explanation:
Given that (x-2) is a factor of polynomial
P(x)=x^3+ax^2+bx+6P(x)=x
3
+ax
2
+bx+6
Also when divided by (x - 3) leaves a remainder 3.
we have to find the value of a and b.
As 2 is the zero of the polynomial therefore by remainder theorem
P(2)=0P(2)=0
2^3+a(2)^2+2b+6=02
3
+a(2)
2
+2b+6=0
8+4a+2b+6=08+4a+2b+6=0
4a+2b+14=04a+2b+14=0
2a+b=-72a+b=−7 → (1)
\text{Also the polynomial }x^3+ax^2+bx+6\text{ when divided by (x - 3) leaves a remainder 3}Also the polynomial x
3
+ax
2
+bx+6 when divided by (x - 3) leaves a remainder 3
∴ P(3)=3P(3)=3
3^3+a(3)^2+3b+6=33
3
+a(3)
2
+3b+6=3
27+9a+3b+6=327+9a+3b+6=3
9a+3b+33=39a+3b+33=3
3a+b=-103a+b=−10 → (2)
Solving (1) and (2), we get
Subtracting equation (2) from (1)
2a+b-3a-b=-7-(-10)2a+b−3a−b=−7−(−10)
-a=3−a=3
a=-3a=−3
2a+b=-72a+b=−7
2(-3)+b=-72(−3)+b=−7
b=-7+6=-1b=−7+6=−1
Hence, the value of a and b are -3 and -1 respectively.