(x-2) is a factor of (x^3+ax^2+bx+6) leaves remainder 3 on divided by (x-3)
Answers
★ Answer :
given :
(x - 2) is a factor of x³ + ax² + bx + 6 and leaves 3 as remainder when divided by (x - 3)
Solution :
if x - 2 is a factor then, f(2) should be 0
f(x) = x³ + ax² + bx + 6
f(2) = (2)³ + a(2)² + 2b + 6
= 8 + 4a + 2b + 6
14 + 4a + 2b = 0
4a + 2b = -14
2(2a + b) = -14
2a + b = -7 ..... (1)
given : f(3) = 3
f(3) = (3)³ + a(3)² + 3b + 6
= 27 + 9a + 3b + 6
= 33 + 9a + 3b = 3
9a + 3b = 3-33
9a + 3b = -30
3(3a + b) = -30
3a + b = -10 ..... (2)
(1) - (2)
(2a + b) - (3a + b) = -7-(-10)
2a + b - 3a -b = -7 +10
- a = 3
→ a = -3
by putting value of a in equation 1...
2a + b = -7
2(-3) + b = -7
-6 + b = -7
b = -7+6
→ b = -1
therefore,
a = -3 and b= -1..
equation =
x³ + (-3)x² + (-1)x + 6
= x³ - 3x² - x + 6
★Answer:
Valueof
a=−3
b=−1
★Solution:
Let p(x)=x³+ax²+bx+6
i ) it is given that , (x-2) is a factor of p(x) ,then
p(x)=0
2³+a(2)²+b×2+6=0
8+4a+2b+6=0
4a+2b+14=0
Divide each term by 2 , we get
2a+b+7=0 ---(1)
ii)It is given that, if p(x) divided by (x-3) leaves a remainder 3
p(3)=3
3³+a(3)²+b×3+6=3
27+9a+3b+6-3=0
9a+3b+30=0
Divide each term by 3 , we get
3a+b+10=0 ---(2)
Subtract equation (1) from equation (2) , we get
a + 3 = 0
a = -3
Now,
Substitute a=-3 in equation (1) , we get
2(-3)+b+7=0
-6+b+7=0
b+1=0
b = -1
Therefore,
Valueof
a=−3
b=−1