let p(x) be a cubic polynomial such that p(x)+x is divisible by (x-2) and (x-3). if p(1)=9 and p(-1) = -11, find the value of p(0)
Answers
Given : p(x) a cubic polynomial such that p(x)+x is divisible by (x-2) and (x-3).
p(1)=9 and p(-1) = -11,
To find : the value of p(0)
Solution:
p(x) be a cubic polynomial
= ax³ + bx² + cx + d
p(x)+x is divisible by (x-2) and (x-3)
=> p(2) + 2 = 0 => p(2) = - 2
P(3) + 3 = 0 => p(3) = - 3
P(1) = 9
p(-1) = - 11 given
p(0) = d
P(1) = 9 => a + b + c + d = 9
p(-1) = -11 => -a + b - c + d = - 11
=> b + d = - 1 and a + c = 10
=> b = -d - 1 c = 10 - a
p(2) = - 2 ,
=> 8a + 4b + 2c + d = - 2
=> 8a + 4(-d - 1) + 2(10 - a) + d = - 2
=> 6a -3d = -18
=> 2a - d = - 6
p(3) = - 3
27a + 9b + 3c + d = - 3
=> 27a + 9(-d - 1) + 3(10 - a) + d = - 3
=> 24a - 8d = - 24
=> 3a - d = - 3
on solving
a= 3 and d = 12
value of p(0) = 12
finding b and c using b = -d - 1 c = 10 - a
p(x) = 3x³ -13x² + 7x + 12
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