Advertisements
Advertisements
प्रश्न
Apply division algorithm to find the quotient q(x) and remainder r(x) on dividing f(x) by g(x) in the following f(x) = 4x3 + 8x2 + 8x + 7, g(x) = 2x2 − x + 1
Advertisements
उत्तर
we have
f(x) = 4x3 + 8x2 + 8x + 7
g(x) = 2x2 − x + 1
Here, Degree (f(x)) = 3 and
degree (g(x)) = 2
Therefore, quotient q(x) is of degree 3 - 2 = 1 and Remainder r(x) is of degree less than 2
Let q(x) = ax + b and
r(x) = cx + d
Using division algorithm, we have
f(x) = g(x) x q(x) + r(x)
4x3 + 8x2 + 8x + 7 = (2x2 - x + 1)(ax + b) + cx + d
4x3 + 8x2 + 8x + 7 = 2ax3 - ax2 + ax + 2bx2 - xb + b + cx + d
4x3 + 8x2 + 8x + 7 = 2ax3 - ax2 + 2bx2 + ax - xb + cx + b + d
4x3 + 8x2 + 8x + 7 = 2ax3 + x2(-a + 2b) + x(a - b + c) + b + d
Equating the co-efficient of various Powers of x on both sides, we get
On equating the co-efficient of x3
2a = 4
`a=4/2`
a = 2
On equating the co-efficient of x2
8 = -a + 2b
Substituting a = 2 we get
8 = -2 + 2b
8 + 2 = 2b
10 = 2b
`10/2=b`
5 = b
On equating the co-efficient of x
a - b + c = 8
Substituting a = 2 and b = 5 we get
2 - 5 + c = 8
-3 + c = 8
c = 8 + 3
c = 11
On equating the constant term, we get
b + d = 7
Substituting b = 5, we get
5 + d = 7
d = 7 - 5
d = 2
Therefore, quotient q(x) = ax + b
= 2x + 5
Remainder r(x) = cx + d
= 11x + 2
Hence, the quotient and remainder are q(x) = 2x + 5 and r(x) = 11x + 2
