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Question
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) = 15x3 − 20x2 + 13x − 12; g(x) = x2 − 2x + 2
Solution
Given
f(x) = 15x3 − 20x2 + 13x − 12
g(x) = x2 − 2x + 2
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)
15x3 − 20x2 + 13x − 12 = (x2 − 2x + 2)(ax + b) + cx + d
15x3 − 20x2 + 13x − 12 = ax3 - 2ax2 + 2ax + bx2 - 2bx + 2b + cx + d
15x3 − 20x2 + 13x − 12 = ax3 - 2ax2 + bx2 + 2ax - 2bx + cx + 2b + d
15x3 − 20x2 + 13x − 12 = ax3 - x2(2a - b) + x(2a - 2b + c) + 2b + d
Equating the co-efficients of various powers of x on both sides, we get
On equating the co-efficient of x3
ax3 = 15x3
a = 15
On equating the co-efficient of x2
2a - b = 20
Substituting a = 15, we get
2(15) - b = 20
30 - b = 20
-b = 20 - 30
-b = -10
b = 10
On equating the co-efficient of x
2a - 2b + c = 13
Substituting a = 15 and b = 10, we get
2(15) - 2(10) + c = 13
30 - 20 + c = 13
10 + c = 13
c = 13 - 10
c = 3
On equating constant term
2b + d = -12
Substituting b = 10, we get
2(10) + d = -12
20 + d = -12
d = -12 - 20
d = -32
Therefore, quotient q(x) = ax+ b
= 15x + 10
Remainder r(x) = cx + d
= 3x - 32
Hence, the quotient and remainder are q(x) = 15x + 10 and r(x) = 3x - 32
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