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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) = 10x4 + 17x3 − 62x2 + 30x − 3, g(x) = 2x2 + 7x + 1
उत्तर
We have
f(x) = 10x4 + 17x3 − 62x2 + 30x − 3
g(x) = 2x2 + 7x + 1
Therefore, quotient q(x) is of degree 4 - 2 = 2 and remainder r(x) is of degree less than 2
Let g(x) = ax2 + bx + c and
r(x) = px + q
Using division algorithm, we have
f(x) = g(x) x q(x) + r(x)
1Ox4+ 17x3 - 62x2 + 30x - 3 = (2x2 + 7x+ 1)(ax2 +bx+c)+ px+q
1Ox4+ 17x3 - 62x2 + 30x - 3 = 2ax4 + 7ax3 + ax2 + 2bx3 + 7bx2 + bx + 2cx2 + 7xc + c + px + q
1Ox4+ 17x3 - 62x2 + 30x - 3 = 2ax4 + 7ax3 + 2bx3 + ax2 + 7bx2 + 2cx2 + bx + 7xc + px + c + q
a + 1Ox4+ 17x3 - 62x2 + 30x - 3 = 2ax4 + x3 (7a + 2b) + x2 (a + 7b + 2c) + x(b + 7c + p) + c + q
Equating the co-efficients of various powers x on both sides, we get
On equating the co-efficient of x4
2a = 10
`a=10/2`
a = 5
On equating the co-efficient of x3
7a + 2b = 17
Substituting a = 5 we get
7 x 5 2b = 17
35 + 2b = 17
2b = 17 - 35
2b = -18
`b=(-18)/2`
b = -9
On equating the co-efficient of x2
a + 7b + 2c = -62
Substituting a = 5 and b = -9, we get
-9 + 7 x -2 + p = 30
-9 - 14 + p = 30
-23 + p = 30
p = 30 + 23
p = 53
On equating constant term, we get
c + q = -3
Substituting c = -2, we get
-2 + q = -3
q = -3 + 2
q = -1
Therefore, quotient q(x) = ax2 + bx + c
= 5x2 - 9x - 2
Remainder r(x) = px + q
= 53x - 1
Hence, the quotient and remainder are q(x) = 5x2 - 9x - 2 and r(x) = 53x - 1
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