(i) g(t) = t2 - 3 and f(t) = 2t4 + 3t2 - 9t - 12
Degree of f(t) is 4 and degree of g(t) is 2; therefore degree of and degree of q(t) is 4 - 2 = 2 remainder is of degree 1 or less,
Let q(t) = at2 + bt + c and r(t) = pt + q
By applying division algorithm:
Dividend = Quotient× Divisor + Remainder
f(t) = q(t) x g(t) + r(t)
On substituting values in the above relation we get,
On comparing coefficients we get,
On solving above equations we get,
a = 2, b = 3,c = - 4,p = 0,q = 0
On substituting these values for q(t) and r(t)
Since remainder is zero, therefore g(t) is a factor of f(t)
(ii) g(x) = x3 - 3x + 1 and f(x) = x5 - 4x3 + x2 + 3x + 1
Degree of f(x) is 5 and degree of g(x) is 3; therefore degree of q(x) is 5 - 3 = 2 and degree of remainder is of degree 1 or less,
Let q(x) = ax2 + bx + c and r(x) = px + q
By applying division algorithm:
Dividend = Quotient× Divisor + Remainder
f(x) = q(x) x g(x) + r(x)
On substituting values in the above relation we get,
On comparing coefficients we get,
On solving above equations we get,
a = 1, b = 0, c = -1, p = 0,q = 2
On substituting these values for q(x) and r(x)
Since remainder is 2, therefore g(x) is not a factor of f(x)
(iii) g(x) = 2x2 - x + 3 and f(x) = 6x5 - x4 + 4x3 - 5x2 - x - 15
Degree of f(x) is 5 and degree of g(x) is 2; therefore degree of q(x) is 5 - 2 = 3 and degree of remainder is of degree 2 or less,
Let q(x) = ax3 + bx2 + cx + d and r(x) = px + q
By applying division algorithm:
Dividend = Quotient× Divisor + Remainder
f(x) = q(x) x g(x) + r(x)
On substituting values in the above relation we get,
On comparing coefficients we get,
On solving above equations we get,
a = 3, b = 1,c = - 2, d = 5, p = 10q = - 30
On substituting these values for q(x) and r(x)
Since remainder is 10x - 3, therefore g(x) is not a factor of f(x0
