EXERCISE 2.3
1.Divide the polynomial p(x) by the polynomial g(x) and find the quotient and remainder in each of the following:
(i) p(x) = x3-3x2+ 5x – 3, g (x) = x2 _ 2
(ii) p(x) =x3 + 3x2+ 4x, g(x) = x2 + 1-x
(iii) p(x) = x4– 5x+ 6, g(x) = 2 – x2
Sol. (i) We have,
=> The quotient is x – 3 and the remainder is 7x – 9.
(ii) Here, the dividend is already in the standard form
and the divisor is also in the standard form.
We have,
=> The quotient is x2+ x – 3 and the remainder is 8.
(iii) To carry out the division, we first write divisor in
the standard form.
So, divisor = – x2 + 2
We have,
=> The quotient is – x2 – 2 and the remainder is – 5x + 10.
2.Check whether the first polynomial is a factor of the second polynomial by dividing the second polynomial by the first polynomial :
(i) t2 – 3, 2t4+ 3t3– 2t2 – 9t – 12
(ii) x2 + 3x + 1, 3x4 + 5x3 – 7x2 + 2x + 2
(iii) x3-3x + 1, x5 – 4x3 + x2+ 3x + 1
Sol.(i) t2 – 3, 2t4+ 3t3– 2t2 – 9t – 12
We have,
Since the remainder is zero, therefore, t2-3 is a
factor of 2t4+ 3t3– 2t2 – 9t – 12.
(ii) Let us divide 3x4 + 5x3 – 7x2 + 2x + 2 by x2 + 3x + 1
We get
Since the remainder is zero, therefore, x2 + 3x + 1 is a factor of
3x4 + 5x3 – 7x2 + 2x + 2
(iii) Let us divide x5 – 4x3 + x2+ 3x + 1 by x3-3x + 1.
we get
Here, remainder is 2
Therefore, x3-3x + 1 is
not a factor of x5 – 4x3 + x2+ 3x + 1.
3.Obtain all the zeroes of 3x4 + 6x3 – 2x2 – 10x – 5,
=> (3x2-5) is a factors of the given polynomial.
Applying the division algorithm to the given polynomial and 3x2-5,
we have
4.On dividing x3 – 3x2 + x + 2 by a polynomial g(x), the quotient and remainder were x – 2 and – 2x + 4 respectively. Find g(x).
Sol. Since on dividing x3 – 3x2 + x + 2 by a polynomial g(x), the quotient and remainder were (x – 2) and (- 2x +4)respectively, therefore,
=> Quotient × Divisor + Remainder = Dividend
= >(x-2) x g(x) + (-2x+4) = x3 – 3x2 + x + 2
= > ( – 2) x g(x) = x3 – 3x2 + x + 2+2x-4
5.Give examples of polynomials p(x), g(x), q(x) and r(x), which satisfy the division algorithm and
(i) deg p(x) = deg q(x) (ii) deg q(x) = deg r(x)
(iii) deg r(x) = 0
Sol. There can be several examples for each of (i), (ii) and (iii).
However, one example for each case may be taken as under:
(i) p(x) = 2x2 – 2x + 14, g(x) = 2,
g(x) =x2-x+7,r(x) = 0
(ii) p(x)=X3+x2+x+1,g(x) = X2– 1,
q(x) = x+1, r(x) = 2x + 2
(iii) p(r) = x3 + 2x2 – x + 2,
g(x) =x2 -1 q(x)= x+ 2, r(x) = 4