## Divide the polynomial p(x) by the polynomial g(x) and find the quotient and remainder in each given of the following :

Question : Divide the polynomial p(x) by the polynomial g(x) and find the quotient and remainder in each given of the following : (i)  p(x) = $$x^3 – 3x^2 + 5x – 3$$,  g(x) = $$x^2 – 2$$ (ii)  p(x) = $$x^4 – 3x^2 + 4x + 5$$, g(x) = $$x^2 + 1 – x$$ …

## Find a quadratic polynomial whose sum of zeroes and product of zeroes are respectively.

Question : Find a quadratic polynomial whose sum of zeroes and product of zeroes are respectively. (i)  $$1\over 4$$, -1 (ii)  $$\sqrt{2}$$, $$1\over 3$$ (iii)  0, $$\sqrt{5}$$ (iv)  1, 1 (v)  $$-1\over 4$$, $$1\over 4$$ (vi)  4, 1 Solution : Let the polynomial be $$ax^2 + bx + c$$ and its zeroes be $$\alpha$$ and …

## Write down the decimal expansion of these given rational numbers which have terminating decimal expansions.

Question : Write down the decimal expansion of these given rational numbers which have terminating decimal expansions. (i) $$13\over 3125$$ (ii)  $$17\over 8$$ (iii)  $$64\over 455$$ (iv)  $$15\over 1600$$ (v)  $$29\over 343$$ (vi)  $$23\over {2^3 5^2}$$ (vii)  $$129\over {2^2 5^7 7^5}$$ (viii)  $$6\over 15$$ (ix)  $$35\over 50$$ (x)  $$77\over 210$$ Solution : (i)  $$13\over 3125$$ …

## Without actually performing the long division, state whether the following rational numbers will have a terminating decimal expansion or a non-terminating repeating decimal expansion :

Question : Without actually performing the long division, state whether the following rational numbers will have a terminating decimal expansion or a non-terminating repeating decimal expansion : (i) $$13\over 3125$$ (ii)  $$17\over 8$$ (iii)  $$64\over 455$$ (iv)  $$15\over 1600$$ (v)  $$29\over 343$$ (vi)  $$23\over {2^3 5^2}$$ (vii)  $$129\over {2^2 5^7 7^5}$$ (viii)  $$6\over 15$$ (ix)  …

Question : Prove that the following are irrationals : (i)  $$1\over \sqrt{2}$$ (ii)  $$7\sqrt{5}$$ (iii)  $$6 + \sqrt{2}$$ Solution : (i)  Let us assume, to the contrary, that $$1\over \sqrt{2}$$ is rational. That is, we can find co-prime integers p and q (q $$\ne$$ 0) such that $$1\over \sqrt{2}$$ = $$p\over q$$  or   $$1\times \sqrt{2}\over … ## Prove that \(3 + 2\sqrt{5}$$ is irrational.

Solution : Let us assume, to the contrary, that $$3 + 2\sqrt{5}$$ is an irrational number. Now, let $$3 + 2\sqrt{5}$$ = $$a\over b$$, where a and b are coprime and b $$ne$$ 0. So, $$2\sqrt{5}$$ = $$a\over b$$ – 3  or  $$\sqrt{5}$$ = $$a\over 2b$$ – $$3\over 2$$ Since a and b are integers, …

## Prove that $$\sqrt{5}$$ is an irrational number by contradiction method.

Solution : Suppose that $$\sqrt{5}$$ is an irrational number. Then $$\sqrt{5}$$ can be expressed in the form $$p\over q$$ where p, q are integers and have no common factor, q $$ne$$ 0. $$\sqrt{5}$$ = $$p\over q$$ Squaring both sides, we get 5 = $$p^2\over q^2$$   $$\implies$$  $$p^2$$ = $$5q^2$$              …