Ncert Solutions For Class 10 Maths Ex 2.3

Ncert Solutions For Class 10 Maths Chapter 2 Ex 2.3

Q.1 If a polynomial x3 -3x2+x+2 is divided by a polynomial g(x), the quotient and remainder obtained are (x-2) and (-2x+4), respectively. Find the equation of g(x).

Sol.

Since, Dividend = Divisor × Quotient + Remainder

Therefore, x³ -3x2+x+2 = g(x) × (x-2) + (-2x+4)

(x3 -3x2+x+2) – (-2x+4) = g(x) × (x-2)

Therefore, g(x) × (x-2) = x3 -3x2+3x-2

Now, for finding g(x) we will divide “x3 -3x2+3x-2” with (x-2)

Therefore, g(x) = (x2 – x +1)

Q.2 Find the quotient and remainder by dividing the polynomial f(x) by the polynomial g(x).

(i) f(x) = x+ 2x– 9x + 5,   g(x) = x2+5

Therefore, Quotient is (x+2) and Remainder is (-14x − 5)

(ii) f(x) = x5+2x4-9x3+5x2-2x+1, g(x) = x3 + x2 – x+1

Therefore, Quotient is (x2 + x − 9) and Remainder is (14x2 – 12x +10).

(iii) f(x) = 2x+ 7x+ 5x+ 8x + 5, g(x) = 11− 2x3 + x2

Therefore, Quotient is  – (x + 4) and Remainder is (9x2 + 19x +49).

Q3. Find all the zeroes of the polynomial equation 2x4-3x3-3x2+6x-2, if two of its zeroes are 2and2$\sqrt{2}\;\; and \;\; -\sqrt{2}$.

Sol.

Since this is a polynomial equation of degree 4, there will be a total of 4 roots.

Let, f(x) = 2x– 3x– 3x+ 6x – 2

2and2$\;\;\sqrt{2}\;\;and\;\;-\sqrt{2}$ are zeroes of f(x).

(x2)(x+2)$\;\;(x-\sqrt{2})\;\;(x+\sqrt{2})$= (x2− 2) = g(x), is a factor of given polynomial f(x).

If we divide f(x) by g(x), the quotient will also be a factor of f(x) with remainder =0.

So, 2x4 − 3x3 − 3x2 + 6x – 2 = (x2 – 2) (2x2 – 3x +1).

Now, on further factorizing (2x2 – 3x +1) we get,

2x2 – 3x +1 = 2x2 – 2x − x +1 = 0

2x (x − 1) – 1(x−1) = 0

(2x−1) (x−1) = 0

So, its zeroes are given by:  x= 12$\frac{1}{2}$ and x = 1

Therefore, all four zeroes of the given polynomial equation are:

2,2,12and1$\sqrt{2}\;,-\sqrt{2}\;,\frac{1}{2}\;and\;1$.

Q4. Find all zeroes of a polynomial equation x4-6x3-26x2+138x-35, if two of its zeroes are 2+3$2+{\sqrt{3}}$ and 23$2-\sqrt{3}$.

Sol.

Since this is a polynomial equation of degree 4, hence there will be total 4 roots.

Let, f(x) = x– 6x– 26x2+ 138x – 35

Since, (2+3)and(23)$(2+\sqrt{3})\;and\;(2-\sqrt{3})$ are zeroes of given polynomial f(x).

[x(2+3)][x(23)]=0$\;\;[x-(2+\sqrt{3})]\;\;[x-(2-\sqrt{3})]=0$

(x23)(x2+3)$(x-2-\sqrt{3})\;\;(x-2+\sqrt{3})$

On multiplying the above equation we get,

x2 −4x + 1, this is a factor of a given polynomial f(x).

Now, if we will divide f(x) by g(x), the quotient will also be a factor of f(x) and the remainder will be 0.

So, x4 − 6x− 26x2 + 138x – 35 = (x2 – 4x + 1) (x2 – 2x −35).

Now, on further factorizing (x2 – 2x −35) we get,

x2 – (7−5)x − 35 = x2 – 7x + 5x +35 = 0

x(x − 7) + 5 (x−7) = 0

(x+5) (x−7) = 0

So, its zeroes are given by:              x= −5 and x = 7.

Therefore, all four zeroes of given polynomial equation are:

2+3,23$2+{\sqrt{3}}\;, \; 2-\sqrt{3}$, −5 and 7.

Q5. Find all zeroes of a polynomial equation 3x4 + 6x3 – 2x2 – 10x – 5, if two of its zeroes are 53$\sqrt{\frac{5}{3}}$ and 53$-\sqrt{\frac{5}{3}}$.

Sol.

Since this is a polynomial equation of degree 4, hence there will be total 4 roots.

Let, f(x)= 3x4+6x3-2x2-10x-5

53$\sqrt{\frac{5}{3}}$ and 53$-\sqrt{\frac{5}{3}}$ are zeroes of polynomial f(x).

(x53)(x+53)$\;\;(x-\sqrt{\frac{5}{3}})\;\;(x+\sqrt{\frac{5}{3}})$

(x253)=0$(x^{2}-\frac{5}{3}) = 0$

(3x2−5)=0, is a factor of given polynomial f(x).

Now, when we will divide f(x) by (3x2−5) the quotient obtained will also be factor of f(x) and remainder will be 0.

Therefore, 3x4 + 6x3 − 2x2 − 10x – 5 = (3x2 – 5) (x2 + 2x +1).

Now, on further factorizing (x2 + 2x +1) we get,

x2 + 2x +1 = x2 + x + x +1 = 0

x(x + 1) + 1(x+1) = 0

(x+1) (x+1) = 0

So, its zeroes are given by: x= −1 and x = −1

Therefore, all four zeroes of given polynomial equation are:

53$\sqrt{\frac{5}{3}}$, 53$-\sqrt{\frac{5}{3}}$, −1 and −1.