NCERT Solutions for Exercise 2.4 Class 10 Maths Chapter 2 - Polynomials

Polynomials Class 10 Maths Chapter 2 Excercise: 2.4

Q1 (1) Verify that the numbers given alongside the cubic polynomials below are their zeroes.
Also verify the relationship between the zeroes and the coefficients in each case:
$2x^3 + x^2 - 5x +2 ; \frac{1}{2} , 1 , -2$

Answer:

p(x) = 2x ³ + x ² -5x + 2

$\\p(\frac{1}{2})=2\times \left ( \frac{1}{2} \right )^{3}+\left ( \frac{1}{2} \right )^{2}-5\times \frac{1}{2}+2\\ p(\frac{1}{2})=\frac{1}{4}+\frac{1}{4}-\frac{5}{2}+2\\ p(\frac{1}{2})=0$

p(1) = 2 x 1 ³ + 1 ² - 5 x 1 + 2

p(1) =2 + 1 - 5 + 2

p(1) = 0

p(-2) = 2 x (-2) ³ + (-2) ² - 5 x (-2) +2

p(-2) = -16 + 4 + 10 + 2

p(-2) = 0

Therefore the numbers given alongside the polynomial are its zeroes

Verification of relationship between the zeroes and the coefficients

Comparing the given polynomial with ax ³ + bx ² + cx + d, we have

a = 2, b = 1, c = -5, d = 2

The roots are $\alpha ,\beta \ and\ \gamma$

$\\\alpha=\frac{1}{2}\\ \beta =1\\ \gamma =-2$

$\\\alpha+\beta +\gamma \\ =\frac{1}{2}+1+(-2)\\ =-\frac{1}{2}\\ =-\frac{b}{a}$

Verified

$\\\alpha\beta +\beta \gamma +\gamma \alpha \\ =\frac{1}{2}\times 1+1\times (-2)+(-2)\times \frac{1}{2}\\ =\frac{-5}{2}\\ =\frac{c}{a}$

Verified

$\\\alpha\beta\gamma \\=\frac{1}{2}\times 1\times -2\\ =-1 \\=-\frac{2}{2}\\ =-\frac{d}{a}$

Verified

Q1 (2) Verify that the numbers given alongside the cubic polynomials below are their zeroes. Also, verify the relationship between the zeroes and the coefficients in each case:

$x ^ 3- 4x ^ 2 + 5x - 2 ; 2,1,1$

Answer:

p(x) = x ³ - 4x ² + 5x - 2

p(2) = 2 ³ - 4 x 2 ² + 5 x 2 - 2

p(2) = 8 - 16 + 10 - 2

p(-2) = 0

p(1) = 1 ³ - 4 x 1 ² + 5 x 1 - 2

p(1) = 1 - 4 + 5 - 2

p(1) = 0

Therefore the numbers given alongside the polynomial are its zeroes

Verification of relationship between the zeroes and the coefficients

Comparing the given polynomial with ax ³ + bx ² + cx + d, we have

a = 1, b = -4, c = 5, d = -2

The roots are $\alpha ,\beta \ and\ \gamma$

$\\\alpha=2\\ \beta =1\\ \gamma =1$

$\\\alpha+\beta +\gamma \\ =2+1+1\\ =4\\ =-\frac{-4}{1}\\=-\frac{b}{a}$

Verified

$\\\alpha\beta +\beta \gamma +\gamma \alpha \\ =2\times 1+1\times 1+1\times 2\\ =5\\ =\frac{5}{1}\\ =\frac{c}{a}$

Verified

$\\\alpha\beta\gamma \\=2\times 1\times 1\\=2 \\=-\frac{-2}{1}\\=-\frac{d}{a}$

Verified

Q2 Find a cubic polynomial with the sum, sum of the product of its zeroes taken two at a time, and the product of its zeroes as 2, –7, –14 respectively.

Answer: Let the roots of the polynomial be $\alpha ,\beta \ and\ \gamma$

$\\\alpha +\beta +\gamma =2\\ \alpha \beta +\beta \gamma +\gamma \alpha =-7\\ \alpha \beta \gamma =-14$

$\\x^{3}-(\alpha +\beta +\gamma )x^{2}+(\alpha \beta +\beta \gamma +\gamma \alpha)x-\alpha \beta \gamma =0\\ x^{3}-2x^{2}-7x+14=0$

Hence the required cubic polynomial is x ³ - 2x ² - 7x + 14 = 0

Q3 If the zeroes of the polynomial $x^3 - 3 x^2+ x +1$ are a – b, a, a + b, find a and b.

Answer:

$x^3 - 3 x^2+ x +1$

The roots of the above polynomial are a, a - b and a + b

Sum of the roots of the given plynomial = 3

a + (a - b) + (a + b) = 3

3a = 3

a = 1

The roots are therefore 1, 1 - b and 1 + b

Product of the roots of the given polynomial = -1

1 x (1 - b) x (1 + b) = - 1

1 - b ² = -1

b ² - 2 = 0

$b=\pm \sqrt{2}$

Therefore a = 1 and $b=\pm \sqrt{2}$ .

Q4 If two zeroes of the polynomial $x^4 - 6x ^3 - 26 x^2 + 138 x - 35$ $x^4 - 6x ^3 - 26 x^2 + 138 x - 35$ are $2 \pm \sqrt 3$ , find other zeroes .

Answer: Given the two zeroes are

$2+\sqrt{3}\ and\ 2-\sqrt{3}$

therefore the factors are

$[x-(2+\sqrt{3})]\ and[x-(\ 2-\sqrt{3})]$

We have to find the remaining two factors. To find the remaining two factors we have to divide the polynomial with the product of the above factors

$[x-(2+\sqrt{3})]\[x-(\ 2-\sqrt{3})]=x^2-(2+\sqrt{3})x-(\ 2-\sqrt{3})x+(\ 2+\sqrt{3})(\ 2-\sqrt{3})\\=x^2-2x-\sqrt{3}x-2x+\sqrt{3}x+1\\=x^2-4x+1$

Now carrying out the polynomial division

Now we get $x^2-2x -35 \ is \ also \ a\ factor$

$\\x^2-2x -35 =x^2-7x+5x-35\\=x(x-7)+5(x-7)\\=(x-7)(x+5)$

So the zeroes are $2+\sqrt{3}\ ,\ 2-\sqrt{3},7\ and\ -5$

Q5 If the polynomial $x ^4 - 6x ^3 + 16 x^2 - 35 x + 10$ is divided by another polynomial $x^2 - 2x + k$ , the remainder comes out to be x + a, find k and a.

Answer: The polynomial division is carried out as follows

Given the remainder =x+a

The obtained remainder after division is $(2k-9)x+10-k(8-k)$

now equating the coefficient of x

$2k-9=1$

which gives the value of $k=5$

now equating the constants

$a=10-k(8-k)=10-5(8-5)=-5$

Therefore k=5 and a=-5