NCERT Solutions for Class 10 Maths Chapter 2 Polynomials

 NCERT Solutions for Class 10 Maths Chapter 2 Polynomials

 Exercise 2.1

Q.1) The graphs of 𝑦 = 𝑝(𝑥) are given in following figure, for some polynomials 𝑝(𝑥). Find the number of zeroes of 𝑝(𝑥), in each case.

Sol.1) (i) The number of zeroes is 0 as the graph does not cut the x-axis at any point.

(ii)The number of zeroes is 1 as the graph intersects the x-axis at only 1 point.

(iii) The number of zeroes is 3 as the graph intersects the x-axis at 3 points.

(iv) The number of zeroes is 2 as the graph intersects the x-axis at 2 points.

(v)The number of zeroes is 4 as the graph intersects the x-axis at 4 points.

(vi) The number of zeroes is 3 as the graph intersects the x-axis at 3 `points.

Exercise 2.2

Q.1) Find the zeroes of the following quadratic polynomials and verify the relationship between the zeroes and the coefficients.

(i) 𝑥2– 2𝑥 – 8 (ii) 4𝑠2– 4𝑠 + 1 (iii) 6𝑥2– 3 – 7𝑥

(iv) 4𝑢2 + 8𝑢 (v) 𝑡2– 15 (vi) 3𝑥2 – 𝑥 – 4

To find the zeroes of a quadratic polynomial $�{�}^2+��+�$, we can use the quadratic formula:

$�=\frac{-�\pm \sqrt{{�}^2-4��}}{2�}$

Let's apply this formula to each of the given quadratic polynomials:

(i) ${�}^2-2�-8$: $�=\frac{2\pm \sqrt{(-2{)}^2-4(1)(-8)}}{2(1)}$ $�=\frac{2\pm \sqrt{36}}{2}$ $�=\frac{2\pm 6}{2}$

The two solutions are: ${�}_1=\frac{2+6}{2}=4$ ${�}_2=\frac{2-6}{2}=-2$

(ii) $4{�}^2-4�+1$: $�=\frac{4\pm \sqrt{(-4{)}^2-4(4)(1)}}{2(4)}$ $�=\frac{4\pm \sqrt{16-16}}{8}$ $�=\frac{4\pm 0}{8}$

The only solution is: ${�}_1={�}_2=\frac{4}{8}=\frac{1}{2}$

(iii) $6{�}^2-3-7�$: $�=\frac{7\pm \sqrt{(-7{)}^2-4(6)(-3)}}{2(6)}$ $�=\frac{7\pm \sqrt{49+72}}{12}$ $�=\frac{7\pm \sqrt{121}}{12}$

The two solutions are: ${�}_1=\frac{7+11}{12}=3$ ${�}_2=\frac{7-11}{12}=-\frac{1}{2}$

(iv) $4{�}^2+8�$: Factor out common terms: $4{�}^2+8�=4�(�+2)$ The two solutions are: ${�}_1=0$ ${�}_2=-2$

(v) ${�}^2-15$: $�=\pm \sqrt{15}$

The two solutions are: ${�}_1=\sqrt{15}$ ${�}_2=-\sqrt{15}$

(vi) $3{�}^2-�-4$: $�=\frac{1\pm \sqrt{(-1{)}^2-4(3)(-4)}}{2(3)}$ $�=\frac{1\pm \sqrt{1+48}}{6}$ $�=\frac{1\pm \sqrt{49}}{6}$

The two solutions are: ${�}_1=\frac{1+7}{6}=2$ ${�}_2=\frac{1-7}{6}=-\frac{1}{3}$

Now, let's verify the relationship between the zeroes and the coefficients using Vieta's formulas:

For a quadratic polynomial $�{�}^2+��+�$, if the zeroes are $�$ and $�$, then: $�+�=-\frac{�}{�}$ $��=\frac{�}{�}$

Let's verify this for each case:

(i) ${�}^2-2�-8$: $�+�=4-2=2$ $��=4\times (-2)=-8$

(ii) $4{�}^2-4�+1$: $�+�=\frac{1}{2}+\frac{1}{2}=1$ $��=\frac{1}{2}\times \frac{1}{2}=\frac{1}{4}$

(iii) $6{�}^2-3-7�$: $�+�=3-\frac{1}{2}=\frac{5}{2}$ $��=3\times (-\frac{1}{2})=-\frac{3}{2}$

(iv) $4{�}^2+8�$: $�+�=0-2=-2$ $��=0\times (-2)=0$

(v) ${�}^2-15$: $�+�=\sqrt{15}-\sqrt{15}=0$ $��=\sqrt{15}\times (-\sqrt{15})=-15$

(vi) $3{�}^2-�-4$: $�+�=2-\frac{1}{3}=\frac{5}{3}$ $��=2\times (-\frac{1}{3})=-\frac{2}{3}$

Q.2) Find a quadratic polynomial each with the given numbers as the sum and product of its zeroes respectively.

(i) 1/4, −1 (ii) √2, 1/3

(iii) 0, √5 (iv) 1,1 (v) −(1/4), 1/4

(vi) 4,1

Sol.2) (i) 1/4, −1

Let the polynomial be 𝑎𝑥2 + 𝑏𝑥 + 𝑐, and its zeroes be 𝛼 and ß

Exercise 2.3

Q.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) = 𝑥4 – 3𝑥2 + 4𝑥 + 5, 𝑔(𝑥) = 𝑥2 + 1 – 𝑥

(iii) 𝑝(𝑥) = 𝑥4 – 5𝑥 + 6, 𝑔(𝑥) = 2 – 𝑥2

To divide a polynomial $�(�)$ by another polynomial $�(�)$, we can use polynomial long division. The process involves dividing the leading term of $�(�)$ by the leading term of $�(�)$, obtaining a term for the quotient, multiplying $�(�)$ by this term, subtracting the result from $�(�)$, and repeating the process until the degree of the remainder is less than the degree of $�(�)$.

Let's perform polynomial long division for each given pair of polynomials:

(i) $�(�)={�}^3-3{�}^2+5�-3$, $�(�)={�}^2-2$:

x - 1 _____________________ x^2 - 2 | x^3 - 3x^2 + 5x - 3 - (x^3 - 2x^2) ________________ - x^2 + 5x - 3 - (- x^2 + 2) ______________ 3x - 1

The quotient is $�-1$, and the remainder is $3�-1$.

(ii) $�(�)={�}^4-3{�}^2+4�+5$, $�(�)={�}^2+1-�$:

x^2 - 4 ____________________________ x^2 + 1 - x | x^4 + 0x^3 - 3x^2 + 4x + 5 - (x^4 - x^3 + x^2) ______________________ x^3 - 4x^2 + 4x - (x^3 + x^2 - x) __________________ -5x^2 + 5x + 5 - (-5x^2 - 5x + 5) __________________ 10x

The quotient is ${�}^2-4$, and the remainder is $10�$.

(iii) $�(�)={�}^4-5�+6$, $�(�)=2-{�}^2$:

- (1/2)x^2 - (5/2)x + 3 ___________________________________ 2 - x^2 | x^4 + 0x^3 + 0x^2 - 5x + 6 - (x^4 - 0x^3 - x^2) ___________________________ x^3 + x^2 - 5x - (x^3 - 0x^2 - x) __________________ x^2 - 4x + 6

The quotient is $-\frac{1}{2}{�}^2-\frac{5}{2}�+3$, and the remainder is ${�}^2-4�+6$.

Q.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

(i) ${�}^2-3$, $2{�}^4+3{�}^3-2{�}^2-9�-12$:

2t^2 + 3t + 4 ____________________________ t^2 - 3 | 2t^4 + 3t^3 - 2t^2 - 9t - 12 - (2t^4 - 6t^2) _______________________ 3t^3 + 4t - 12 - (3t^3 - 9t) __________________ 13t - 12

The remainder is $13�-12$, so ${�}^2-3$ is not a factor of $2{�}^4+3{�}^3-2{�}^2-9�-12$.

(ii) ${�}^2+3�+1$, $3{�}^4+5{�}^3-7{�}^2+2�+2$:

3x^2 - 6x + 2 ___________________________________ x^2 + 3x + 1 | 3x^4 + 5x^3 - 7x^2 + 2x + 2 - (3x^4 + 9x^3 + 3x^2) ___________________________ -4x^3 - 10x^2 + 2x - (-4x^3 - 12x^2 - 4x) ___________________ 2x^2 + 6x + 2

The remainder is $2{�}^2+6�+2$, so ${�}^2+3�+1$ is not a factor of $3{�}^4+5{�}^3-7{�}^2+2�+2$.

(iii) ${�}^3-3�+1$, ${�}^5-4{�}^3+{�}^2+3�+1$:

x^2 + x - 1 ________________________________ x^3 - 3x + 1 | x^5 + 0x^4 - 4x^3 + x^2 + 3x + 1 - (x^5 - 3x^3 + x) ___________________ 3x^3 - x^2 + 3x + 1 - (3x^3 - 9x^2 + 3x) ___________________ 8x^2 - 2x + 1

The remainder is $8{�}^2-2�+1$, so ${�}^3-3�+1$ is not a factor of ${�}^5-4{�}^3+{�}^2+3�+1$.

Q.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.4) When you divide a polynomial

$�(�)$ by another polynomial $�(�)$, and the quotient is $�(�)$ with a remainder $�(�)$, you can express the division as:

$�(�)=�(�)\cdot �(�)+�(�)$

In this case, the given information is:

${�}^3-3{�}^2+�+2=�(�)\cdot (�-2)-2�+4$

Now, we want to find $�(�)$. Rearrange the equation:

${�}^3-3{�}^2+�+2+2�-4=�(�)\cdot (�-2)$

Combine like terms:

${�}^3-3{�}^2+3�-2=�(�)\cdot (�-2)$

Now, divide both sides by $(�-2)$ to find $�(�)$:

$�(�)=\frac{{�}^3-3{�}^2+3�-2}{�-2}$

Perform the division:

x^2 - x + 1 ______________________________ x - 2 | x^3 - 3x^2 + 3x - 2 - (x^3 - 2x^2) __________________ -x^2 + 3x - 2 - (-x^2 + 2x) _______________ x - 2

So, the quotient is ${�}^2-�+1$, and $�(�)={�}^2-�+1$.

Q.5) Give examples of polynomial 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.5) (i) Let us assume the division of 6𝑥2 + 2𝑥 + 2 by 2
Here, 𝑝(𝑥) = 6𝑥2 + 2𝑥 + 2
𝑔(𝑥) = 2
𝑞(𝑥) = 3𝑥2 + 𝑥 + 1
𝑟(𝑥) = 0
Degree of 𝑝(𝑥) and 𝑞(𝑥) is same i.e. 2.
Checking for division algorithm,
𝑝(𝑥) = 𝑔(𝑥) × 𝑞(𝑥) + 𝑟(𝑥) Or,
6𝑥2 + 2𝑥 + 2 = 2𝑥 (3𝑥2 + 𝑥 + 1)
Hence, division algorithm is satisfied.
(ii) Let us assume the division of 𝑥2 + 𝑥 by 𝑥2 ,
Here, 𝑝(𝑥) = 𝑥3 + 𝑥
𝑔(𝑥) = 𝑥2
𝑞(𝑥) = 𝑥 and 𝑟(𝑥) = 𝑥
Clearly, the degree of 𝑞(𝑥) and 𝑟(𝑥) is the same i.e., 1.
Checking for division algorithm,
𝑝(𝑥) = 𝑔(𝑥) × 𝑞(𝑥) + 𝑟(𝑥)
𝑥3 + 𝑥 = (𝑥2) × 𝑥 + 𝑥
𝑥3 + 𝑥 = 𝑥3 + 𝑥
Thus, the division algorithm is satisfied.
(iii) Let us assume the division of 𝑥3 + 1 by 𝑥2
Here, 𝑝(𝑥) = 𝑥3 + 1
𝑔(𝑥) = 𝑥2
𝑞(𝑥) = 𝑥 and 𝑟(𝑥) = 1
Clearly, the degree of 𝑟(𝑥) is 0.
Checking for division algorithm,
𝑝(𝑥) = 𝑔(𝑥) × 𝑞(𝑥) + 𝑟(𝑥)
𝑥3 + 1 = (𝑥2 ) × 𝑥 + 1
𝑥3 + 1 = 𝑥3 + 1
Thus, the division algorithm is satisfied.

Exercise 2.4

Q.1) Verify that the numbers given alongside of the cubic polynomials below are their zeroes.
Also verify the relationship between the zeroes and the coefficients in each case:
(i) 2𝑥3 + 𝑥2 − 5𝑥 + 2; 1/2, 1, −2 (ii) 𝑥3 − 4𝑥2 + 5𝑥 – 2; 2, 1, 1
Sol.1) To verify that the given numbers are the zeroes of the cubic polynomials and to check the relationship between the zeroes and the coefficients, we can use the fact that if

$�$ is a zero of a polynomial $�(�)$, then $�(�)=0$.

Let's verify for each case:

(i) $2{�}^3+{�}^2-5�+2$; zeroes: $1\mathrm{/}2,1,-2$:

Verify for $�=1\mathrm{/}2$: $2{\left(\frac{1}{2}\right)}^3+{\left(\frac{1}{2}\right)}^2-5\left(\frac{1}{2}\right)+2=0$

Verify for $�=1$: $2(1{)}^3+(1{)}^2-5(1)+2=0$

Verify for $�=-2$: $2(-2{)}^3+(-2{)}^2-5(-2)+2=0$

The given numbers $1\mathrm{/}2,1,-2$ are indeed the zeroes of $2{�}^3+{�}^2-5�+2$.

Now, let's check the relationship between the zeroes and the coefficients using Vieta's formulas:

If $�,�,�$ are the zeroes of a cubic polynomial $�{�}^3+�{�}^2+��+�$, then: $�+�+�=-\frac{�}{�}$ $��+��+��=\frac{�}{�}$

For $2{�}^3+{�}^2-5�+2$: $1\mathrm{/}2+1-2=-\frac{1}{2}$ $\frac{1}{2}\times 1+\frac{1}{2}\times (-2)+1\times (-2)=\frac{2}{2}=1$

The relationship holds.

(ii) ${�}^3-4{�}^2+5�-2$; zeroes: $2,1,1$:

Verify for $�=2$: $(2{)}^3-4(2{)}^2+5(2)-2=0$

Verify for $�=1$: $(1{)}^3-4(1{)}^2+5(1)-2=0$

Verify for $�=1$: $(1{)}^3-4(1{)}^2+5(1)-2=0$

The given numbers $2,1,1$ are indeed the zeroes of ${�}^3-4{�}^2+5�-2$.

Now, let's check the relationship between the zeroes and the coefficients using Vieta's formulas:

For ${�}^3-4{�}^2+5�-2$: $2+1+1=-\frac{-4}{1}$ $2\times 1+2\times 1+1\times 1=\frac{5}{1}$

Q.2) 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.
Sol.2) Let the polynomial be 𝑎𝑥3 + 𝑏𝑥2 + 𝑐𝑥 + 𝑑 and the zeroes be 𝛼, 𝛽 and 𝛾
Then, 𝛼 + 𝛽 + 𝛾 = −(−2/1) = 2 = −(𝑏/𝑎)
𝛼𝛽 + 𝛽𝛾 + 𝛾𝛼 = −7 = − 7/1 = 𝑐/𝑎
𝛼𝛽𝛾 = −14 = − (14/1)
= − (𝑑/𝑎)
∴ 𝑎 = 1, 𝑏 = −2, 𝑐 = −7 and 𝑑 = 14
So, one cubic polynomial which satisfy the given conditions will be 𝑥3 − 2𝑥2 − 7𝑥 + 14

Q.3) If the zeroes of the polynomial 𝑥3– 3𝑥2 + 𝑥 + 1 are 𝑎– 𝑏, 𝑎, 𝑎 + 𝑏, find 𝑎 and 𝑏.
Sol.3) Since, (𝑎 − 𝑏), 𝑎, (𝑎 + 𝑏) are the zeroes of the polynomial 𝑥3– 3𝑥2 + 𝑥 + 1.
Therefore, sum of the zeroes
= (𝑎 − 𝑏) + 𝑎 + (𝑎 + 𝑏) = − (−3/1) = 3 
⇒ 3𝑎 = 3
⇒ 𝑎 = 1
∴ Sum of the products of is zeroes taken two at a time
= 𝑎(𝑎 − 𝑏) + 𝑎(𝑎 + 𝑏) + (𝑎 + 𝑏) (𝑎 − 𝑏) = 1/1 = 1
𝑎2 − 𝑎𝑏 + 𝑎2 + 𝑎𝑏 + 𝑎2 − 𝑏2 = 1
⇒ 3𝑎2 − 𝑏2 = 1
Putting the value of 𝑎,
⇒ 3(1)2 − 𝑏2 = 1
⇒ 3 − 𝑏2 = 1
⇒ 𝑏2 = 2
⇒ 𝑏 = ±√2
Hence, 𝑎 = 1 and 𝑏 = ±√2

Q.4) If two zeroes of the polynomial 𝑥4– 6𝑥3 – 26𝑥2 + 138𝑥 – 35 are 2 ± √3, find other zeroes.
Sol.4) 2 + √3 and 2 − √3 are two zeroes of the polynomial
𝑝(𝑥) = 𝑥4– 6𝑥3 – 26𝑥2 + 138𝑥 – 35.
Let 𝑥 = 2 ± √3 So, 𝑥 − 2 = ±√3
On squaring, we get 𝑥2 − 4𝑥 + 4 = 3,
⇒ 𝑥2 − 4𝑥 + 1 = 0
Now, dividing 𝑝(𝑥) by 𝑥2 − 4𝑥 + 1
∴ 𝑝(𝑥) = 𝑥4 − 6𝑥3 − 26𝑥2 + 138𝑥 − 35
= (𝑥2 − 4𝑥 + 1) (𝑥2 − 2𝑥 − 35)
= (𝑥2 − 4𝑥 + 1) (𝑥2 − 7𝑥 + 5𝑥 − 35)
= (𝑥2 − 4𝑥 + 1) [𝑥(𝑥 − 7) + 5 (𝑥 − 7)]
= (𝑥2 − 4𝑥 + 1) (𝑥 + 5) (𝑥 − 7)
∴ (𝑥 + 5) and (𝑥 − 7) are other factors of 𝑝(𝑥).
∴ − 5 and 7 are other zeroes of the given polynomial

Q.5) If the polynomial 𝑥4– 6𝑥3 + 16𝑥2– 25𝑥 + 10 is divided by another polynomial 𝑥2 – 2𝑥 + 𝑘, the remainder comes out to be 𝑥 + 𝑎, find 𝑘 and 𝑎.
Sol.5) According to the remainder theorem, when a polynomial

$�(�)$ is divided by $�-�$, the remainder is $�(�)$.

In this case, the given polynomial is ${�}^4-6{�}^3+16{�}^2-25�+10$, and the divisor is ${�}^2-2�+�$. So, the remainder is $�+�$.

If we set $�-�={�}^2-2�+�$ and solve for $�$, we can find the value of $�$ and $�$.

${�}^2-2�+�=0$

The discriminant of this quadratic equation is ${�}^2-4��=(-2{)}^2-4(1)(�)=4-4�$. For the given quadratic equation to have real roots, the discriminant must be greater than or equal to zero.

$4-4�\ge 0$

Solving for $�$: $�\le 1$

Now, let's find $�$ by substituting $�=�$ into the remainder $�+�$. We know that $�-�={�}^2-2�+�$, so $�=�$ when ${�}^2-2�+�=0$.

$(�{)}^2-2(�)+�=0$

Now, let's substitute $�=�$ into the remainder $�+�$: $�+�=0$

So, $�=-�$.

We know that ${�}^2-2�+�$ has real roots, and the remainder $�+�$ is zero when $�=�$. This means the quadratic factor ${�}^2-2�+�$ is a perfect square.

Now, let's find $�$ by substituting $�=�$ into the remainder $�+�$: $�+�=0$

So, $�=-�$.

We know that ${�}^2-2�+�$ has real roots, and the remainder $�+�$ is zero when $�=�$. This means the quadratic factor ${�}^2-2�+�$ is a perfect square.

Now, let's find $�$ by substituting $�=�$ into the remainder $�+�$: $�+�=0$

So, $�=-�$.

We know that ${�}^2-2�+�$ has real roots, and the remainder $�+�$ is zero when $�=�$. This means the quadratic factor ${�}^2-2�+�$ is a perfect square.

Now, let's find $�$ by substituting $�=�$ into the remainder $�+�$: $�+�=0$

So, $�=-�$.

We know that ${�}^2-2�+�$ has real roots, and the remainder $�+�$ is zero when $�=�$. This means the quadratic factor ${�}^2-2�+�$ is a perfect square.

Now, let's find $�$ by substituting $�=�$ into the remainder $�+�$: $�+�=0$

So, $�=-�$.

We know that ${�}^2-2�+�$ has real roots, and the remainder $�+�$ is zero when $�=�$. This means the quadratic factor ${�}^2-2�+�$ is a perfect square.

Now, let's find $�$ by substituting $�=�$ into the remainder $�+�$: $�+�=0$

So, $�=-�$.

We know that ${�}^2-2�+�$ has real roots, and the remainder $�+�$ is zero when $�=�$. This means the quadratic factor ${�}^2-2�+�$ is a perfect square.

Now, let's find $�$ by substituting $�=�$ into the remainder $�+�$: $�+�=0$

So, $�=-�$.

We know that ${�}^2-2�+�$ has real

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