# NCERT Solutions for Class 10 Maths Chapter 2 - Polynomials

## NCERT Solutions for Class 10 Mathematics Chapter 2 Free PDF Download

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Exercise 2.1

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

Sol. (i) The graph do not intersects the x-axis.
Then Number of zeroes = No zeroes.
(ii) The graph intersects the x-axis at one point only.
Then Number of zeroes = 1
(iii) The graph intersects the x-axis at three points.
Then Number of zeroes = 3
(iv) The graph intersects the x-axis at two points.
Then Number of zeroes = 2
(v) The graph intersects the x-axis at four points.
Then Number of zeroes = 4
(vi) The graph intersects the x-axis at three points.
Then Number of zeroes = 3

Exercise 2.2

1. Find the zeroes of the following quadratic polynoials and verify the relationship between the zeroes and the coefficients.
(i) x2 – 2x – 8
(ii) 4s2 – 4s + 1
(iii) 6x2 – 3 – 7x
(vi) 4u2 + 8u
(v) t2 – 15
(vi) 3x2 – x – 4

Sol. (i) Let be
F(x) = x2 – 2x – 8
F(x) = x2 – 4x + 2x – 8
F(x) = x(x – 4) + 2(x – 4)
F(x) = (x – 4)(x + 2)
So, the value of x2 – 2x – 8 is zero.
F(x) = 0
(x – 4)(x + 2) = 0
When
x – 4 = 0, x = 4
x + 2 = 0, x = – 2
Therefore, the zeroes of x2 – 2x – 8 are 4
and – 2
Now,
Sum of zeroes = 4 + (– 2)
= 2

$$\text{Sum of zeroes}=\frac{−(coefficient of x)}{−(coefficient of x^2)}\\=\frac{-(-2)}{1}\\=2\\\text{Product of zeroes} = 4 × – 2 = – 8\\\text{Product of zeroes}=\frac{\text{constant term}}{\text{termcoefficient of}\space x^2}\\=\frac{-8}{1}\\=-8$$

(ii) Let be F(s) = 4s2 – 4s + 1
F(s) = 4s2 – 2s – 2s + 1
= 2s(2s – 1) – 1(2s – 1)
= (2s – 1)(2s – 1) The value of F(s) = 0
(2s – 1)(2s – 1) = 0

$$(2s – 1) = 0, s =\frac{1}{2}\\(2s – 1) = 0, s =\frac{1}{2}\\\text{Therefore, the zeroes of}\space4s^2– 4s + 1\space are\space and\space\frac{1}{2}\\\text{Sum of zeroes} =\frac{1}{2}+\frac{1}{2}=1\\\text{Sum of zeroes}=\frac{−(coefficient of S)}{(coefficient of S^2)}\\=\frac{-(-4)}{4}=1\\\text{Product of zeroes}=\frac{1}{2}×\frac{1}{2}=\frac{1}{4}\\\text{Product of zeroes}=\frac{\text{constant term}}{\text{coefficient of} \space S^2}\\=\frac{1}{4}\\\text{Relationship between the zeroes and the coefficients is true.}$$

(iii) Let be
F(x) = 6x2 – 3 – 7x
⇒ F(x) = 6x2 – 7x – 3
⇒ F(x) = 6x2 – 9x + 2x – 3
⇒ F(x) = 3x(2x – 3) + 1(2x – 3)
⇒ F(x) = (2x – 3)(3x + 1)
Value of F(x) = 0
⇒ (2x – 3)(3x + 1) = 0
⇒ 2x – 3 = 0, 3x + 1 = 0
⇒ 2x = 3, 3x = – 1

$$x=\frac{3}{2},\space\space\space\space\space\space\space\space\space\space x=\frac{-1}{3},\\\text{Therefore, the zeroes of}\space 6x^2 – 3 – 7x\space\text{are}\frac{3}{2}\space\text{and}\space\frac{-1}{3}\\\text{Sum of zeroes}=\frac{3}{2}+(\frac{-1}{3})\\=\frac{9-2}{6}=\frac{7}{6}\\\text{Sum of zeroes}=\frac{\text{−(coefficient of x)}}{\text{(coefficient of}\space x^2)}\\=\frac{-(-7)}{6}\\=\frac{7}{6}\\\text{Product of zeroes}=\frac{3}{2}×\frac{-1}{3}\\=-\frac{1}{2}\\\text{Relationship between the zeroes and the coefficients is true.}$$

(iv) Let be
F(u) = 4u2 + 8u
⇒ F(u) = 4u (u + 2)
The value of F(u) = 0
⇒ 4u(u + 2) = 0
⇒ 4u (u + 2) = 0
⇒ 4u = 0, u + 2 = 0
⇒ u = 0, u = – 2
Therefore, the zeroes of 4u2 + 8u are 0 and – 2.
Now
sum of zeores = 0 + (– 2) = – 2

$$\text{Sum of zeroes}=\frac{\text{−(coefficient of u)}}{\text{(coefficient of}\space u^2)}\\=\frac{8}{4}=-2\\\text{Sum of zeroes}= 0 × – 2 = 0\\\text{Sum of zeroes}=\frac{\text{constant term}}{\text{term coefficient of}\space u^2}\\=\frac{0}{4}=0\\\text{Relationship between the zeroes and the coefficients is true.}$$

(v) Let be
F(t) = t2 – 15
The value of F(t) = 0
t2 – 15 = 0
(t2- 152)=0
(t- 15)(t+ 15)=0
Then,
t− 15 = 0
t+ 15 = 0
t = 15
t = -15
Therefore, the zeroes of t2 – 15 are 15 and −15.
Sum of zeroes = 15−15= 0
$$\text{Sum of zeroes}=\frac{\text{−(coefficient of t)}}{\text{(coefficient of}\space t^2)}\\=\frac{-0}{1}=0\\\text{Product of zeroes}= \sqrt[]{15}×-\sqrt[]{15}=-15\\\text{Product of zeroes}=\frac{\text{constant term}}{\text{coefficient of}\space t^2}\\=\frac{-15}{1}=-15\\\text{Relationship between the zeroes and the coefficients is true.}$$

(vi) Let be
F(x) = 3x2 – x – 4
F(x) = 3x2 – 4x + 3x – 4
F(x) = x(3x – 4) + 1(3x – 4)
F(x) = (3x – 4)(x + 1)
So, the value of F(x) is zero.
F(x) = 0
(3x – 4)(x + 1) = 0
3x – 4 = 0, x + 1 = 0
3x = 4, x = – 1
x=4/3
$$\text{Therefore, the zeroes of the}\space 3x^2 – x – 4\text{are}\frac{4}{3}\text{and}-1\\\text{Sum of zeroes}=\frac{4}{3}+(-1)\\\frac{4-3}{3}\\=\frac{1}{3}\\\text{Sum of zeroes}=\frac{\text{−(coefficient of x)}}{\text{(coefficient of}\space x^2)}\\=\frac{-(-1)}{3}=\frac{1}{3}\\\text{Product of zeroes}=\frac{4}{3}×-1\\=\frac{-4}{3}\\\text{Product of zeroes}=\frac{\text{(coefficient of x)}}{\text{(coefficient of}\space x^2)}\\\text{Product of zeroes}=\frac{-4}{3}\\\text{Relationship between zeroes and coefficients are true.}$$

2. Find a quadratic polynomial each with the given numbers as the sum and product of its zeroes respectively.
$$(i)\frac{1}{4},-1\\(ii)\sqrt{2},\frac{1}{3}\\(iii)0,\sqrt{5}\\(iv)1,1\\(v)\frac{-1}{4},\frac{1}{4}\\(vi)4,1$$

Sol. (i) Given that,
Sum of zoeres = 1/4
Product of zeroes = – 1
x2 – (sum of zeroes)x + product of zeroes = 0
$$x^2-\frac{1}{4}×x+(-1)=0\\\frac{4x^2-x-4}{4}=0\\4x^2-x-4=0\\\text{Hence, the Quadratic polynomial is}\\F(x) =4x^2-x-4.$$

$$(ii)\text{Given that,}\\\text{Sum of zeroes}=\sqrt{2}\space\text{and}\\\text{Product of zeroes}=\frac{1}{3}\\\text{Quadratic polynomial is :}\\x^2 –\text{(sum of zeroes)x + product of zeroes = 0}\\x^2-\sqrt{2x}+\frac{1}{3}=0\\{3x^2-3\sqrt{2x}+1=0}\\\text{Hence, the polynomial is :}\\F(x) ={3x^2-3\sqrt{2x}+1=0}$$

(iii) Given that,
Sum of zeroes = 0
Product of zeroes =  5
x2 – (sum of zeroes)x + product of zeroes = 0
x2-0 × X+ 5 0
x2+ 5=0
Hence, the polynomial is
F(x) =x2+ 5

(iv) Given that,
Sum of zeroes = 1
Product of zeroes = 1
x2 – (sum of zeroes)x + product of zeroes = 0
x2 – x + 1 = 0
F(x) = x2 – x +1

$$(v) \text{Given that,}\\\text{Sum of zeroes}=\frac{-1}{4}\\\text{Product of zeroes}=\frac{1}{4}\\\text{Quadratic polynomial is}\\\text{x2 – (sum of zeroes)x + product of zeroes = 0}\\x^2-(\frac{-1}{4})x+\frac{1}{4}=0\\x^2+\frac{x}{4}+\frac{1}{4}=0\\\frac{4x^2+x+1}{4}=0\\4x^2 + x + 1 = 0\\4x^2 + x + 1 = 0\\\text{Hence, the polynomial is}\\ F(x) = 4x^2 + x + 1$$

(vi) Given that,
Sum of zeroes = 4
Product of zeroes = 1
x2 – (sum of zeroes)x + product of zeroes = 0
x2 – 4x + 1 = 0
Hence, the polynomial is
F(x) = x2 – 4x + 1.

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) = x4 – 3x2 + 4x + 5, g(x) = x2 + 1 – x
(iii) p(x) = x4 – 5x + 6, g(x) = 2 – x2

Sol. (i) P(x) = x3 – 3x2 + 5x – 3, g(x) = x2 – 2
x2 – 2 ) x3 – 3x2 + 5x – 3 (x – 3
x3 – 2x
$$x^3-2x\\\frac{-\space\space\space\space\space\space\space +}{– 3x^2 + 7x – 3}\text{By subtract}\\– 3x^2 + 6\\\frac{+\space\space\space\space\space\space\space -}{7x – 9}\text{By subtract}$$
Quotient q(x) = x – 3
Remainder R(x) = 7x – 9

(ii) p(x) = x4 – 3x2 + 4x + 5
g(x) = x2 + 1 – x
Rearranging g(x)
g(x) = x2 – x + 1
x2 – x + 1 ) x4 – 3x2 + 4x + 5 (x2 + x – 3
x4 – x3 + x2

$$\frac{-\space\space\space+\space\space\space\space -}{x^3 – 4x^2 + 4x + 5}\text{By subtract}\\x^3 – x^2 + x\\\frac{-\space\space\space\space+\space\space\space\space -}{– 3x^2 + 3x + 5}\\– 3x^2 + 3x + 5\\\frac{+\space\space\space\space-\space\space\space\space +}{8}\text{By subtract}\\\text{Quotient q(x)} = x^2 + x – 3\\\text{Remainder r(x)} = 8$$

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

$$\frac{- \space\space\space\space+}{2x^2–5x+6}\\2x^2–4\\\frac{-\space\space\space\space +}{–5x+10}\\\text{Quotient q(x)} = – x^2 – 2\\\text{Remainder R(x)} = – 5x + 10$$

2. Check whether the first polynomial is a factor of the second polynoial 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) Let be
I Polynomial p(t) = t2 – 3
II Polynomial q(t) = 2t4 + 3t3 – 2t2 – 9t – 12
t2 – 3) 2t4 + 3t3 – 2t2 – 9t – 12 (2t2 + 3t + 4
2t4– 6t

$$\frac{- \space\space\space\space+}{3t^3 + 4t^2 – 9t – 12}\\3t^3 – 9t\\\frac{-\space\space\space\space +}{4t^2 – 12}\\4t^2 – 12\\\frac{-\space\space\space\space +}{0}\\\text{Remainder }= 0\\\text{Then First polynomial is a factor of the second polynomial.}$$

(ii) Let be
I polynomial p(x) = x2 + 3x + 1
II polynomial q(x) = 3x4 + 5x3 – 7x2 + 2x + 2
x2 + 3x + 1) 3x4 + 5x3 – 7x2 + 2x + 2 (3x2 – 4x + 2
3x4 + 9x3 + 3x

$$\frac{- \space\space-\space\space-}{-4x^3 + 10x^2 +2x +2}\\-4x^3 – 12x^2-4x\\\frac{-\space\space-\space\space -}{2x^2 +6x+2}\\2x^2 +6x+2\\\frac{-\space\space-\space\space -}{0}\\\text{Remainder }= 0\\\text{Therefore First polynomial is a factor of the second polynomial.}$$

(iii) Let be
I Polynomial p(x) = x3 – 3x + 1
II Polynomial q(x) = x5 – 4x3 + x2 + 3x + 1
x3 – 3x + 1) x5 – 4x3 + x2 + 3x + 1 (x2 – 1
x5 – 3x3 + x

$$\frac{- \space\space+\space\space-}{-x^3 +3x +1}\\-x^3 +3x+1\\\frac{+\space\space-\space\space +}{2}\\\text{Remainder }= 2\\\text{Therefore,}\\\text{ First polynomial is not a factor of the second polynomial.}$$

3. Obtain all other zoroes of 3x4 + 6x3 – 2x2 – 10x – 5, if two of its zeroes are √5/3 and -√ 5/3

$$\text{Sol. Since two zeroes are}\space\sqrt{\frac{5}{3}}\space\text{and}-\space\sqrt{\frac{5}{3}}\\\Biggl(x-\sqrt{\frac{5}{3}}\Biggr)\Biggl(x+\sqrt{\frac{5}{3}}\Biggr)\\=x^2-\frac{5}{3}\\\text{is a factor of the given polynomial. Now, we divide the given polynomial by}\space x^2-\frac{5}{3}\\ x^2-\frac{5}{3}\Biggr)3x^4+6x^3-2x^2-10x-5\Biggl(3x^2+6x+3\\3x^2-5x^2\\\frac{- \space\space\space\space+}{6x^3 +3x^2 -10x-5}\\6x^3 -10x\\\frac{-\space\space\space\space +}{3x^2-5}\\3x^2-5\\\frac{- +}{0}\\\text{so,}3x^4 + 6x^3 – 2x^2 – 10x – 5\\=\Biggl(x^2-\frac{5}{3}\Biggr)(3x^2+6x+3)\\\text{We factorise}(3x^2+6x+3)\\3x^2 + 6x + 3 = 3x^2 + 3x + 3x + 3\\= 3x (x + 1) + 3(x + 1)\\= (x + 1)(3x + 3)\\\text{So, its zeroes are given by x = – 1 and x = – 1. Therefore, the zeroes of the given} \\\text{polynomial are}\space\sqrt{\frac{5}{3}},-\sqrt{\frac{5}{3}}\space– 1 and – 1 .$$

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. We know that
Dividend = Divisor × Quotient + Remainder
Given
Dividend = x3 – 3x2 + x + 2 Divisor = g(x) = ?
Quotient = x – 2
Remainder = – 2x + 4
Then
x3 – 3x2 + x + 2 = g(x) × (x – 2) + (– 2x + 4)
x3 – 3x2 + x + 2 = g(x) × (x – 2) – 2x + 4
g(x)(x – 2) = x3 – 3x2 + x + 2 + 2x – 4
g(x) = x3 – 3x2 + 3x – 2

$$g(x)=\frac{x^3-3x^2+3x-2}{x-2}\\x – 2) x^3 – 3x^2 + 3x – 2 (x^2 – x + 1\\x^3-2x^2\\\frac{-\space\space\space\space+}{-x^2 +3x-2}\\-x^2 +2x\\\frac{+\space\space\space\space -}{3x^2-5}\\3x^2-5\\\frac{+ -}{x-2}\\x-2\\\frac{- +}{0}\\g(x) = x^2 – x + 1$$

5. Given 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. (i) p(x) = 3x3+ 6x2 + 9x + 12
g(x) = 3
3) 3x3 + 6x2 + 9x + 12 (x3 + 2x2 + 3x + 4
3x2

$$\frac{-}{6x^2 +3x+12}\\6x^2\\\frac{-}{9x+12}\\9x\\\frac{-}{12}\\12\\\frac{-}{0}\\q(x) = x^3 + 2x^2 + 3x + 4, r(x) = 0\\\text{deg p(x) = deg q(x)}$$

(ii) p(x) = x4 + x3 + x + 4, g(x) = x3
x3 ) x4 + x3 + x + 4 ( x + 1
x

$$\frac{-}{x^3 +x+4}\\x^3\\\frac{-}{x+4}\\q(x) = x + 1, r(x) = x + 4\\\text{deg q(x) = deg r(x)}$$

(iii) p(x) = x4 + 2, g(x) = x + 1
x + 1 ) x4 + 2 ( x3 – x2 + x – 1
x4 + x

$$\frac{-\space-}{-x^3 +2}\\-x^3-x^2\\\frac{+\space-}{x^2+2}\\x^2+x\\\frac{-\space-}{-x+2}\\-x-1\\\frac{+ +}{3}\\q(x) = x3 – x2 + x – 1\\r(x) = 3\\\text{deg r(x) = 0}$$

Play Video about Chapter 2 Polynomials

Exercise 2.4

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; 1212,,−
(ii) x3 – 4x2 + 5x – 2; 2, 1, 1

Sol. (i) Let
P(x) = 2x3 + x2 – 5x + 2

$$p\biggl(\frac{1}{2}\biggr)=2×\biggl(\frac{1}{2}\biggr)^3+\biggl(\frac{1}{2}\biggr)^2-5×\frac{1}{2}+2\\=2×\frac{1}{8}+\frac{1}{4}-\frac{5}{2}+2\\=\frac{1}{2}-\frac{5}{2}\\=0\\P(1) = 2 × 1^3 + 1^2 – 5 × 1 + 2\\= 2 × 1 + 1 – 5 + 2\\= 5 – 5\\=0\\P(– 2) = 2 × (– 2)^3 + (– 2)^2 – 5 × (– 2) + 2\\= – 16 + 4 + 10 + 2\\= – 16 + 16\\= 0\\Hence,\frac{1}{2},1\text{and – 2 are zeroes of polynomial P(x). According to question,}\\a = 2, b = 1, c = – 5, d = 2\\\text{Zeroes}, α =\frac{1}{2},β= 1, γ = – 2.\\\text{Therefore,}\\α + β + γ =\frac{-b}{a}\\\frac{1}{2}+1-2=\frac{-1}{2}\\\frac{-1}{2}=\frac{-1}{2}\\αβ + βγ + γα =\frac{c}{a}\\\frac{1}{2}×1+1×(-2)+(-2)×\frac{1}{2}=\frac{-5}{2}\\\frac{1}{2}-2-1=\frac{-5}{2}\\\frac{1}{2}-3=\frac{-5}{2}\\\frac{-5}{2}=\frac{-5}{2}\\\text{And,}\\αβγ =\frac{-d}{a}\\\frac{1}{2}×1×(-2)=\frac{-2}{2}\\– 1 = – 1\\\text{Relationship between the zeroes and the coefficients in each case is true.}$$

(ii) Let P(x) = x3 – 4x2 + 5x – 2
P(2) = 23 – 4 × 22 + 5 × 2 – 2
P(2) = 8 – 16 + 10 – 2
P(2) = 0
P(1) = 13 – 4 × 12 + 5 × 1 – 2
P(1) = 1 – 4 + 5 – 2
= 0
Hence, 2, 1 and 1 are zeroes of polynomial P(x).
According to question,
a = 1, b = – 4, c = 5, d = – 2
α= 2, β = 1, γ = 1
Now

$$α + β + γ =\frac{-b}{a}\\2 + 1 + 1 =\frac{-(-4)}{1}\\4 = 4\\\text{Again}\\αβ + βg + γα=\frac{c}{a}\\2 × 1 + 1 × 1 + 1 × 2=\frac{5}{1}\\2 + 1 + 2\\5 = 5\\\text{And,}\\αβγ =\frac{-d}{a}\\2 × 1 ×1 =\frac{-(-2)}{1}\\2 = 2\\\text{Relationship betw een the zeroes and the coefficients in each case is true.}$$

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. Let be cubic polynomial
P(x) = ax3 + bx2 + cx + d and its zeroes are α,βandγ .
Then according to question
α + β +γ = 2
$$\frac{-b}{a}=2\\a = 1, b = – 2\\αβ + βγ + γα =-7\\\frac{c}{a}=-7\\c = – 7a\\a = 1, c = – 7\\αβγ = – 14\\\frac{-d}{a}=-14\\d = 14\\\text{Put the value of a, b, c and d in polynomial P(x).}\\P(x) = x^3 – 2x^2 – 7x + 14$$

3. If the zeroes of the polynomial x3 – 3x2 + x + 1 are a – b, a, a + b find a and b.

Sol. Given Polynomial
P(x) = x3 – 3x2 + x + 1
Compare
with Ax3 + Bx2 + Cx + D
Then
A = 1, B = – 3, C = 1, D = 1
Given,
α = a – b, β= a, γ = a + b
We know that

$$α + β + γ =\frac{-B}{A}\\a – b + a + a + b =\frac{+3}{1}\\3a = 3\\a = 1\\α β γ =\frac{-D}{A}\\(a – b) × a × (a + b) =\frac{-1}{1}\\(a^2 – b^2) × a = – 1\text{Put the value of a}\\(1^2 – b^2) × 1 = – 1\\1 – b^2 = – 1\\– b^2 = – 2\\b^2 = 2\\b = ±\sqrt{2}$$

4. If two zeroes of the polynomial x4 – 6x3 – 26x2 + 138x – 35 are 2± 3, find other zeroes.

Sol. Since, two zeroes are 23+ and 2+ 3 and 2- 3,(x-2- 3 )(x-2+ 3 )=x2 – 4x + 1 is a factor of the given polynomial. Now, we divide the given polynomial x2 – 4x + 1.
x2 – 4x + 1 ) x4 – 6x3 – 26x2 + 138x – 35 (x2 – 2x – 35
x4 – 4x3 + x

$$\frac{-+-}{– 2x^3 – 27x^2 + 138x – 35}Subtract\\– 2x^3 + 8x^2 – 2x\\\frac{+-+}{– 35x^2 + 140x – 35}Subtract\\– 35x^2 + 140x – 35\\\frac{+-+}{0}Subtract\\So,\\ x^4 – 6x^3 – 26x^2 + 138x – 35 = (x^2 – 4x + 1)(x^2 – 2x – 35)\\\text{We factorise}x^2 – 2x – 35 as (x – 7)(x + 5).\\\text{So, its zeroes are given by x = 7 and x = – 5.}\\\text{Therefore,the zeroes of the given polynomial are}\\2+\sqrt{3},2- \sqrt{3}\space\text{and} – 5.$$

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. x2 – 2x + k ) x4 – 6x3 + 16x2 – 25x + 10 ( x2 – 4x + (8 – k)
x4 – 2x3 + kx2

$$\frac{-+-}{– 4x^3 + (16 – k)x^2 – 25x + 10}\\– 4x^3 + 8x^2 – 4kx\\\frac{+-+}{(16 – k – 8)x^2 + (4k – 25)x + 10}\\(8 – k)x^2 + (4k – 25)x + 10\\(8 – k)x^2 – 2(8 – k)x + k(8 – k)\\\frac{-+-}{(4k – 25 + 16 – 2k)x + 10 – k(8 – k)}\\$$

(2k – 9)x + 10 – k(8 – k)
Quotient q(x) = x2 – 4x + (8 – k)
Remainder r(x) = (2k – 9)x + 10 – k(8 – k)
Given remainder r(x) = x + a
Compare both remainder
2k – 9 = 1
2k = 10
k = 5
but
a = 10 – k(8 – k)
Put k = 5
⇒ a = 10 – 5(8 – 5)
⇒ a = 10 – 5 × 3
⇒ a = 10 – 15
⇒ a = – 5.
Hence, k = 5 and a = – 5.