# RS Aggarwal Solutions Chapter 2 Polynomials MCQ Class 10 Maths

 Chapter Name RS Aggarwal Chapter 2 Polynomials Book Name RS Aggarwal Mathematics for Class 10 Other Exercises Exercise 2AExercise 2BExercise 2C Related Study NCERT Solutions for Class 10 Maths

### MCQ for Polynomials Class 10 Maths

1. Which of the following is a polynomial?

(a) x– 5x + 6x + 3

(b) x3⁄2 – x + x1⁄2 + 1

(c) √x + 1/√x

(d) None of these

Solution

(d) none of these

A polynomial in x of degree n is an expression of the form p(x) = a+ a1x + a2x+ ...…+ anxn, where an ≠ 0.

2. Which of the following is not a polynomial?

(a) √3x2 - 2√3x + 5

(b) 9x2 – 4x + √2

(c) 3/2.x+ 6x- 1√2.x – 8

(d) x + 3/x

Solution

(d) x + 3/x is not a polynomial.

It is because in the second term, the degree of x is –1 and an expression with a negative degree is not a polynomial.

3. The Zeroes of the polynomial x2– 2x – 3 are

(a) -3, 1

(b) -3, -1

(c) 3, -1

(d) 3, 1

Solution

(c) 3, –1

Let f(x) = x– 2x – 3= 0

= x– 3x + x – 3= 0

= x(x – 3) + 1(x – 3) = 0

= (x – 3) (x + 1) = 0

⇒ x = 3 or x = –1

4. The zeroes of the polynomial x- √2x – 12 are

(a) √2, -√2

(b) 3√2, -2√2

(c) -3√2, 2√2

(d) 3√2, 2√2

Solution

(b) 3√2, –2√2

Let f(x) = x2– √2x – 12 = 0

⇒ x2– 3√2x +2√2x – 12 = 0

⇒ x(x – 3√2) + 2√2(x – 3√2) = 0

⇒ (x – 3√2) (x + 2√2) = 0

⇒ x = 3√2 or x = –2√2

5. The zeroes of the polynomial 4x+ 5√2x – 3 are

(a) -3√2, √2

(b) -3√2, √2/2

(c) (−3/√2), √2/4

(d) none of these

Solution

(c) –3√2, √2/4

Let f(x) = 4x2 + 5√2x – 3 = 0

⇒ 4x2 + 6√2x – √2x – 3 = 0

⇒ 2√2x(√2x + 3) –1 (√2x + 3) = 0

⇒ (√2x + 3) (2√2x – 1) = 0

⇒ x = –(3√2) or x = 1/(2√2)

⇒ x = –(3√2) or x = 1/(2√2) × √2/√2 = √2/4

6. The zeros of the polynomial x2 + 1/6.x – 2 are

(a) -3, 4

(b) −3/2, 4/3

(c) −4/3, 3/2

(d) none of these

Solution

(b) −3/2, 4/3

Let f(x) = x2 + 1/6.x – 2 = 0

⇒ 6x+ x – 12 = 0

⇒ 6x+ 9x – 8x – 12 = 0

⇒ 3x (2x + 3) – 4(2x + 3) = 0

⇒ (2x + 3) (3x – 4) = 0

∴ x = −3/2 or x = 4/3

7. The zeros of the polynomial 7x– 11/3.x – 2/3 are

(a) 2/7, −1/7

(b) 2/7, -1/3

(c) −2/3, 1/7

(d) none of these

Solution

(a) 2/3, −1/7

Let f(x) = 7x– 113x – 23 = 0

⇒ 21x– 11x – 2 = 0

⇒ 21x2 – 14x + 3x – 2 = 0

⇒ 7x (3x – 2) + 1(3x – 2) = 0

⇒ (3x – 2) (7x + 1) = 0

⇒ x = 2/3 or x = −1/7

8. The sum and product of the zeroes of a quadratic polynomial are 3 and -10 respectively. The quadratic polynomial is

(a) x– 3x + 10

(b) x+ 3x – 10

(c) x2– 3x – 10

(d) x+ 3x + 10

Solution

(c) x– 3x – 10

Given: Sum of zeroes, Î± + Î² = 3

Also, product of zeroes, Î±Î² = –10

∴ Required polynomial = x– Î± + Î² = x– 3x – 10

9. A quadratic polynomial whose zeroes are 5 and -3, is

(a) x+ 2x - 15

(b) x- 2x + 15

(c) x– 2x – 15

(d) none of these

Solution

(c) x– 2x – 15

Here, the zeroes are 5 and –3.

Let Î± = 5 and Î² = –3

So, sum of the zeroes, Î± + Î² = 5 + (–3) = 2

Also, product of the zeroes, Î±Î² = 5 × (–3) = –15

The polynomial will be x– (Î± + Î²) x + Î±Î²

∴ The required polynomial is x– 2x – 15.

10. A quadratic polynomial whose zeroes are 3/5 and −1/2, is

(a) 10x2 + x + 3

(b) 10x2 + x – 3

(c) 10x– x + 3

(d) x- 110x – 310

Solution

(d) x2 – 1/10.x – 3/10

Here, the zeroes are 3/5 and −1/2

Let Î± = 3/5 and Î² = −1/2

So, sum of the zeroes, Î± + Î² = 3/5 + (−1/2) = 1/10

Also, product of the zeroes, Î±Î² = 3/5 × (−1/2) = −3/10

The polynomial will be x2 – (Î± + Î²)x + Î±Î²

∴ The required polynomial is x2 – 1/10.x - 3/10.

11. The zeroes of the quadratic polynomial x+ 88x + 125 are

(a) both positive

(b) both negative

(c) one positive and one negative

(d) both equal

Solution

(b) both negative

Let Î± and Î² be the zeroes of x+ 88x + 125.

Then Î± + Î² = –88 and Î± × Î² = 125

This can only happen when both the zeroes are negative.

12. If Î± and Î² are the zeros of x+ 5x + 8, then the value of (Î± + Î²) is

(a) 5

(b) -5

(c) 8

(d) -8

Solution

(b) –5

Given: Î± and Î² be the zeroes of x+ 5x + 8.

If Î± + Î² is the sum of the roots and Î±Î² is the product, then the required polynomial will be x– (Î± + Î²) x + Î±Î²

∴ Î± + Î² = –5

13. If Î± and Î² are the zeroes of 2x+ 5x - 9, then the value of Î±Î² is

(a) −5/2

(b) 5/2

(c) −9/2

(d) 9/2

Solution

(c) −9/2

Given: Î± and Î² be the zeroes of 2x+ 5x – 9.

If Î± + Î² are the zeroes, then x– (Î± + Î²)x + Î±Î² is the required polynomial.

The polynomial will be x2 – 5/2.x – 9/2.

∴ Î±Î² = −9/2.

14. If one zero of the quadratic polynomial kx+ 3x + k is 2, then the value of k is

(a) 5/6

(b) −5/6

(c) 6/5

(d) −6/5

(d) -6/5

Since 2 is a zero of kx+ 3x + k, we have:

k × (2)+ 3(2) + k = 0

⇒ 4k + k + 6 = 0

⇒ 5k = -6

⇒ k = -6/5

15. If one zero of the quadratic polynomial (k – 1)x– kx + 1 is - 4, then the value of k is

(a) −5/4

(b) 5/4

(c) −4/3

(d) 4/3

Solution

(b) 5/4

Since –4 is a zero of (k – 1) x+ kx + 1, we have:

(k – 1) × (-4)+ k × (-4) + 1 = 0

⇒ 16k – 16 – 4k + 1 = 0

⇒ 12k – 15 = 0

⇒ k = 15/12 = 5/4

⇒ k = 5/4

16. If -2 and 3 are the zeroes of the quadratic polynomial x+ (a + 1)x + b, then

(a) a = -2, b = 6

(b) a = 2, b = -6

(c) a = -2, b = -6

(d) a = 2, b = 6

Solution

(c) a = –2, b = –6

Given: –2 and 3 are the zeroes of x+ (a + 1) x + b.

Now, (–2)+ (a + 1) × (–2) + b = 0 ⇒ 4 – 2a – 2 + b = 0

⇒ b – 2a = –2 ….(1)

Also, 3+ (a + 1) × 3 + b = 0 ⇒ 9 + 3a + 3 + b = 0

⇒ b + 3a = –12 ….(2)

On subtracting (1) from (2), we get a = –2

∴ b = –2 – 4 = –6 [From (1)]

17. If one zero of 3x– 8x + k be the reciprocal of the other, then k = ?

(a) 3

(b) -3

(c) 1/3

(d) −1/3

Solution

(a) k = 3

Let Î± and 1/Î± be the zeroes of 3x– 8x + k.

Then the product of zeroes = k/3

⇒ Î± × 1/Î± = k/3

⇒ 1 = k/3

⇒ k = 3

18. If the sum of the zeroes of the quadratic polynomial kx+ 2x + 3k is equal to the product of its zeroes, then k = ?

(a) 1/3

(b) −1/3

(c) 2/3

(d) −2/3

Solution

(d) -2/3

Let Î± and Î² be the zeroes of kx+ 2x + 3k.

Then Î± + Î² = -2/k and Î±Î² = 3

⇒ Î± + Î² = Î±Î²

⇒ -2/k = 3

⇒ k = -2/3

19. If Î±, Î² are the zeroes of the polynomial x+ 6x + 2, then (1/Î± + 1/Î²) = ?

(a) 3

(b) -3

(c) 12

(d) -12

Solution

(b) –3

Since Î± and Î² be the zeroes of x+ 6x + 2, we have:

Î± + Î² = –6 and Î±Î² = 2

∴ (1/Î± + 1/Î²) = (Î± + Î²)/Î±Î² = -6/2 = -3

20. If Î±, Î², Î³ be the zeroes of the polynomial x– 6x– x + 30, then (Î±Î² + Î²Î³ + Î³Î±) = ?

(a) -1

(b) 1

(c) -5

(d) 30

Solution

(a) -1

It is given that Î±, Î² and Î³ are the zeroes of x– 6x– x + 30.

∴ (Î±Î² + Î²Î³ + Î³Î±) = (co-efficient of x)/(co-efficient of x3) = -1/1 = -1

21. If Î±, Î², Î³ be the zeroes of the polynomial 2x+ x– 13x + 6, then Î±Î²Î³ = ?

(a) -3

(b) 3

(c) -1/2

(d) -13/2

Solution

(a) –3

Since, Î±, Î² and Î³ are the zeroes of 2x+ x– 13x + 6, we have:

Î±Î²Î³ = (-constant term)/(coefficient of x3) = -6/2 = -3

22. If Î±, Î², Î³ be the zeroes of the polynomial p(x) such that (Î± + Î² + Î³) = 3, (Î±Î² + Î²Î³ + Î³Î±) = –10 and Î±Î²Î³ = –24, then p(x) = ?

(a) x+ 3x– 10x + 24

(b) x+ 3x+ 10x – 24

(c) x– 3x– 10x + 24

(d) none of these

Solution

(c) x– 3x– 10x + 24

Given: Î±, Î² and Î³ are the zeroes of polynomial p(x).

Also, (Î± + Î² + Î³) = 3, (Î±Î² + Î²Î³ + Î³Î±) ) = –10 and Î±Î²Î³ = –24

∴ p(x) = x3– (Î± + Î² + Î³)x+ (Î±Î² + Î²Î³ + Î³Î±)x – Î±Î²Î³

= x– 3x– 10x + 24

23. If two of the zeroes of the cubic polynomial ax+ bx+ cx + d are 0, then the third zero is

(a) -b/a

(b) b/a

(c) c/a

(d) -d/a

Solution

(a) -b/a

Let Î±, 0 and 0 be the zeroes of ax+ bx+ cx + d = 0

Then the sum of zeroes = -b/a

⇒ Î± + 0 + 0 = -b/a

⇒ Î± = -b/a

Hence, the third zero is -b/a.

24. If one of the zeroes of the cubic polynomial ax+ bx+ cx + d is 0, then the product of the other two zeroes is

(a) -c/a

(b) c/a

(c) 0

(d) -b/a

Solution

(b) c/a

Let Î±, Î² and 0 be the zeroes of ax+ bx+ cx + d.

Then, sum of the products of zeroes taking two at a time is given by

(Î±Î² + Î² × 0 + Î± × 0) = c/Î±

⇒ Î±Î² = c/Î±

∴ The product of the other two zeroes is c/Î±.

25. If one of the zeroes of the cubic polynomial x+ ax+ bx + c is -1, then the product of the other two zeroes is

(a) a – b – 1

(b) b – a – 1

(c) 1 – a + b

(d) 1 + a – b

Solution

(c) 1 – a + b

Since –1 is a zero of x+ ax+ bx + c, we have:

(–1)+ a × (–1)+ b × (–1) + c = 0

⇒ a – b + c + 1 = 0

⇒ c = 1 – a + b

Also, product of all zeroes is given by

Î±Î² × (–1) = – c

⇒ Î±Î² = c

⇒ Î±Î² = 1 – a + b

26. If Î±, Î² be the zeroes of the polynomial 2x+ 5x + k such that (Î± + Î²)– Î±Î² = 21/4, then k = ?

(a) 3

(b) -3

(c) -2

(d) 2

Solution

(d) 2

Since Î± and Î² are the zeroes of 2x+ 5x + k, we have:

Î± + Î² = -5/2 and Î±Î² = k/2

Also, it is given that Î±+ Î²+ Î±Î² = 21/4.

⇒ (Î± + Î²)– Î±Î² = 21/4

⇒ (−5/2)2 - k/2 = 21/4

⇒ 25/4 - k/2 = 21/4

⇒ k/2 = 25/4 - 21/4 = 4/4 = 1

⇒ k = 2

27. On dividing a polynomial p(x) by a non-zero polynomial q(x), let g(x) be the quotient and r(x) be the remainder, then p(x) = q(x). g(x) + r(x), where

(a) r(x) = 0 always

(b) deg r(x) ˂ deg g(x) always

(c) either r(x) = 0 or deg r(x) ˂ deg g(x)

(d) r(x) = g(x)

Solution

(c) either r(x) = 0 or deg r(x) ˂ deg g(x)

By division algorithm on polynomials, either r(x) = 0 or deg r(x) ˂ deg g(x).

28. Which of the following is a true statement?

(a) x+ 5x – 3 is a linear polynomial.

(b) x+ 4x – 1 is a binomial

(c) x + 1 is a monomial

(d) 5x2is a monomial

Solution

(d) 5xis a monomial.

5xconsists of one term only. So, it is a monomial.

### Exercise – Assesment

1. The zeroes of the polynomial P(x) = x– 2x – 3 are

(a) -3, 1

(b) -3, -1

(c) 3, -1

(d) 3, 1

Solution

(c) 3, -1

Here, p(x) = x– 2x – 3

Let x– 2x – 3 = 0

⇒ x2– (3 – 1)x – 3 = 0

⇒ x2– 3x + x – 3 = 0

⇒ x(x – 3) + 1(x – 3) = 0

⇒ (x – 3) (x + 1) = 0

⇒ x = 3, –1

2. If Î±, Î², Î³ be the zeroes of the polynomial x– 6x– x + 3, then the values of (Î±Î² + Î²Î³ + Î³Î±) = ?

(a) -1

(b) 1

(c) -5

(d) 3

Solution

(a) -1

Here, p(x) = x– 6x– x + 3

Comparing the given polynomial with x– (Î± + Î² + Î³)x2 + ((Î±Î² + Î²Î³ + Î³Î±)x – Î±Î²Î³, we get: (Î±Î² + Î²Î³ + Î³Î±) = -1

3. If Î±, Î² are the zeros of kx– 2x + 3k is equal Î± + Î² = Î±Î² then k = ?

(a) 1/3

(b) -1/3

(c) 2/3

(d) -2/3

Solution

(c) 2/3

Here, p(x) = x– 2x + 3k

Comparing the given polynomial with ax+ bx + c, we get:

a = 1, b = – 2 and c = 3k

It is given that Î± and Î² are the roots of the polynomial.

∴ Î± + Î² = -b/a

⇒ Î± + Î² = – (-2)/1

⇒ Î± + Î² = 2 ….(i)

Also, Î±Î² = c/Î±

⇒ Î±Î² = 3k/1

⇒ Î±Î² = 3k ….(ii)

Now, Î± + Î² = Î±Î²

⇒ 2 = 3k  [Using (i) and (ii)]

⇒ k = 2/3

4. It is given that the difference between the zeroes of 4x– 8kx + 9 is 4 and k > 0. Then, k = ?

(a) 1/2

(b) 3/2

(c) 5/2

(d) 7/2

Solution

(c) 5/2

Let the zeroes of the polynomial be Î± and Î± + 4

Here, p(x) = 4x– 8kx + 9

Comparing the given polynomial with ax+ bx + c, we get:

a = 4, b = -8k and c = 9

Now, sum of the roots = −b/a

⇒ Î± + Î± + 4 = -(-8)/4

⇒ 2Î± + 4 = 2k

⇒ Î± + 2 = k

⇒ Î± = (k – 2)  ….(i)

Also, product of the roots, Î±Î² = c/Î±

⇒ Î±(Î± + 4) = 9/4

⇒ (k – 2) (k – 2 + 4) = 9/4

⇒ (k – 2) (k + 2) = 9/4

⇒ k– 4 = 9/4

⇒ 4k– 16 = 9

⇒ 4k= 25

⇒ k= 25/4

⇒ k = 5/2  (∵ k > 0)

5. Find the zeroes of the polynomial x+ 2x – 195.

Solution

Here, p(x) = x+ 2x – 195

Let p(x) = 0

⇒ x+ (15 – 13)x – 195 = 0

⇒ x+ 15x – 13x – 195 = 0

⇒ x (x + 15) – 13(x + 15) = 0

⇒ (x + 15) (x – 13) = 0

⇒ x = –15, 13

Hence, the zeroes are –15 and 13.

6. If one zero of the polynomial (a+ 9)x– 13x + 6a is the reciprocal of the other, find the value of a.

Solution

(a + 9)x– 13x + 6a = 0

Here, A = (a+ 9), B = 13 and C = 6a

Let Î± and 1/Î± be the two zeroes.

Then, product of the zeroes = C/A

⇒ Î±.1/Î± = 6Î±/(Î±2 + 9)

⇒ 1 = 6Î±/(Î±2 + 9)

⇒ a+ 9 = 6a

⇒ a– 6a + 9 = 0

⇒ a– 2 × a × 3 + 3= 0

⇒ (a – 3)= 0

⇒ a – 3 = 0

⇒ a = 3

7. Find a quadratic polynomial whose zeroes are 2 and -5.

Solution

It is given that the two roots of the polynomial are 2 and -5.

Let Î± = 2 and Î² = -5

Now, the sum of the zeroes, Î± + Î² = 2 + (-5) = -3

Product of the zeroes, Î±Î² = 2 × (-5) = -15

∴ Required polynomial = x– (Î± + Î²)x + Î±Î²

= x– (-3)x + 10

= x+ 3x – 10

8. If the zeroes of the polynomial x3– 3x+ x + 1 are (a – b), a and (a + b), find the values of a and b.

Solution

The given polynomial = x– 3x+ x + 1 and its roots are (a – b), a and (a + b).

Comparing the given polynomial with Ax+ Bx+ Cx + D, we have:

A = 1, B = -3, C = 1 and D = 1

Now, (a – b) + a + (a + b) = -B/A

⇒ 3 a = -3/1

⇒ a = 1

Also, (a – b) × a × (a + b) = -D/A

⇒ a (a– b2) = -1/1

⇒ 1 (1– b2) = -1

⇒ 1 – b= -1

⇒ b= 2

⇒ b = ±√2

∴ a = 1 and b = ±√2

9. Verify that 2 is a zero of the polynomial x+ 4x– 3x – 18.

Solution

Let p(x) = x+ 4x– 3x – 18

Now, p(2) = 2+ 4 × 2– 3 × 2 – 18 = 0

∴ 2 is a zero of p(x).

10. Find the quadratic polynomial, the sum of whose zeroes is -5 and their product is 6.

Solution

Given:

Sum of the zeroes = -5

Product of the zeroes = 6

∴ Required polynomial = x– (sum of the zeroes) x + product of the zeroes

= x– (-5) x + 6

= x+ 5x + 6

11. Find a cubic polynomial whose zeroes are 3, 5 and -2.

Solution

Let Î±, Î² and Î³ are the zeroes of the required polynomial.

Then we have:

Î± + Î² + Î³ = 3 + 5 + (-2) = 6

Î±Î² + Î²Î³ + Î³Î± = 3×5 + 5×(-2) + (-2)×3 = -1

and Î±Î²Î³ = 3 × 5 × -2 = -30

Now, p(x) = x3– x(Î± + Î² + Î³) + x (Î±Î² + Î²Î³ + Î³Î±) – Î±Î²Î³

= x– x× 6 + x × (-1) – (-30)

= x– 6x– x + 30

So, the required polynomial is p(x) = x– 6x– x + 30.

12. Using remainder theorem, find the remainder when p(x) = x+ 3x– 5x + 4 is divided by (x – 2).

Solution

Given:

p(x) = x+ 3x– 5x + 4

Now, p(2) = 2+ 3(22) – 5(2) + 4

= 8 + 12 – 10 + 4

= 14

13. Show that (x + 2) is a factor of f(x) = x+ 4x+ x – 6.

Solution

Given: f(x) = x+ 4x+ x – 6

Now, f(-2) = (-2)+ 4(-2)+ (-2) - 6

= - 8 + 16 – 2 – 6

= 0

∴ (x + 2) is a factor of f(x) = x+ 4x+ x – 6.

14. If Î±Î², Î³ are the zeroes of the polynomial p(x) = 6x+ 3x– 5x + 1, find the value of (1/Î± + 1/Î² + 1/Î³).

Solution

Given: p(x) = 6x+ 3x– 5x + 1

= 6x– (–3) x+ (–5) x – 1

Comparing the polynomial with x– x(Î± + Î² + Î³) + x(Î±Î² + Î²Î³ + Î³Î±) – Î±Î²Î³, we get: Î±Î² + Î²Î³ + Î³Î± = –5

and Î±Î²Î³ = – 1

∴ (1/Î± + 1/Î² + 1/Î³

= (Î²Î³ + Î±Î³ + Î±Î²)/Î±Î²Î³

= (-5/-1)

= 5

15. If Î±, Î² are the zeroes of the polynomial f(x) = x– 5x + k such that Î± – Î² = 1, find the value of k.

Solution

Given: x– 5x + k

The co-efficients are a = 1, b = -5 and c = k.

∴ Î± + Î² = -b/a

⇒ Î± + Î² = (-5)/1

⇒ Î± + Î² = 5 ...(1)

Also, Î± – Î² = 1 ...(2)

From (1) and (2), we get:

2Î± = 6

⇒ Î± = 3

Putting the value of Î± in (1), we get Î² = 2.

Now, Î±Î² = c/Î±

⇒ 3 × 2 = k/1

∴ k = 6

16. Show that the polynomial f(x) = x+ 4x + 6 has no zero.

Solution

Let t = x2

So, f(t) = t+ 4t + 6

Now, to find the zeroes, we will equate f(t) = 0

⇒ t+ 4t + 6 = 0

The zeroes of a polynomial should be real numbers.

∴ The given f(x) has no zeroes.

17. If one zero of the polynomial p(x) = x– 6x+ 11x – 6 is 3, find the other two zeroes.

Solution

p(x) = x– 6x+ 11x – 6 and its factor, x + 3

Let us divide p(x) by (x – 3).

Here, x– 6x+ 11x – 6 = (x – 3) (x– 3x + 2)

= (x – 3) [(x– (2 + 1)x + 2]

= (x – 3) (x– 2x – x + 2)

= (x – 3) [x (x – 2) – 1(x – 2)]

= (x – 3) (x – 1) (x – 2)

∴ The other two zeroes are 1 and 2.

18. If two zeroes of the polynomial p(x) = 2x– 3x– 3x+ 6x – 2 are √2 and – √2, find its other two zeroes.

Solution

Given: p(x) = 2x– 3x– 3x+ 6x – 2 and the two zeroes, √2 and – √2

So, the polynomial is (x + √2) (x – √2) = x– 2.

Let us divide p(x) by (x– 2)

Here, 2x– 3x– 3x+ 6x – 2 = (x– 2) (2x– 3x + 1)

= (x– 2) [(2x– (2 + 1) x + 1]

= (x– 2) (2x– 2x – x + 1)

= (x– 2) [(2x (x – 1) –1(x – 1)]

= (x– 2) (2x – 1) (x – 1)

The other two zeroes are 1/2 and 1.

19. Find the quotient when p(x) = 3x+ 5x3– 7x+ 2x + 2 is divided by (x+ 3x + 1)

Solution

Given: p(x) = 3x+ 5x– 7x+ 2x + 2

Dividing p(x) by (x+ 3x + 1), we have:

∴ The quotient is 3x– 4x + 2

20. Use remainder theorem to find the value of k, it being given that when x+ 2x+ kx + 3 is divided by (x – 3), then the remainder is 21.

Solution

Let p(x) = x+ 2x+ kx + 3

Now, p(3) = (3)+ 2(3)+ 3k + 3

= 27 + 18 + 3k + 3

= 48 + 3k

It is given that the reminder is 21

∴ 3k + 48 = 21

⇒3k = –27

⇒ k = –9