Frank Chapter 10 Remainder and Factor Theorems Solutions Class 10 Maths

Frank Solutions for Chapter 10 Remainder and Factor Theorems Class 10 ICSE Mathematics

Exercise 10

1. Find without division, the remainder in each of the following:

(i) 5x² – 9x + 4 is divided by (x – 2)

(ii) 5x³ – 7x² + 3 is divided by (x – 1)

(iii) 8x² – 2x + 1 is divided by (2x + 1)

(iv) x³ + 8x² + 7x – 11 is divided by (x + 4)

(v) 2x³ – 3x² + 6x – 4 is divided by (2x – 3)

Answer

(i) 5x² – 9x + 4 is divided by (x – 2)

From the question it is given that, 5x² – 9x + 4 is divided by (x – 2)

Let us assume x – 2 = 0, x = 2

Now, substitute the value of x in given expression,

= 5×(2)² – (9 × 2) + 4

= (5 × 4) – 18 + 4

= 20 – 18 + 4

= 24 – 18

= 6

Therefore, the remainder of the given expression is 6.

(ii) 5x³ – 7x² + 3 is divided by (x – 1)

From the question it is given that, 5x³ – 7x² + 3 is divided by (x – 1)

Let us assume x – 1 = 0, x = 1

Now, substitute the value of x in given expression,

= 5 × (1)² – 7 × (1)² + 3

= (5 × 1) – (7 × 1) + 3

= 5 – 7 + 3

= 8 – 7

= 1

Therefore, the remainder of the given expression 1.

(iii) 8x² – 2x + 1 is divided by (2x + 1)

From the question it is given that, 8x² – 2x + 1 is divided by (2x + 1)

Let us assume 2x + 1 = 0, x = -½

Now, substitute the value of x in given expression,

= 8(-½)² – 2(-½) + 1

= (8 × 1/4) + 1 + 1

= 2 + 1 + 1

= 4

Therefore, the remainder of the given expression 4.

(iv) x³ + 8x² + 7x – 11 is divided by (x + 4)

From the question it is given that, x³ + 8x² + 7x – 11 is divided by (x + 4)

Let us assume x + 4 = 0, x = -4

Now, substitute the value of x in given expression,

= (-4)³ + 8(-4)² + 7(-4) – 11

= – 64 + 8(-16) – 28 – 11

= -64 + 128 – 28 – 11

= 25

Therefore, the remainder of the given expression 25.

(v) 2x³ – 3x² + 6x – 4 is divided by (2x – 3)

From the question it is given that, 2x³ – 3x² + 6x – 4 is divided by (x + 4)

Let us assume 2x – 3 = 0, x = 3/2

Now, substitute the value of x in given expression,

= 2(3/2)³ – 3(3/2)² + 6(3/2) – 4

= (2 × 27/8) – 3(9/4) + (3 × 3) – 4

= 27/4 – 27/4 + 9 – 4

= 9 – 4

= 5

Therefore, the remainder of the given expression 5.

2. Prove by factor theorem that,

(i) (x – 2) is a factor of 2x³ – 7x – 2

(ii) (2x + 1) is a factor of 4x³ + 12x² + 7x + 1

(iii) (3x – 2) is a factor of 18x³ – 3x² + 6x – 8

(iv) (2x – 1) is a factor of 6x³ – x² – 5x + 2

(v) (x – 3) is a factor of 5x² – 21x + 18

Answer

(i) (x – 2) is a factor of 2x³ – 7x – 2

From the question it is given that, f(x) = 2x³ – 7x – 2

Let us assume, x – 2 = 0, x = 2

Then, substitute the value of x,

f(2) = 2(2)³ – 7(2) – 2

= 2(8) – 14 – 2

= 16 – 14 – 2

= 16 – 16

= 0

Now, it is clear that (x – 2) is a factor of 2x³ – 7x – 2.

(ii) (2x + 1) is a factor of 4x³ + 12x² + 7x + 1

From the question it is given that, f(x) = 4x³ + 12x² + 7x + 1

Let us assume, 2x + 1 = 0, x = -½

Then, substitute the value of x,

f(-½) = 4 (-½)³ + 12 (-½)² + 7 (-½) + 1

= 4(-1/8) + 12(1/4) – 7/2 + 1

= – ½ + 3 – 7/2 + 1

= – ½ – 7/2 + 4

= -8/2 + 4

= – 4 + 4

= 0

Now, it is clear that (2x + 1) is a factor of 4x³ + 12x² + 7x + 1.

(iii) (3x – 2) is a factor of 18x³ – 3x² + 6x – 8

From the question it is given that, f(x) = 18x³ – 3x² + 6x – 8

Let us assume, 3x – 2 = 0, x = 2/3

Then, substitute the value of x,

f(2/3) = 18 (2/3)³ – 3 (2/3)² + 6 (2/3) – 8

= 18(8/27) – 3(4/9) + (2 × 2) – 8

= 2(8/3) – (4/3) + 4 – 8

= 16/3 – 4/3 + 4 – 8

= 12/3 + 4 – 8

= 4 + 4 – 8

= 8 – 8

= 0

Now, it is clear that (3x – 2) is a factor of 18x³ – 3x² + 6x – 8.

(iv) (2x – 1) is a factor of 6x³ – x² – 5x + 2

From the question it is given that, f(x) = 6x³ – x² – 5x + 2

Let us assume, 2x – 2 = 0, x = ½

Then, substitute the value of x,

f(½) = 6 (½)³ – (½)² – 5 (½) + 2

= 6(1/8) – (1/4) – (5/2) + 2

= 3(1/4) – (1/4) – 5/2 + 2

= 3/4 – 1/4 – 5/2 + 2

= 2/4 – 5/2 + 2

= 1/2 – 5/2 + 2

= -4/2 + 2

= -2 + 2

= 0

Now, it is clear that (2x – 1) is a factor of 6x³ – x² – 5x + 2.

(v) (x – 3) is a factor of 5x² – 21x + 18

From the question it is given that, f(x) = 5x² – 21x + 18

Let us assume, x – 3 = 0, x = 3

Then, substitute the value of x,

f(3) = 5(3)² – 21(3) + 18

= 5(9) – 63 + 18

= 45 – 63 + 18

= 63 – 63

= 0

Now, it is clear that (x – 3) is a factor of 5x² – 21x + 18.

3. Find the values of a and b in the polynomial f(x) = 2x³+ ax²+ bx + 10, if it is exactly divisible by (x + 2) and (2x – 1)

Answer

From the question it is given that,

f(x) = 2x³ + ax² + bx + 10

let us assume x + 2 = 0

x = -2

⇒ 2x – 1 = 0

⇒ x = ½

Now, substitute the value of x,

x = -2

⇒ f(-2) = 2(-2)³ + a(-2)² + b(-2) + 10 = 0

⇒ 2(-8) + a(4) – 2b + 10 = 0

⇒ –16 + 4a – 2b + 10 = 0

⇒ –6 + 4a – 2b = 0

Divide both side by 2 we get,

-6/2 + 4a/2 – 2b/2 = 0

⇒ –3 + 2a – b = 0

⇒ 2a = b + 3

⇒ a = b/2 + 3/2 …[equation (i)]

Then,

f(½) = 2(½)³ + a(½)² + b(½) + 10 = 0

⇒ 2(1/8) + a(1/4) + b/2 + 10 = 0

⇒ ¼ + a/4 + b/2 + 10 = 0

Multiply by 4 for each terms we get,

1 + a + 2b + 40 = 0

⇒ 41 + a + 2b = 0

⇒ a = –2b – 41 …[equation (ii)]

By combining both equation (i) and equation (ii) we get,

b/2 + 3/2 = -2b – 41

⇒ (b + 3)/2 = -2b – 41

⇒ b + 3 = – 4b – 82

By transposing we get,

4b + b = -82 – 3

⇒ 5b = –85

⇒ b = -85/5

⇒ b = –17

So, a = – 2b – 41

= -2(-17) – 41

= 34 – 41

= –7

Therefore, the value of a is –7 and b is –17.

4. Using remainder theorem, find the value of m if the polynomial f(x) = x³+ 5x²– mx + 6 leaves a remainder 2m when divided by (x – 1).

Answer

From the question it is given that,

f(x) = x³ + 5x² – mx + 6

Remainder = 2m

Let us assume that x – 1 = 0, x = 1

Now, substitute the value of x in f(x) we get,

f(1) = 1³ + 5(1)² – m(1) + 6 = 2m

⇒ 1 + 5 – m + 6 = 2m

⇒ 6 – m + 6 = 2m

By transposing we get,

12 = 2m + m

⇒ 12 = 3m

⇒ m = 12/3

⇒ m = 4

Therefore, the value of m is 4.

5. Find the value of m when x³+ 3x²– mx + 4 is exactly divisible by (x – 2)

Answer

From the question it is given that,

f(x) = x³ + 3x² – mx + 4

Let us assume that x – 2 = 0, x = 2

Now, substitute the value of x in f(x) we get,

f(2) = 2³ + 3(2)² – m(2) + 4 = 0

⇒ 8 + 3(4) – 2m + 4 = 0

⇒ 8 + 12 – 2m + 4 = 0

⇒ 24 – 2m = 0

By transposing we get,

24 = 2m

⇒ m = 24/2

⇒ m = 12

Therefore, the value of m is 12.

6. Find the values of p and q in the polynomial f(x) = x³– px²+ 14x – q, if it is exactly divisible by (x – 1) and (x – 2).

Answer

From the question it is given that, f(x) = x³ – px² + 14x – q

Let us assume that, x – 1= 0, x = 1

x – 2 = 0, x = 2

Now, substitute the value of x in f(x) we get,

f(1) = 1³ – p(1)² + 14(1) – q = 0

⇒1 – p + 14 – q = 0

⇒ 15 – p – q = 0

⇒ p = 15 – q …[equation (i)]

Then, f(2) = 2³ – p(2)² + 14(2) – q = 0

8 – 4p + 28 – q = 0

⇒ 36 – 4p – q = 0

⇒ q = 36 – 4p …[equation (ii)]

So, substitute the value of q in equation (i) we get,

p = 15 – (36 – 4p)

⇒ p = 15 – 36 + 4p

By transposing we get,

36 – 15 = 4p – p

⇒ 21 = 3p

⇒ p = 21/3

⇒ p = 7

Consider the equation (ii) to find the value of q,

q = 36 – 4(7)

⇒ q = 36 – 28

⇒ q = 8

Therefore, the value of p is 7 and q is 8.

7. Find the values of a and b when the polynomial f(x) = ax³+ 3x²+ bx – 3 is exactly divisible by (2x + 3) and leaves a remainder – 3 when divided by (x + 2).

Answer

From the question it is given that, f(x) = ax³ + 3x² + bx – 3

Remainder = -3

Let us assume that, 2x + 3 = 0, x = -3/2

x + 2 = 0, x = -2

Now, substitute the value of x in f(x) we get,

f(-3/2) = a(-3/2)³ + 3(-3/2)² + b(-3/2) – 3= 0

⇒ a(-27/8) + 3(9/4) – b(3/2) – 3 = 0

⇒ a(-27/8) + (27/4) – 3 – b(3/2) = 0

⇒ a(-27/8) + (27- 12)/4 – b(3/2) = 0

⇒ a(-27/8) + 15/4 – b(3/2) = 0

⇒ -27a + 30 – 12b = 0

⇒ 27a = – 12b + 30

Then,

f(- 2) = a(-2)³ + 3(-2)² + b(-2) – 3= -3

⇒ a(-8) + 3(4) – 2b – 3 = -3

⇒ -8a + 12 – 2b – 3 + 3= 0

⇒ -4a + 6 – b = 0

⇒ b = 6 – 4a …[equation (ii)]

By combining both equation (i) and equation (ii) we get,

27a = – 12(6 – 4a) + 30

⇒ 27a = – 72 + 48a + 30

⇒ 27a – 48a = – 42

⇒ –21a = -42

⇒ a = -42/-21

⇒ a = 2

Then,

b = 6 – 4a

⇒ b = 6 – 4(2)

⇒ b = 6 – 8

⇒ b = -2

Therefore, the value of a is 2 and b is –2.

8. Find the values of m and n when the polynomial f(x) = x³– 2x²+ mx + n has a factor (x + 2) and leaves a remainder 9 when divided by (x + 1).

Answer

From the question it is given that,

f(x) = x³ – 2x² + mx + n

Remainder = 9

Let us assume that, x + 2 = 0, x = – 2

x + 1 = 0, x = -1

Now, substitute the value of x in f(x) we get,

f(-2) = (-2)³ – 2(-2)² + m(-2) + n = 0

⇒ –8 – 8 – 2m + n = 0

⇒ –16 – 2m + n = 0

⇒ n = 2m + 16 …[equation (i)]

Then, f(-1) = (-1)³ – 2(-1)² + m(-1) + n = 9

⇒ –1 – 2 – m + n = 9

⇒ –3 – m + n = 9

⇒ m = – 3 – 9 + n

⇒ m = n – 12 …[equation (ii)]

Now, combining both equation (i) and equation (ii) we get,

n = 2(n – 12) + 16

⇒ n = 2n – 24 + 16

⇒ n = 2n – 8

⇒ 2n – n = 8

⇒ n = 8

Consider the equation (ii) to find out the value of m,

m = n – 12

⇒ m = 8 – 12

⇒ m = -4

Therefore, the value of n is 8 and m is – 4.

9. Find the values of a and b when the polynomials f(x) = 2x²– 5x + a and g(x) = 2x²+ 5x + b both have a factor (2x + 1)

Answer

From the question it is given that,

f(x) = 2x² – 5x + a

g(x) = 2x² + 5x + b

Let us assume 2x + 1 = 0, x = -½

Now, substitute the value of x in f(x) we get,

f(-½) = 2(-½)² – 5(-½) + a = 0

⇒ 2(1/4) + 5/2 + a = 0

⇒ ½ + 5/2 + a = 0

⇒ 6/2 + a = 0

⇒ 3 + a = 0

⇒ a = –3

Then,

g(-½) = 2(-½)² + 5(-½) + b = 0

⇒ 2(1/4) – 5/2 + b = 0

⇒ ½ – 5/2 + b = 0

⇒ – 4/2 + b = 0

⇒ –2 + b = 0

⇒ b = 2

Therefore, the value of a is – 3 and b is 2.

10. Find the values of a and b when the factors of the polynomial f(x) = ax³+ bx²+ x – a are (x + 3) and (2x – 1). Factorize the polynomial completely.

Answer

From the question it is given that, f(x) = ax³ + bx² + x – a

Let us assume, x + 3 = 0, x = -3

2x – 1 = 0, x = ½

Now, substitute the value of x in f(x) we get,

f(-3) = a(-3)³ + b(-3)² + (-3) – a = 0

⇒ -27a + 9b – 3 – a = 0

⇒ -28a + 9b – 3 = 0

By transposing we get,

-28a = -9b + 3

⇒ a = -9b/-28 + 3/-28

⇒ a = 9b/28 – 3/28 …[equation (i)]

Then, f(½) = a(½)³ + b(½)² + (½) – a = 0

⇒ a(1/8) + b(1/4) + ½ – a= 0

⇒ (a – 8a)/8 + b(1/4) + ½ = 0

⇒ -7a/8 + b(¼) + ½ = 0

⇒ b(¼) = -½ + 7a/8

⇒ b = -4/2 + 28a/8

⇒ b = -2 + 7a/2 …[equation (ii)]

Now, combining equation (i) and equation (ii) we get,

a = 9/28 × (7a/2 – 2) – 3/28

By simplification we get,

56a = 63a – 42

⇒ a = 6

Consider the equation (ii) to find out the value of b,

b = 7a/2 – 2

⇒ b = (7 × 6)/2 – 2

⇒ b = 42/2 – 2

⇒ b = 21 – 2

⇒ b = 19

Substitute the value of a and b in f(x) we get,

f(x) = 6x³ + 19x² + x – 6

Therefore, equation becomes (x + 3) (2x – 1) (3x + 2) = 0

11. What number should be subtracted from x²+ x + 1 so that the resulting polynomial is exactly divisible by (x – 2)?

Answer

From the question it is given that, f(x) = x² + x + 1

Then, x – 2 = 0, x = 2

Let us assume the number should be subtracted from x² + x + 1 be b,

f(2) = 2² + 2 + 1 – b = 0

⇒ 4 + 2 + 1 – b = 0

⇒ 7 – b = 0

⇒ b = 7

Therefore, the number is 7.

12. What number should be added to 2x³– 3x²+ 7x – 8 so that the resulting polynomial is exactly divisible by (x – 1)?

Answer

From the question it is given that, f(x) = 2x³ – 3x² + 7x – 8

Then, x – 1 = 0, x = 1

Let us assume the number should be added to 2x³ – 3x² + 7x – 8 be b,

f(1) = 2(1)³ – 3(1)² + 7(1) – 8 + b = 0

⇒ 2 – 3 + 7 – 8 + b = 0

⇒ 9 – 11 + b = 0

⇒ –2 + b = 0

⇒ b = 2

Therefore, the number is 2.

13. What number should be subtracted from polynomial f(x) 2x³– 5x²+ 8x – 17 so that the resulting polynomial is exactly divisible by (2x – 5)?

Answer

From the question it is given that, f(x) = 2x³ – 5x² + 8x – 17

Then, 2x – 5 = 0, x = 5/2

Let us assume the number should be subtracted from 2x³ – 5x² + 8x – 17 be b,

f(5/2) = 2(5/2)³ – 5(5/2)² + 8(5/2) – 17 – b = 0

⇒ 2(125/8) – 5(25/4) + 40/2 – 17 – b = 0

⇒ 125/4 – 125/4 + 20 – 17 – b = 0

⇒ 3 – b = 0

⇒ b = 3

Therefore, the number is 3.

14. What number should be added to polynomial f(x) 12x³+ 16x²– 5x – 8 so that the resulting polynomial is exactly divisible by (2x – 1)?

Answer

From the question it is given that, f(x) = 12x³ + 16x² – 5x – 8

Then, 2x – 1 = 0, x = ½

Let us assume the number should be added to 2x³ – 3x² + 7x – 8 be b,

f(½) = 12(½)³ + 16(½)² – 5(½) – 8 + b = 0

⇒ 12(1/8) + 16(1/4) – 5/2 – 8 + b = 0

⇒ 3/2 + 4 – 5/2 – 8 + b = 0

⇒ –4 – 2/2 + b = 0

⇒ –4 – 1 + b = 0

⇒ -5 + b = 0

⇒ b = 5

Therefore, the number is 5.

15. Use the remainder theorem to find the factors of (a – b)³+ (b – c)³+ (c – a)³

Answer

From the question it is given that, f(x) = (a – b)³ + (b – c)³ + (c – a)³

We know the formula, (a – b)³ = a³ – 3a²b + 3ab² – b³ …[equation (i)]

Let us assume that a – b = 0, a = b

Now substitute the above value in f(x), we get,

f(x) = 0 + (a – c)³ + (c – a)³ = 0

⇒ (a – c)³ – (a – c)³ = 0

⇒ 0 = 0

Therefore, (a – b) is a factor. …[equation (ii)]

Again, f(x) = 0 + (b³ – 3b²c + 3bc² – c³) + (c³ – 3c²a + 3ca² – a²)

= – 3b²c + 3bc² – 3ca² + 3ca²

= 3(-b²c + bc² – ca² + ca²)

So, now we put b – c = 0, b = c

Substitute the above value in f(x), we get,

Then, f(b = c), 3((-c² × c) + (c × c²) – (c × c²) + (c × c²)) = 0

Factors are 3(a – b) (b – c) …[equation (iii)]
similarly if we put c = a,

(c – a) is a factor …[equation (iv)]

By combining equation (ii), equation (iii) and (iv), we get,

3(a – b)(b – c)(c – a).

16. Prove that (p – q) is a factor of (q – r)³ + (r – p)³

Answer

If p – q is assumed to be factor, then p = q. Substituting this in problem polynomial, we get:

17. Prove that (x – y) is a factor of yz(y² – z²) + zx(z² – x²) + xy(x² – y²)

Answer

If x – y is assumed to be factor, then x = y. Substituting this in problem polynomial, we get:

18. Prove that (x – 3) is a factor of x³ – x² – 9x + 9 and hence factorize it completely.

Answer

If (x – 3) is assumed to be factor, then x = 3. Substituting this in problem polynomial, we get:

f(3) = 3 × 3 × 3 – 3 × 3 – 9 × 3 + 9 = 0

Hence its proved that x – 3 is a factor of the polynomial.

19. Prove that (x + 1) is a factor of x³ – 6x² + 5x + 12 and hence factorize it completely.

Answer

If x + 1 is assumed to be factor, then x = - 1. Substituting this in problem polynomial, we get:

20. Prove that (5x – 4) is a factor of the polynomial f(x) = 5x³ – 4x² – 5x + 4. Hence factorize it completely.

Answer

If 5x – 4 is assumed to be factor, then x = 4/5, Substituting this in problem polynomial, we get:

 

21. A polynomial f(x) when divided by (x – 1) leaves a remainder 3 and when divided by (x – 2) leaves a remainder of 1. Show that when its divided by (x – 1)(x – 2), the remainder is (- 2x + 5).

Answer

Given f(x) = (x – 1)(x – 2) + (- 2x + 5)

= (x² – 3x + 2) + (- 2x + 5)

22. The polynomial f(x) = ax⁴ + x³ + bx² – 4x + c has (x + 1), (x – 2) and (2x – 1) as its factors. Find the values of a, b, c and remaining factor.

Answer

When x + 1 is a factor, we can substitute x = - 1 to evaluate values ...(i)