# 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)

(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

(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)

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).

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)

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).

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).

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).

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)

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.

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)?

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)?

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)?

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)?

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

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((-c× 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

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)

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.

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.

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.

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).