ML Aggarwal Solutions for Class 10 Maths Chapter 6 Factorization are provided here. Our expert faculty team has prepared solutions to help students in their exam preparation to obtain good marks in Maths. If you wish to secure an excellent score, solving ML Aggarwal Solutions is an utmost necessity. Scoring high marks requires a good amount of practice on each and every topic. These solutions will help you in gaining knowledge and a strong command over the subject. Practising the textbook questions will help you in analyzing your level of preparation and knowledge of the concept.

Chapter 6 – Factorization solutions are available for download in pdf format and provide solutions to all the questions provided in ML Aggarwal Solutions for Class 10 Maths Chapter 6. Factorization is when you break a number down into smaller numbers that, multiplied together, give you that original number. When you split a number into its factors or divisors, it is called factorization. Now, let us have a look at some of the important concepts discussed in this chapter.

- Polynomial and Related terms
- Polynomial Equations
- Equality of Two Polynomials
- Division Algorithm for Polynomials
- Factors of Polynomial
- Root of a Polynomial Equations

- Remainder Theorem
- Factor Theorem
- Use of Factor Theorem

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Exercise 6.1

**1. Find the remainder (without division) on dividing f(x) by (x – 2) where**

**(i) f(x) = 5x ^{2} â€“ 7x + 4**

**Solutions:-**

Let us assume x â€“ 2 = 0

Then, x = 2

Given, f(x) = 5x^{2} â€“ 7x + 4

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

f(2)= (5 Ã— 2^{2}) â€“ (7 Ã— 2) + 4

= (5 Ã— 4) â€“ 14 + 4

= 20 â€“ 14 + 4

= 24 â€“ 14

= 10

Therefore, the remainder is 10.

**(ii) f(x) = 2x ^{3} â€“ 7x^{2} + 3**

**Solution:-**

Let us assume x â€“ 2 = 0

Then, x = 2

Given, f(x) = 2x^{3} â€“ 7x^{2} + 3

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

f(2)= (2 Ã— 2^{3}) â€“ (7 Ã— 2^{2}) + 3

= (2 Ã— 8) â€“ (7 Ã— 4) + 3

= 16 â€“ 28 + 3

= 19 â€“ 28

= -9

Therefore, the remainder is -9.

**2. Using the remainder theorem, find the remainder on dividing f(x) by (x + 3) where**

**(i) f(x) = 2x ^{2} â€“ 5x + 1**

**Solution:-**

Let us assume x + 3 = 0

Then, x = -3

Given, f(x) = 2x^{2} â€“ 5x + 1

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

f(-3)= (2 Ã— -3^{2}) â€“ (5 Ã— (-3)) + 1

= (2 Ã— 9) â€“ (-15) + 1

= 18 + 15 + 1

= 34

Therefore, the remainder is 34.

**(ii) f(x) = 3x ^{3} + 7x^{2} â€“ 5x + 1**

Solution:-

Let us assume x + 3 = 0

Then, x = -3

Given, f(x) = 3x^{3} + 7x^{2} â€“ 5x + 1

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

f(-3)= (3 Ã— -3^{3}) + (7 Ã— -3^{2}) â€“ (5 Ã— -3) + 1

= (3 Ã— -27) + (7 Ã— 9) â€“ (-15) + 1

= – 81 + 63 + 15 + 1

= -81 + 79

= -2

Therefore, the remainder is -2.

**3. Find the remainder (without division) on dividing f(x) by (2x + 1) where,**

**(i) f(x) = 4x ^{2} + 5x + 3**

**Solution:-**

Let us assume 2x + 1 = 0

Then, 2x = -1

X = -Â½

Given, f(x) = 4x^{2} + 5x + 3

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

f (-Â½) = 4 (-Â½)^{2} + 5 (-Â½) + 3

= (4 Ã— Â¼) + (-5/2) + 3

= 1 â€“ 5/2 + 3

= 4 â€“ 5/2

= (8 – 5)/2

= 3/2 = 1Â½

Therefore, the remainder is 1Â½.

**(ii) f(x) = 3x ^{3} â€“ 7x^{2} + 4x + 11**

**Solution:-**

Let us assume 2x + 1 = 0

Then, 2x = -1

X = -Â½

Given, f(x) = 3x^{3} â€“ 7x^{2} + 4x + 11

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

f(-Â½) = (3 Ã— (-Â½)^{3}) â€“ (7 Ã— (-Â½)^{2} + (4 Ã— -Â½) + 11

= 3 Ã— (-1/8) â€“ (7 Ã— Â¼) + (- 2) + 11

= -3/8 â€“ 7/4 â€“ 2 + 11

= – 3/8 â€“ 7/4 + 9

= (-3 â€“ 14 + 72)/8

= 55/8

=

**4. Using remainder theorem, find the value of k if on dividing 2x ^{3} + 3x^{2} â€“ kx + 5 by x â€“ 2 leaves a remainder 7.**

**Solution:-**

Let us assume, x â€“ 2 = 0

Then, x = 2

Given, 2x^{3} + 3x^{2} â€“ kx + 5

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

f(2) = (2 Ã— 2^{3}) + (3 Ã— 2^{2}) â€“ (k Ã— 2) + 5

= (2 Ã— 8) + (3 Ã— 4) â€“ 2k + 5

= 16 + 12 â€“ 2k + 5

= 33 â€“ 2k

Form the question it is given that, remainder is 7.

So, 7 = 33 â€“ 2k

2k = 33 â€“ 7

2k = 26

K = 26/2

K = 13

Therefore, the value of k is 13.

**5. Using remainder theorem, find the value of â€˜aâ€™ if the division of x ^{3} + 5x^{2} â€“ ax + 6 by (x – 1) leaves the remainder 2a.**

**Solution:-**

Let us assume x -1 = 0

Then, x = 1

Given, f(x) = x^{3} + 5x^{2} â€“ ax + 6

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

f(1) = 1^{3}_{ }+ (5 Ã— 1^{2}) â€“ (a Ã— 1) + 6

= 1 + 5 â€“ a + 6

= 12 â€“ a

From the question it is given that, remainder is 2a

So, 2a = 12 â€“ a

2a + a = 12

3a = 12

a = 12/3

a = 4

Therefore, the value of a is 4.

**6. (i) What number must be divided be subtracted from 2x ^{2} â€“ 5x so that the resulting polynomial leaves the remainder 2, when divided by 2x + 1?**

**Solution:-**

let us assume â€˜pâ€™ be subtracted from 2x^{2} â€“ 5x

So, dividing 2x^{2} â€“ 5x by 2x + 1,

Hence, remainder is 3 â€“ p

From the question it is given that, remainder is 2.

3 â€“ p = 2

p = 3 â€“ 2

p = 1

Therefore, 1 is to be subtracted.

**(ii) What number must be added to 2x ^{3} â€“ 7x^{2} + 2x so that the resulting polynomial leaves the remainder â€“ 2 when divided by 2x â€“ 3?**

**Solution:-**

let us assume â€˜pâ€™ be subtracted from 2x^{3} â€“ 7x^{2} + 2x,

So, dividing it by 2x â€“ 3,

Hence, remainder is p â€“ 6

From the question it is given that, remainder is – 2.

P â€“ 6 = – 2

P = -2 + 6

P = 4

Therefore, 4 is to be added.

**7. (i) When divided by x â€“ 3 the polynomials x ^{3} â€“ px^{2} + x + 6 and 2x^{3} â€“ x^{2} â€“ (p + 3) x â€“ 6 leave the same remainder. Find the value of â€˜pâ€™.**

**Solution:-**

From the question it is given that, by dividing x^{3} â€“ px^{2} + x + 6 and 2x^{3} â€“ x^{2} â€“ (p + 3)x â€“ 6 by x â€“ 3 = 0, then x = 3.

Let us assume p(x) = x^{3} â€“ px^{2} + x + 6

Now, substitute the value of x in p(x),

p(3) = 3^{3} â€“ (p Ã— 3^{2}) + 3 + 6

= 27 â€“ 9p + 9

= 36 â€“ 9p

Then, q(x) = 2x^{3} â€“ x^{2} â€“ (p + 3)x â€“ 6

Now, substitute the value of x in q(x),

q(3) = (2 Ã— 3^{3}) â€“ (3)^{2} â€“ (p + 3) Ã— 3 â€“ 6

= (2 Ã— 27) â€“ 9 â€“ 3p – 9 â€“ 6

= 54 â€“ 24 â€“ 3p

= 30 â€“ 3p

Given, the remainder in each case is same,

So, 36 â€“ 9p = 30 â€“ 3p

36 â€“ 30 = 9p â€“ 3p

6 = 6p

p = 6/6

p = 1

Therefore, value of p is 1.

**(ii) Find â€˜aâ€™ if the two polynomials ax ^{3} + 3x^{2} â€“ 9 and 2x^{3} + 4x + a, leaves the same remainder when divided by x + 3.**

**Solution:-**

Let us assume p(x) = ax^{3} + 3x^{2} â€“ 9 and q(x) = 2x^{3} + 4x + a

From the question it is given that, both p(x) and q(x) leaves the same remainder when divided by x + 3.

Let us assume that, x + 3 = 0

Then, x = -3

Now, substitute the value of x in p(x) and in q(x),

So, p(-3) = q(-3)

a(-3)^{3} + 3(-3)^{2} â€“ 9 = 2(-3)^{3} + 4(-3) + a

-27a + 27 â€“ 9 = – 54 â€“ 12 + a

-27a + 18 = – 66 + a

-27a â€“ a = -66 â€“ 18

-28 a = -84

a = 84/28

Therefore, a = 3

**(iii) The polynomials ax ^{3} + 3x^{2} â€“ 3 and 2x^{3} â€“ 5x + a when divided by x â€“ 4 leave the remainder r_{1} and r_{2} respectively. If 2r_{1} = r_{2}, then find the value of a. **

**Solution: **

Let us assume p(x) = ax^{3} + 3x^{2} â€“ 3 and q(x) = 2x^{3} â€“ 5x + a

From the question it is given that, both p(x) and q(x) leaves the remainder r_{1} and r_{2} respectively when divided by x – 4.

Also, given relation 2r_{1} = r_{2}

Let us assume that, x – 4 = 0

Then, x = 4

Now, substitute the value of x in p(x) and in q(x),

By factor theorem, r_{1} = p(x) and r_{2} = q(x)

So, 2 Ã— p(4) = q(4)

2[a(4)^{3} + 3(4)^{2} â€“ 3] = 2(4)^{3} â€“ 5(4) + a

2[64a + 48 – 3] = 128 â€“ 20 + a

128a + 96 â€“ 6 = 128 â€“ 20 + a

128a + 90 = 108 + a

128a â€“ a = 108 â€“ 90

127a = 18

a = 18/127

Therefore, the value of a = 18/127.

**8. Using remainder theorem, find the remainders obtained when x ^{3} + (kx + 8)x + k Is divided by x + 1 and x â€“ 2. Hence, find k if the sum of two remainders is 1.**

**Solution: **

Let us assume p(x) = x^{3} + (kx + 8)x + k

From the question it is given that, the sum of the remainders when p(x) is divided by (x + 1) and (x â€“ 2) is 1.

Let us assume that, x + 1 = 0

Then, x = -1

Also, when x â€“ 2 = 0

Then, x = 2

Now, by remainder theorem we have

p(-1) + p(2) = 1

(-1)^{3} + [k(-1) + 8](-1) + k + (2)^{3} + [k(2) + 8](2) + k = 1

-1 + k â€“ 8 + k + 8 + 4k + 16 + k = 1

7k + 15 = 1

7k = 1 â€“ 15

k = -14/7

k = -2

Therefore, k = -2.

**9. By factor theorem, show that (x + 3) and (2x – 1) are factors of 2x ^{2} + 5x â€“ 3.**

**Solution:-**

Let us assume, x + 3 = 0

Then, x = – 3

Given, f(x) = 2x^{2} + 5x â€“ 3

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

f(-3) = (2 Ã— (-3)^{2}) + (5 Ã— -3) â€“ 3

= (2 Ã— 9) + (-15) â€“ 3

= 18 â€“ 15 â€“ 3

= 18 â€“ 18

= 0

Now, 2x â€“ 1 = 0

Then, 2x = 1

x = Â½

Given, f(x) = 2x^{2} + 5x â€“ 3

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

f(Â½) = (2 Ã— (Â½)^{2}) + (5 Ã— Â½) â€“ 3

= (2 Ã— (Â¼)) + 5/2 â€“ 3

= Â½ + 5/2 â€“ 3

= (1 + 5)/2 â€“ 3

= 6/2 â€“ 3

= 3 -3

= 0

Hence, it is proved that, (x + 3) and (2x – 1) are factors of 2x^{2} + 5x â€“ 3.

**10. Without actual division, prove that x ^{4} + 2x^{3} â€“ 2x^{2} + 2x + 3 is exactly divisible by x^{2} + 2x â€“ 3.**

**Solution:-**

Consider x^{2} + 2x – 3

By factor method, x^{2} + 3x â€“ x â€“ 3

= x (x + 3) â€“ 1(x + 3)

= (x – 1) (x + 3)

So, f(x) = x^{4} + 2x^{3} â€“ 2x^{2} + 2x + 3

Now take, x + 3 = 0

X = -3

Then, f(-3) = (-3)^{4} + 2 Ã— -(3^{3}) â€“ (2 Ã— (-3)^{2}) + (2 Ã— -3) + 3

= 81 â€“ 54 â€“ 18 â€“ 6 â€“ 3

= 0

Therefore, (x + 3) is a factor of f(x)

And also, take x â€“ 1 = 0

X = 1

Then, f(1) = 1^{4} + 2(1)^{3} â€“ 2(1)^{2} + 2(1) â€“ 3

= 0

Therefore, (x – 1) is a factor of f(x)

By comparing both results, p(x) is exactly divisible by x^{2} + 2x â€“ 3.

**11. Show that (x – 2) is a factor of 3x ^{2} â€“ x â€“ 10. Hence factories 3x^{2} â€“ x â€“ 10.**

**Solution:-**

Let us assume x â€“ 2 = 0

Then, x = 2

Given, f(x) = 3x^{2} â€“ x â€“ 10

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

f(2) = (3 Ã— 2^{2}) â€“ 2 â€“ 10

= (3 Ã— 4) â€“ 2 â€“ 10

= 12 â€“ 2 â€“ 10

= 12 â€“ 12

= 0

Therefore, (x – 2) is a factor of f(x)

Then, dividing (3x^{2} â€“ x – 10) by (x – 2), we get

Therefore, 3x^{2} â€“ x â€“ 10 = (x – 2) (3x + 5)

**12. Using the factor theorem, show that (x – 2) is a factor of x ^{3} + x^{2} – 4x â€“ 4. Hence factorize the polynomial completely.**

**Solution:-**

Let us assume, x â€“ 2 = 0

Then, x = 2

Given, f(x) = x^{3} + x^{2} – 4x â€“ 4

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

f(2) = (2)^{3} + (2)^{2} â€“ 4(2) â€“ 4

= 8 â€“ 4 â€“ 8 â€“ 4

= 0

Therefore, by factor theorem (x – 2) is a factor of x^{3} + x^{2} – 4x â€“ 4

Then, dividing f(x) by (x – 2), we get

Therefore, x^{3} + x^{2} – 4x â€“ 4 = (x – 2) (x^{2} + 3x + 2)

= (x â€“ 2) (x^{2} + 2x + x + 2)

= (x â€“ 2) (x(x + 2) + 1(x + 2))

= (x â€“ 2) (x + 2) (x + 1)

**13. Show that 2x + 7 is a factor of 2x ^{3} + 5x^{2} â€“ 11x â€“ 14. Hence factorize the given expression completely, using the factor theorem.**

**Solution:-**

Let us assume 2x + 7 = 0

Then, 2x = -7

X = -7/2

Given, f(x) = 2x^{3} + 5x^{2} â€“ 11x â€“ 14

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

f(-7/2) = 2(-7/2)^{3} + 5(-7/2)^{2} + 11(-7/2) â€“ 14

= 2(-343/8) + 5(49/4) + (-77/2) â€“ 14

= -343/4 + 245/4 â€“ 77/2 â€“ 14

= (-343 + 245 + 154 – 56)/4

= -399 + 399/4

= 0

Therefore, (2x + 7) is a factor of 2x^{3} + 5x^{2} â€“ 11x â€“ 14

Then, dividing f(x) by (2x + 1), we get

Therefore, 2x^{3} + 5x^{2} â€“ 11x â€“ 14 = (2x + 7) (x^{2} â€“ x – 2)

= (2x + 7) (x^{2} â€“ 2x + x – 2)

= (2x + 7) (x(x – 2) + 1 (x – 2))

= (x + 1) (x – 2) (2x + 7)

**14. Use factor theorem to factorize the following polynomials completely.**

**(i) x ^{3} + 2x^{2} â€“ 5x â€“ 6**

**Solution:-**

Let us assume x = -1,

Given, f(x) = x^{3} + 2x^{2} â€“ 5x â€“ 6

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

f(-1) = (-1)^{3} + 2(-1)^{2} â€“ 5(-1) â€“ 6

= -1 +2 (1) + 5 â€“ 6

= -1 +2 + 5 â€“ 6

= -7 + 7

= 0

Then, dividing f(x) by (x + 1), we get

Therefore, x^{3} + 2x^{2} â€“ 5x â€“ 6 = (x + 1) (x^{2} + 3x â€“ 2x – 6)

= (x + 1) (x(x + 3) â€“ 2(x + 3))

= (x + 1) (x – 2) (x + 3)

**(ii) x ^{3} â€“ 13x â€“ 12**

**Solution:-**

Let us assume x = -1,

Given, f(x) = x^{3} â€“ 13x â€“ 12

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

f(-1) = (-1)^{3} – 13(-1) â€“ 12

= -1 + 13 – 12

= – 13 + 13

= 0

Then, dividing f(x) by (x + 1), we get

Therefore, x^{3} â€“ 13x â€“ 12 = (x + 1) (x^{2} â€“ x – 12)

= (x + 1) (x^{2} â€“ 4x + 3x – 12)

= (x + 1) (x(x – 4)) + 3(x – 4))

= (x + 1) (x + 3) (x – 4)

**15. Use the remainder theorem to factorize the following expression.**

**(i) 2x ^{3} + x^{2} â€“ 13x + 6**

**Solution:-**

Let us assume x = 2,

Then, f(x) = 2x^{3} + x^{2} â€“ 13x + 6

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

f(2) = (2 Ã— 2^{3}) + 2^{2}_{ }â€“ 13 Ã— 2 + 6

= (2 Ã— 8) + 4 â€“ 26 + 6

= 16 + 4 â€“ 26 + 6

= 26 â€“ 26

= 0

Then, dividing f(x) by (x – 2), we get

Therefore, 2x^{3} + x^{2} â€“ 13x + 6 = (x – 2) (2x^{2} + 5x – 3)

= (x – 2)(2x^{2} + 6x â€“ x â€“ 3)

= (x – 2) (2x(x + 3) â€“ 1 (x + 3))

= (x – 2) (x + 3) (2x – 1)

**(ii) 3x ^{2} + 2x^{2} â€“ 19x + 6**

**Solution:-**

Given, f(x) = 3x^{3} + 2x^{2} â€“ 19x + 6

Let us assume x = 1

Then, f(1) = 3(1)^{3} + 2(1)^{2} â€“ (19 Ã— 1) + 6

= 3 + 2 â€“ 19 + 6

= 11 â€“ 19

= – 8

So, – 8 â‰ 0

Let us assume x = -1

Then, f(-1) = 3(-1)^{3} + 2(-1)^{2} â€“ (19 Ã— (-1)) + 6

= – 3 + 2 + 19 + 6

= – 3 + 27

= 24

So, 24 â‰ 0

Now, assume x = 2

Then, f(2) = 3(2)^{3} + 2(2)^{2} â€“ (19 Ã— (2)) + 6

= 24 + 8 – 38 + 6

= 38 – 38

= 0

So, 0 = 0

Therefore, (x – 2) is a factor of f(x).

f(x) = 3x^{3} + 2x^{2} â€“ 19x + 6

= 3x^{3} â€“ 6x^{2} + 8x^{2} â€“ 16x â€“ 3x + 6

= 3x^{2} (x – 2) + 8x (x – 2) â€“ 3(x – 2)

= (x – 2) (3x^{2} + 8x – 3)

= (x – 2) (3x^{2} + 9x â€“ x â€“ 3)

= (x – 2) (3x(x + 3) â€“ 1(x + 3)

= (x – 2) (x + 3) (3x – 1)

**(iii) 2x ^{3} + 3x^{2} â€“ 9x â€“ 10**

**Solution:-**

Given, f(x) = 2x^{3} + 3x^{2} â€“ 9x â€“ 10

Let us assume, x = -1

= 2(-1)^{3} + 3(-1)^{2} â€“ 9 (-1) â€“ 10

= -2 + 3 + 9 â€“ 10

= 12 â€“ 12

= 0

Therefore, (x + 1) is the factor of 2x^{3} + 3x^{2} â€“ 9x â€“ 10

Then, dividing f(x) by (x + 1), we get

Therefore, 2x^{3} + 3x^{2} â€“ 9x â€“ 10 = 2x^{2} + 5x â€“ 4x â€“ 10

= x(2x + 5) â€“ 2 (2x + 5) â€“ (2x + 5) (x – 2)

Hence the factors are (x + 1) (x – 2) (2x + 5)

**(iv) x ^{3} + 10x^{2} â€“ 37x + 26**

**Solution:-**

Given, f(x) = x^{3} + 10x^{2} â€“ 37x + 26

Let us assume, x = 1

Then, f(1) = 1^{3} + 10(1)^{2} â€“ 37 (1) + 26

= 1 + 10 â€“ 37 + 26

= 37 â€“ 37

= 0

Therefore, x â€“ 1is a factor of x^{3} + 10x^{2} â€“ 37x + 26

Then, dividing f(x) by (x – 1), we get

Therefore, x^{3} + 10x^{2} â€“ 37x + 26 = (x – 1) (x^{2} + 11x – 26)

= (x – 1) (x^{2} + 13x â€“ 2x – 26)

= (x – 1) (x (x + 13) â€“ 2(x + 13))

= (x – 1) ((x – 2) (x + 13))

**16. If (2x + 1) is a factor of 6x ^{3} + 5x^{2} + ax â€“ 2 find the value of a.**

**Solution:-**

Let us assume 2x + 1 = 0

Then, 2x = – 1

X = -Â½

Given, f(x) = 6x^{3} + 5x^{2} + ax â€“ 2

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

f (-Â½) = 6 (-Â½)^{3} + 5 (-Â½)^{2} + a (-Â½) â€“ 2

= 6 (-1/8) + 5 (Â¼) â€“ Â½a â€“ 2

= -3/4 + 5/4 â€“ a/2 â€“ 2

= (-3 + 4 â€“ 2a – 8)/4

= (-6 â€“ 2a)/4

From the question, (2x + 1) is a factor of 6x^{3} + 5x^{2} + ax â€“ 2

Then, remainder is 0.

So, (-6 â€“ 2a)/4 = 0

-6 â€“ 2a = 4 Ã— 0

– 6 â€“ 2a = 0

-2a = 6

a = -6/2

a = – 3

Therefore, the value of a is â€“ 3.

**17. If (3x – 2) is a factor of 3x ^{3} â€“ kx^{2} + 21x â€“ 10, find the value of k.**

**Solution:-**

Let us assume 3x – 2 = 0

Then, 3x = 2

X = 2/3

Given, f(x) = 3x^{3} â€“ kx^{2} + 21x â€“ 10

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

f (2/3) = 3 (2/3)^{3} â€“ k (2/3)^{2} + 21 (2/3) â€“ 10

= 3 (8/27) â€“ k (4/9) + 14 â€“ 10

= 8/9 â€“ 4k/9 + 14 â€“ 10

= 8/9 â€“ 4k/9 + 4

= (8 â€“ 4k + 36)/9

= (44 â€“ 4k)/9

From the question, (3x – 2) is a factor of 3x^{3} â€“ kx^{2} + 21x â€“ 10

Then, remainder is 0

So, (44 â€“ 4k)/9 = 0

44 â€“ 4k = 0 Ã— 9

44 = 4k

K = 44/4

K = 11

**18. If (x â€“ 2) is a factor of 2x ^{3}Â â€“ x^{2}Â + px â€“ 2, then**Â

**(i) find the value of p.**Â

**(ii) with this value of p, factorize the above expression completely.**

**Solution:-**

Let us assume x -2 = 0

Then, x = 2

Given, f(x) = 2x^{3}Â â€“ x^{2}Â + px â€“ 2

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

f(2) = (2 Ã— 2^{3}) â€“ 2^{2} + (p Ã— 2) â€“ 2

= (2 Ã— 8) â€“ 4 + 2p â€“ 2

= 16 â€“ 4 + 2p â€“ 2

= 16 â€“ 6 + 2p

= 10 + 2p

From the question, (x â€“ 2) is a factor of 2x^{3}Â â€“ x^{2}Â + px â€“ 2

Then, remainder is 0.

10 + 2p = 0

2p = – 10

P = -10/2

P = -5

So, (x â€“ 2) is a factor of 2x^{3}Â â€“ x^{2}Â + 5x â€“ 2

Therefore, 2x^{3}Â â€“ x^{2}Â + 5x â€“ 2 = (x – 2) (2x^{2} + 3x + 1)

= (x – 2) (2x^{2} + 2x + x + 1)

= (x – 2) (2x(x + 1) + 1(x + 1))

= (x + 1) (x – 2) (2x + 1)

**19. What number should be subtracted from 2x ^{3}Â â€“ 5x^{2}Â + 5x so that the resulting polynomial has 2x â€“ 3 as a factor?**

**Solution:-**

Let us assume the number to be subtracted from 2x^{3}Â â€“ 5x^{2}Â + 5x be p.

Then, f(x) = 2x^{3}Â â€“ 5x^{2}Â + 5x â€“ p

Given, 2x â€“ 3 = 0

x = 3/2

f(3/2) = 0

So, f(3/2) = 2(3/2)^{3} â€“ 5(3/2)^{2} + 5(3/2) â€“ p = 0

2(27/8) â€“ 5(9/4) + 15/2 â€“ p = 0

27/4 – 45/4 + 15/2 â€“ p = 0 [multiply by 4 for all numerator]

** **27 â€“ 45 + 30 â€“ 4p = 0

57 â€“ 45 â€“ 4p = 0

12 â€“ 4p = 0

P = 12/4

P = 3

Therefore, 3 is the number should be subtracted from 2x^{3}Â â€“ 5x^{2}Â + 5x.

**20. (i) Find the value of the constants a and b, if (x â€“ 2) and (x + 3) are both factors of the expression x ^{3}Â + ax^{2}Â + bx â€“ 12.**

**Solution:-**

Let us assume x â€“ 2 = 0

Then, x = 2

Given, f(x) = x^{3}Â + ax^{2}Â + bx â€“ 12

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

f(2) = 2^{3} + a(2)^{2} + b(2) â€“ 12

= 8 + 4a + 2b â€“ 12

= 4a + 2b â€“ 4

From the question, (x – 2) is a factor of x^{3}Â + ax^{2}Â + bx â€“ 12.

So, 4a + 2b â€“ 4 = 0

4a + 2b = 4

By dividing both the side by 2 we get,

2a + b = 2 â€¦ [equation (i)]

Now, assume x + 3 = 0

Then, x = -3

Given, f(x) = x^{3}Â + ax^{2}Â + bx â€“ 12

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

f(-3) = (-3)^{3} + a(-3)^{2} + b(-3) â€“ 12

= -27 + 9a – 3b â€“ 12

= 9a – 3b â€“ 39

From the question, (x – 3) is a factor of x^{3}Â + ax^{2}Â + bx â€“ 12.

So, 9a – 3b â€“ 39 = 0

9a â€“ 3b = 39

By dividing both the side by 3 we get,

3a â€“ b = 13 â€¦ [equation (ii)]

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

(2a + b) + (3a – b) = 2 + 13

2a + 3a + b â€“ b = 15

5a = 15

a = 15/5

a = 3

Consider the equation (i) to find out â€˜bâ€™.

2a + b = 2

2(3) + b = 2

6 + b = 2

b = 2 â€“ 6

b = -4

**(ii) If (x + 2) and (x + 3) are factors of x ^{3}Â + ax + b, Find the values of a and b.**

**Solution:-**

Let us assume x + 2 = 0

Then, x = -2

Given, f(x) = x^{3}Â + ax + b

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

f(-2) = (-2)^{3} + a(-2) + b

= -8 â€“ 2a + b

From the question, (x + 2) is a factor of x^{3}Â + ax + b.

Therefore, remainder is 0.

f(x) = 0

– 8 â€“ 2a + b = 0

2a – b = – 8 â€¦ [equation (i)]

Let us assume x + 3 = 0

Then, x = -3

Given, f(x) = x^{3}Â + ax + b

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

f(-2) = (-3)^{3} + a(-3) + b

= -27 â€“ 3a + b

From the question, (x + 3) is a factor of x^{3}Â + ax + b.

Therefore, remainder is 0.

f(x) = 0

– 27 â€“ 3a + b = 0

3a – b = – 27 â€¦ [equation (i)]

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

(2a â€“ b) – (3a – b) = -8 â€“ (-27)

2a – 3a â€“ b + b = – 8 + 27

-a = 19

a = -19

Consider the equation (i) to find out â€˜bâ€™.

2a – b = – 8

2(-19) â€“ b = -8

-38 â€“ b = – 8

b = -38 +8

b = -30

**21. If (x + 2) and (x â€“ 3) are factors of x ^{3}Â + ax + b, find the values of a and b. With these values of a and b, factorize the given expression.**

**Solution:-**

Let us assume x + 2 = 0

Then, x = -2

Given, f(x) = x^{3}Â + ax + b

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

f(-2) = (-2)^{3} + a(-2) + b

= -8 â€“ 2a + b

From the question, (x + 2) is a factor of x^{3}Â + ax + b.

Therefore, remainder is 0.

f(x) = 0

– 8 â€“ 2a + b = 0

2a – b = – 8 â€¦ [equation (i)]

Now, assume x â€“ 3 = 0

Then, x = 3

Given, f(x) = x^{3}Â + ax + b

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

f(3) = (3)^{3} + a(3) + b

= 27 + 3a + b

From the question, (x – 3) is a factor of x^{3}Â + ax + b.

Therefore, remainder is 0.

f(x) = 0

27 + 3a + b = 0

3a + b = – 27 â€¦ [equation (ii)]

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

(2a – b) + (3a + b) = – 8 â€“ 27

2a â€“ b + 3a + b = -35

5a = -35

a = -35/5

a = -7

Consider the equation (i) to find out â€˜bâ€™.

2a â€“ b = – 8

2(-7) â€“ b = -8

-14 â€“ b = -8

b = – 14 + 8

b = -6

Therefore, value of a = -7 and b = -6.

Then, f(x) = x^{3} â€“ 7x â€“ 6

(x + 2) (x – 3)

= x(x – 3) + 2(x – 3)

= x^{2} â€“ 3x + 2x â€“ 6

= x^{2} â€“ x â€“ 6

Dividing f(x) by x^{2} â€“ x â€“ 6 we get,

Therefore, x^{3} â€“ 7x â€“ 6 = (x + 1) (x + 2) (x – 3)

**22. (x â€“ 2) is a factor of the expression x ^{3}Â + ax^{2}Â + bx + 6. When this expression is divided by (x â€“ 3), it leaves the remainder 3. Find the values of a and b.**

**Solution:-**

From the question it is given that, (x â€“ 2) is a factor of the expression x^{3}Â + ax^{2}Â + bx + 6

Then, f(x) = x^{3}Â + ax^{2}Â + bx + 6 â€¦ [equation (i)]

Let assume x â€“ 2 = 0

Then, x = 2

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

f(2) = 2^{3}_{ }+ a(2)^{2} + 2b + 6

= 8 + 4a + 2b + 6

= 14 + 4a + 2b

By dividing the numbers by 2 we get,

= 7 + 2a + b

From the question, (x â€“ 2) is a factor of the expression x^{3}Â + ax^{2}Â + bx + 6.

So, remainder is 0.

f(x) = 0

7 + 2a + b = 0

2a + b = -7 â€¦ [equation (ii)]

Now, expression is divided by (x â€“ 3), it leaves the remainder 3.

So, remainder = 33 + 9a + 3b = 3

9a + 3b = 3 â€“ 33

9a + 3b = -30

By dividing the numbers by 3 we get,

= 3a + b = – 10 â€¦ [equation (iii)]

Now, subtracting equation (iii) from equation (ii) we get,

(3a + b) â€“ (2a + b) = – 10 â€“ (-7)

3a â€“ 2a + b â€“ b = – 10 + 7

a = -3

Consider the equation (ii) to find out â€˜bâ€™.

2a + b = – 7

2(-3) + b = – 7

-6 + b = – 7

b = – 7 + 6

b = – 1

**23. If (x â€“ 2) is a factor of the expression 2x ^{3}Â + ax^{2}Â + bx â€“ 14 and when the expression is divided by (x â€“ 3), it leaves a remainder 52, find the values of a and b.**

**Solution:-**

From the question it is given that, (x â€“ 2) is a factor of the expression 2x^{3}Â + ax^{2}Â + bx â€“ 14

Then, f(x) = 2x^{3}Â + ax^{2}Â + bx â€“ 14 â€¦ [equation (i)]

Let assume x â€“ 2 = 0

Then, x = 2

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

f(2) = 2(2)^{3}_{ }+ a(2)^{2} + 2b – 14

= 16 + 4a + 2b – 14

= 2 + 4a + 2b

By dividing the numbers by 2 we get,

= 1 + 2a + b

From the question, (x â€“ 2) is a factor of the expression 2x^{3}Â + ax^{2}Â + bx â€“ 14.

So, remainder is 0.

f(x) = 0

1 + 2a + b = 0

2a + b = -1 â€¦ [equation (ii)]

Now, expression is divided by (x â€“ 3), it leaves the remainder 52.

So, remainder = 9a + 3b + 40 = 52

9a + 3b = 52 â€“ 40

9a + 3b = 12

By dividing the numbers by 3 we get,

= 3a + b = 4 â€¦ [equation (iii)]

Now, subtracting equation (iii) from equation (ii) we get,

(3a + b) â€“ (2a + b) = 4 â€“ (-1)

3a â€“ 2a + b â€“ b = 4 + 1

a = 5

a = 5

Consider the equation (ii) to find out â€˜bâ€™.

2a + b = – 1

2(5) + b = – 1

10 + b = – 1

b = – 1 – 10

b = – 11

**24. If ax ^{3}Â + 3x^{2}Â + bx â€“ 3 has a factor (2x + 3) and leaves remainder â€“ 3 when divided by (x + 2), find the values of a and b. With these values of a and b, factorize the given expression.**

**Solution:-**

Let us assume, 2x + 3 = 0

Then, 2x = -3

x = -3/2

Given, f(x) = ax^{3}Â + 3x^{2}Â + bx â€“ 3

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

f(-3/2) = a(-3/2)^{3} + 3(-3/2)^{2} + b(-3/2) – 3

= a(-27/8) + 3(9/4) â€“ 3b/2 – 3

= -27a/8 + 27/4 â€“ 3b/2 â€“ 3

From the question it is given that,** **ax^{3}Â + 3x^{2}Â + bx â€“ 3 has a factor (2x + 3)**.**

So, remainder is 0.

-27a/8 + 27/4 â€“ 3b/2 â€“ 3 = 0

-27a + 54 â€“ 12b â€“ 24 = 0

-27a â€“ 12b = -30

By dividing the numbers by – 3 we get,

9a + 4b = 10 [equation (i)]

Now, let us assume x + 2 = 0

Then, x = -2

Given, f(x) = ax^{3}Â + 3x^{2}Â + bx â€“ 3

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

f(2) = a(-2)^{3} + 3(-2)^{2} + b(-2) – 3

= -8a + 12 â€“ 2b – 3

= -8a – 2b + 9

Leaves the remainder -3

So, -8a â€“ 2b + 9 = -3

-8a â€“ 2b = -3 â€“ 9

-8a â€“ 2b = -12

By dividing both sides by -2 we get,

4a + b = 6 [equation (ii)]

By multiplying equation (ii) by 4,

16a + 4b = 24

Now, subtracting equation (ii) from equation (i) we get,

(16a + 4b) – (9a + 4b) = 24 â€“ 10

16a – 9a + 4b â€“ 4b = 14

7a = 14

a = 14/7

a = 2

Consider the equation (i) to find out â€˜bâ€™.

9a + 4b = 10

9(2) + 4b = 10

18 + 4b = 10

4b = 10 â€“ 18

4b = -8

b = -8/4

b = -2

Therefore, f(x) = ax^{3}Â + 3x^{2}Â + bx â€“ 3

= 2x^{3} + 3x^{2} â€“ 2x â€“ 3

Given, 2x + 3 is a factor of f(x)

So, divide f(x) by 2x + 3

Therefore, 2x^{3} + 3x^{2} â€“ 2x â€“ 3 = (2x + 3) (x^{2} – 1)

= (2x + 3) (x + 1) (x – 1)

**25. Given f(x) = ax ^{2}Â + bx + 2 and g(x) = bx^{2}Â + ax + 1. If x â€“ 2 is a factor of f(x) but leaves the remainder â€“ 15 when it divides g(x), find the values of a and b. With these values of a and b, factorize the expression. f(x) + g(x) + 4x^{2}Â + 7x.**

**Solution:-**

From the question it is given that, f(x) = ax^{2}Â + bx + 2 and g(x) = bx^{2}Â + ax + 1 and x â€“ 2 is a factor of f(x),

So, x = 2

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

f(2) = 0

a(2)^{2} + b(2) + 2 = 0

4a + 2b + 2 = 0

By dividing both sides by 2 we get,

2a + b + 1 = 0 â€¦ [equation (i)]

Given, g(x) divide by (x – 2), leaves remainder â€“ 15

g(x) = bx^{2}Â + ax + 1

So, g(2) = -15

b(2)^{2} + 2a + 1 = -15

4b + 2a + 1 + 15 = 0

4b + 2a + 16 = 0

By dividing both sides by 2 we get,

2b + a + 8 = 0 â€¦ [equation (ii)]

Now, subtracting equation (ii) from equation (i) multiplied by 2,

(4a + 2b + 2) â€“ (a + 2b + 8) = 0 â€“ 0

4a â€“ a + 2b â€“ 2b + 2 â€“ 8 = 0

3a â€“ 6 = 0

3a = 6

a = 6/3

a = 2

Consider the equation (i) to find out â€˜bâ€™.

2a + b + 1 = 0

2(2) + b = – 1

4 + b = – 1

b = – 1 – 4

b = – 5

Now, f(x) = ax^{2}Â + bx + 2 = 2x^{2} â€“ 5x + 2

g(x) = bx^{2}Â + ax + 1 = -5x^{2} + 2x + 1

then, f(x) + g(x) + 4x^{2}Â + 7x

= 2x^{2} â€“ 5x + 2 â€“ 5x^{2} + 2x + 1 + 4x^{2} + 7x

= x^{2} + 4x + 3

= x^{2} + 3x + x + 3

= x(x + 3) + 1(x + 3)

= (x + 1) (x + 3)

Chapter Test

**1. Find the remainder when 2x ^{3}Â â€“ 3x^{2}Â + 4x + 7 is divided by**Â

**(i) x â€“ 2**Â

**(ii) x + 3**Â

**(iii) 2x + 1**

**Solution:-**

From the question it is given that, f(x) = 2x^{3} â€“ 3x^{2} + 4x + 7

(i) Consider x -2

let us assume x â€“ 2 = 0

Then, x = 2

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

f(2) = 2(2)^{3} â€“ 3(2)^{2} + 4(2) + 7

= 16 â€“ 12 + 8 + 7

= 31 -12

= 19

Therefore, the remainder is 19

(ii) consider x + 3

let us assume x + 3 = 0

Then, x = -3

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

f(2) = 2(-3)^{3} â€“ 3(-3)^{2} + 4(-3) + 7

= 2(-27) â€“ 3(9) – 12 + 7

= – 54 â€“ 27 â€“ 12 + 7

= – 93 + 7

= – 86

Therefore, remainder is -86.

(iii) consider 2x + 1

Let us assume, 2x + 1 = 0

Then, 2x = -1

X = -Â½

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

f (-Â½) = 2 (-Â½)^{3} â€“ 3(-Â½)^{2} + 4 (-Â½) + 7

= 2(-1/8) â€“ 3 (Â¼) + 4 (-Â½) + 7

= -Â¼ – Â¾ – 2 + 7

= -1 â€“ 2 + 7

= 4

Therefore, remainder is 4.

**2. When 2x ^{3}Â â€“ 9x^{2}Â + 10x â€“ p is divided by (x + 1), the remainder is â€“ 24. Find the value of p.**

**Solution:-**

Let us assume x + 1 = 0

Then, x = -1

Given, f(x) = 2x^{3} â€“ 9x^{2} + 10x â€“ p

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

f(-1) = 2(-1)^{3} â€“ 9(-1)^{2} + 10(-1) â€“ p

= -2 – 9 â€“ 10 + p

= -21 + p

From the question it is given that, the remainder is â€“ 24,

So, -21 + p = -24

p = – 24 + 21

p = -3

So, f(x) = 2x^{3} â€“ 9x^{2} + 10x â€“ (-3)

= 2x^{3} â€“ 9x^{2} + 10x + 3

Therefore, the value of p is 3.

**3. If (2x â€“ 3) is a factor of 6x ^{2}Â + x + a, find the value of a. With this value of a, factorise the given expression.**

**Solution:-**

Let us assume 2x â€“ 3 = 0

Then, 2x = 3

X = 3/2

Given, f(x) = 6x^{2}Â + x + a

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

f(3/2) = 6(3/2)^{2} + (3/2) + a

= 6(9/4) + (3/2) + a

= 3(9/2) + (3/2) + a

= 27/2 + 3/2 + a

= 30/2 + a

= 15 + a

From the question, (2x â€“ 3) is a factor of 6x^{2}Â + x + a.

So, remainder is 0.

Then, 15 + a = 0

a = -15

Therefore, f(x) = 6x^{2} + x â€“ 15

Dividing f(x) by 2x â€“ 3 we get,

Therefore, 6x^{2} + x â€“ 15 = (2x – 3) (3x + 5)

**4. When 3x ^{2}Â â€“ 5x + p is divided by (x â€“ 2), the remainder is 3. Find the value of p. Also factorize the polynomial 3x^{2}Â â€“ 5x + p â€“ 3.**

**Solution:-**

Let us assume x â€“ 2 = 0

Then, x = 2

Given, f(x) = 3x^{2} â€“ 5x + p

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

So, f(2) = 3(2)^{2} â€“ 5(2) + p

= 3(4) â€“ 10 + p

= 12 â€“ 10 + p

= 2 + p

From the question it is given that, remainder is 3.

So, 2 + p = 3

p = 3 â€“ 2

p = 1

Therefore, f(x) = 3x^{2} â€“ 5x + 1

Consider the polynomial, 3x^{2}Â â€“ 5x + p â€“ 3

Now, substitute the value of p in polynomial,

= 3x^{2} â€“ 5x + 1 â€“ 3

= 3x^{2} â€“ 5x â€“ 2

Now, by factorizing the polynomial 3x^{2} â€“ 5x â€“ 2,

Dividing 3x^{2} â€“ 5x â€“ 2 by x â€“ 2 we get,

Therefore, 3x^{2} â€“ 5x â€“ 2 = (x – 2) (3x + 1)

**5. Prove that (5x + 4) is a factor of 5x ^{3}Â + 4x^{2}Â â€“ 5x â€“ 4. Hence factorize the given polynomial completely.**

**Solution:-**

Let us assume (5x + 4) = 0

Then, 5x = -4

x = -4/5

Given, f(x) = 5x^{3}Â + 4x^{2}Â â€“ 5x â€“ 4

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

So, f(-4/5) = 5(-4/5)^{3}Â + 4(-4/5)^{2}Â â€“ 5(-4/5) â€“ 4

= 5(-64/125) + 4 (16/25) + 4 â€“ 4

= -64/25 + 64/25

= (-64 + 64)/25

= 0/25

= 0

Hence, (5x + 4) is a factor of 5x^{3}Â + 4x^{2}Â â€“ 5x â€“ 4.

So, dividing 5x^{3}Â + 4x^{2}Â â€“ 5x â€“ 4 by 5x + 4 we get,

Therefore, 5x^{3}Â + 4x^{2}Â â€“ 5x â€“ 4 = (5x + 4) (x^{2} – 1)

= (5x + 4) (x^{2} â€“ 1^{2})

= (5x + 4) (x + 1) (x – 1)

**6. Use factor theorem to factorize the following polynomials completely:**Â

**(i) 4x ^{3}Â + 4x^{2}Â â€“ 9x â€“ 9**Â

**Solution:-**

Let us assume x = -1,

Given, f(x) = 4x^{3}Â + 4x^{2}Â â€“ 9x â€“ 9Â

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

f(-1) = 4(-1)^{3}Â + 4(-1)^{2}Â â€“ 9(-1) â€“ 9

= -4 + 4 + 9 â€“ 9

= 0

Therefore, x + 1 is the factor of 4x^{3}Â + 4x^{2}Â â€“ 9x â€“ 9.

Now, dividing 4x^{3}Â + 4x^{2}Â â€“ 9x â€“ 9 by x + 1 we get,

Therefore, 4x^{3}Â + 4x^{2}Â â€“ 9x â€“ 9 = (x + 1) (4x^{2} – 9)

= (x + 1) ((2x)^{2} â€“ (3)^{2})

= (x + 1) (2x + 3) (2x – 3)

**(ii) x ^{3}Â â€“ 19x â€“ 30**

**Solution:-**

Let us assume x = -2,

Given, f(x) = x^{3}Â â€“ 19x â€“ 30

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

f(-1) = (-2)^{3}Â â€“ 19(-2) â€“ 30

= -8 + 38 â€“ 30

= -38 + 38

= 0

Therefore, x + 2 is the factor of x^{3}Â â€“ 19x â€“ 30.

Now, dividing x^{3}Â â€“ 19x â€“ 30 by x + 2 we get,

Therefore, x^{3}Â â€“ 19x â€“ 30 = (x + 2)(x^{2} â€“ 2x – 15)

= (x + 2) (x^{2} â€“ 5x + 3x – 15)

= (x + 2) (x – 5) (x + 3)

**7. If x ^{3}Â â€“ 2x^{2}Â + px + q has a factor (x + 2) and leaves a remainder 9, when divided by (x + 1), find the values of p and q. With these values of p and q, factorize the given polynomial completely.**

**Solution:-**

From the question it is given that, (x + 2) is a factor of the expression x^{3}Â â€“ 2x^{2}Â + px + q

Then, f(x) = x^{3}Â â€“ 2x^{2}Â + px + q

Let assume x + 2 = 0

Then, x = -2

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

f(-2) = (-2)^{3}Â â€“ 2(-2)^{2}Â + p(-2) + q

= -8 â€“ 8 â€“ 2p + q

= -16 â€“ 2p + q

2p â€“ q = – 16 â€¦ [equation (i)]

Now, consider (x + 1)

Then, f(x) = x^{3}Â â€“ 2x^{2}Â + px + q

Let assume x + 1 = 0

Then, x = -1

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

f(-1) = (-1)^{3}Â â€“ 2(-1)^{2}Â + p(-1) + q

= -1 â€“ 2 â€“p + q

= – 3 â€“ p + q

Given, remainder is 9

So, -3 â€“ p + q = 9

– p + q = 9 + 3

-p + q = 12 â€¦ [equation (ii)]

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

(2p â€“ q) + (-p + q) = – 16 + 12

2p â€“ q â€“ p + q = -4

P = -4

Consider the equation (ii) to find out â€˜bâ€™.

– p + q = 12

-(-4) + q = 12

4 + q = 12

q = 12 â€“ 4

q = 8

Therefore, by substituting the value of p and q f(x) = x^{3} â€“ 2x^{2} â€“ 4x + 8

Dividing f(x) be (x + 2) we get,

x^{3} â€“ 2x^{2} â€“ 4x + 8 = (x + 2) (x^{2} â€“ 4x + 4)

= (x + 2) (x^{2} â€“ 2 Ã— x (-2) + 2^{2})

= (x + 2) (x – 2)^{2}

**8. If (x + 3) and (x â€“ 4) are factors of x ^{3}Â + ax^{2}Â â€“ bx + 24, find the values of a and b: With these values of a and b, factorize the given expression.**

**Solution:-**

Let us assume x + 3 = 0

Then, x = -3

Given, f(x) = x^{3}Â + ax^{2}Â â€“ bx + 24

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

f(-3) = (-3)^{3}Â + a(-3)^{2}Â â€“ b(-3) + 24

= -27 + 9a + 3b + 24

= 9a + 3b â€“ 3

Dividing all terms by 3 we get,

= 3a + b – 1

From the question, (x + 3) is a factor of x^{3}Â + ax^{2}Â â€“ bx + 24.

Therefore, remainder is 0.

f(x) = 0

3a + b â€“ 1 = 0

3a + b = 1 â€¦ [equation (i)]

Now, assume x â€“ 4 = 0

Then, x = 4

Given, f(x) = x^{3}Â + ax^{2}Â â€“ bx + 24

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

f(4) = 4^{3}Â + a(4)^{2}Â â€“ b(4) + 24

= 64 + 16a – 4b + 24

= 88 + 16a â€“ 4b

Dividing all terms by 4 we get,

= 22 + 4a – b

From the question, (x – 4) is a factor of x^{3}Â + ax^{2}Â â€“ bx + 24.

Therefore, remainder is 0.

f(x) = 0

22 + 4a – b = 0

4a – b = – 22 â€¦ [equation (ii)]

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

(3a + b) + (4a – b) = 1 â€“ 22

3a + b + 4a â€“ b = – 21

7a = – 21

a = -21/7

a = -3

Consider the equation (i) to find out â€˜bâ€™.

3a + b = 1

3(-3) + b = 1

-9 + b = 1

b = 1 + 9

b = 10

Therefore, value of a = -3 and b = 10.

Then, by substituting the value of a and b f(x) = x^{3}Â – 3x^{2}Â â€“ 10x + 24

(x + 3) (x – 4)

= x(x – 4) + 3(x – 4)

= x^{2} â€“ 4x + 3x â€“ 12

= x^{2} â€“ x â€“ 12

Dividing f(x) by x^{2} â€“ x â€“ 12 we get,

Therefore, x^{3}Â – 3x^{2}Â â€“ 10x + 24 = (x^{2} â€“ x â€“ 12) (x – 2)

= (x + 3) (x – 4) (x – 2)

**9. If (2x + 1) is a factor of both the expressions 2x ^{2}Â â€“ 5x + p and 2x^{2}Â + 5x + q, find the value of p and q. Hence find the other factors of both the polynomials.**

**Solution:-**

Let us assume 2x + 1 = 0

Then, 2x = -1

x = -Â½

Given, p(x) = 2x^{2}Â â€“ 5x + p

Now, substitute the value of x in p(x),

p (-Â½) = 2 (-Â½)^{2}Â â€“ 5(-Â½) + p

= 2(1/4) + 5/2 + p

= Â½ + 5/2 + p

= 6/2 + p

= 3 + p

From the question it is given that, (2x + 1) is a factor of both the expressions 2x^{2}Â â€“ 5x + p

So, remainder is 0.

Then, 3 + p = 0

p = – 3

Now consider q(x) = 2x^{2}Â + 5x + q

Substitute the value of x in q(x)

q (-Â½) = 2 (-Â½)^{2} + 5(-Â½) + q

= 2(1/4) – 5/2 + q

= Â½ – 5/2 + q

= (1 – 5)/2 + q

= -4/2 + q

= q – 2

From the question it is given that, (2x + 1) is a factor of both the expressions 2x^{2}Â + 5x + q

So, remainder is 0.

q â€“ 2 = 0

q = 2

Therefore, p = – 3 and q = 2

P(x) = 2x^{2}Â â€“ 5x â€“ 3

q(x) = 2x^{2}Â + 5x + 2

Then, divide p(x) by 2x + 1

Therefore, 2x^{2} â€“ 5x â€“ 3 = (2x + 1) (x – 3)

Now, divide q(x) by 2x + 1

Therefore, 2x^{2} + 5x + 2 = (2x + 1) (x + 2)

**10. If a polynomial f(x)= x ^{4}-2x^{3} + 3x^{2}– ax + b leaves reminder 5 and 19 when divided by (x – 1) and (x + 1) respectively, Find the values of a and b. Hence determined the reminder when f(x) is divided by (x-2).**

**Solution:-**

From the question it is given that,

f(x) = x^{4} â€“ 2x^{3} +3x^{2} â€“ ax + b

Factor (x – 1) leaves remainder 5,

Factor (x + 1) leaves remainder 19,

Where x = 1 and x = – 1

f(-1) = (-1)^{4} â€“ 2(-1)^{3} + 3(-1)^{2} â€“ a(-1) + b = 19

1 â€“ 2(-1) + 3(1) â€“ a(-1) + b = 19

1 + 2 + 3 + a + b = 19

6 + a + b = 19

a + b = 19 â€“ 6

a + b = 13 â€¦ [equation (i)]

f(1) = (1)^{4} â€“ 2(1)^{3} + 3(1)^{2} â€“ a(1) + b = 5

1 â€“ 2(1) + 3(1) â€“ a(1) + b = 5

1 – 2 + 3 – a + b = 5

2 – a + b = 5

– a + b = 5 â€“ 2

– a + b = 3 â€¦ [equation (ii)]

Now, subtracting equation (ii) from equation (i) we get,

(a + b) â€“ (- a + b) = 13 â€“ 3

a + b + a â€“ b = 10

2a = 10

a = 10/2

a = 5

To find out the value of b, substitute the value of a in equation (i) we get,

a + b = 13

5 + b = 13

b = 13 â€“ 5

b = 8

Therefore, value of a = 5 and b = 8

**11. When a polynomial f(x) is divided by (x â€“ 1), the remainder is 5 and when it is, divided by (x â€“ 2), the remainder is 7. Find the remainder when it is divided by (x â€“ 1) (x â€“ 2).**

**Solution:-**

From the question it is given that,

Polynomial f(x) is divided by (x â€“ 1),

Remainder = 5

Let us assume x â€“ 1 = 0

x = 1

f(1) = 5

and the divided be (x – 2), remainder = 7

let us assume x â€“ 2 = 0

x = 2

Therefore, f(2) = 7

So, f(x) = (x – 1) (x – 2) q(x) + ax + b

Where, q(x) is the quotient and ax + b is remainder,

Now put x = 1, we get,

f(1) = (1 – 1)(1 – 2)q(1) + (a Ã— 1) + b

a + b = 5 â€¦ [equation (i)]

x = 2,

f(2) = (2 – 1)(2 – 2)q(2) + (a Ã— 2) + b

2a + b = 7 â€¦ [equation (ii)]

Now subtracting equation (i) from equation (ii) we get,

(2a + b) â€“ (a + b) = 7 â€“ 5

2a + b â€“ a â€“ b = 2

a = 2

To find out the value of b, substitute the value of a in equation (i) we get,

a + b = 5

2 + b = 5

b = 5 â€“ 2

b = 3

Therefore, the remainder = ax + b = 2x + 3