Concise Mathematics Selina Solutions Class 10 ICSE Chapter 8 Factorization of Polynomial (Remainder and Factor Theorems)

This page provides detailed ICSE Class 10 Remainder and Factor Theorem Solutions from the Selina Concise Mathematics textbook. In this chapter, we move beyond basic factorization techniques to explore powerful shortcuts for working with polynomials. The Remainder Theorem allows us to find the remainder of a polynomial division without actually performing the long division, which is a significant time-saver. Building on this, the Factor Theorem gives us a quick method to check if a linear polynomial is a factor of another polynomial. Mastering these theorems is crucial for solving cubic and higher-degree equations, a key skill for your board exams. These solutions will guide you through applying these concepts correctly.

If you are stuck on a particular question about finding a remainder, checking for a factor, or solving for an unknown constant in a polynomial, you have come to the right place. This page contains clear, step-by-step solutions for all 56 questions from Exercise 8(A), Exercise 8(B), and the Test Yourself section of the chapter. Each solution is worked out using the exact method and format that the ICSE board expects, ensuring you learn the correct way to present your answers in exams. Here, you will find reliable guidance to verify your methods and master the chapter.

Exercise 8(A)

Question 1(a)

x – 1 is a factor of 8x^2 – 7x + m; the value of m is :

• (a) -1
• (b) 1
• (c) -2
• (d) 2

According to the factor theorem, if x - a is a factor of a polynomial ( f(x) ), then ( f(a) = 0 ). Here, it is given that x - 1 is a factor of the polynomial 8x^2 - 7x + m.

∴ x - 1 = 0 implies x = 1.

Substitute x = 1 into the polynomial:

This simplifies to:

⇒ 1 + m = 0

⇒ m = -1.

Hence, option 1 is the correct option.

Question 1(b)

If (x – a) is a factor of f(x) then the remainder when f(x) – k is divided by (x – a) is :

• (a) 0
• (b) k
• (c) -k
• (d) a

According to the factor theorem, if x - a is a factor of the polynomial ( f(x) ), then the value of ( f(a) ) is 0.

Given that ( (x – a) ) is a factor of ( f(x) ), it follows that ( f(a) = 0 ).

Now, consider the expression ( f(x) – k ). By substituting x = a, we get:

Since ( f(a) = 0 ), the expression simplifies to:

which equals -k.

Hence, Option 3 is the correct option.

Question 1(c)

One factor of x^3 – kx^2 + 11x – 6 is x – 1. The value of k is :

• (a) -6
• (b) 12
• (c) 6
• (d) -12

According to the factor theorem, if x - a is a factor of a polynomial ( f(x) ), then substituting x = a into ( f(x) ) should result in zero. Here, we know x - 1 is a factor of the polynomial x^3 - kx^2 + 11x - 6.

∴ Set x - 1 = 0, which implies x = 1.

Now, substitute x = 1 into the polynomial:

Simplifying, we get:

This simplifies further to:

⇒ k = 6.

Hence, Option 3 is the correct option.

Question 1(d)

If (x – a) is a factor of x^3 – ax^2 + x + 5; the value of a is :

• (a) \dfrac{1}{5}
• (b) 5
• (c) –\dfrac{1}{5}
• (d) -5

According to the factor theorem, if ((x – a)) is a factor of a polynomial (f(x)), then substituting x = a into the polynomial should yield a remainder of zero. Given that ((x – a)) is a factor of x^3 - ax^2 + x + 5, we substitute x = a into the polynomial:

Simplifying this, we get:

Which reduces to:

Solving for a, we find:

Thus, the value of a is -5. Therefore, option 4 is the correct option.

Question 1(e)

(x – 2) is a factor of :

• (a) x^3 – x^2 + x – 6
• (b) x^3 + x^2 + x + 6
• (c) 2x^3 – 6x^2 + 5x – 1
• (d) x^3 – 4x^2 + x – 8

To determine if x - 2 is a factor, we first set x - 2 = 0, which gives us x = 2.

Substitute x = 2 into the polynomial x^3 - x^2 + x - 6:

Calculating each term, we have:

• 2^3 = 8
• 2^2 = 4

Thus, the expression becomes:

Simplifying, we find:

Since the remainder is 0, it confirms that x - 2 is indeed a factor of x^3 - x^2 + x - 6.

Hence, option 1 is the correct option.

Question 2(i)

Find in each case, the remainder when :

x^4 – 3x^2 + 2x + 1 is divided by x – 1

To find the remainder when dividing by x - 1, we first set x - 1 = 0, which gives x = 1.

The remainder is the value of the polynomial x^4 - 3x^2 + 2x + 1 evaluated at x = 1.

∴ Remainder = ( (1)^4 – 3(1)^2 + 2(1) + 1 )

= 1 - 3 + 2 + 1

= 1.

Hence, remainder = 1.

Question 2(ii)

Find in each case, the remainder when :

x^3 + 3x^2 – 12x + 4 is divided by x – 2.

To find the remainder when dividing the polynomial by x - 2, set x - 2 = 0, which gives x = 2.

The remainder is the value of the polynomial x^3 + 3x^2 - 12x + 4 when x = 2.

∴ Remainder = ( (2)^3 + 3(2)^2 – 12(2) + 4 )

= 8 + 12 - 24 + 4

= 0.

Hence, remainder = 0.

Question 2(iii)

Find in each case, the remainder when :

x^4 + 1 is divisible by x + 1.

To find the remainder when dividing the polynomial x^4 + 1 by x + 1, we set x + 1 = 0, which gives x = -1.

The remainder is determined by evaluating the polynomial at x = -1.

∴ Remainder = (-1)^4 + 1.

Calculating this, we have 1 + 1 = 2.

Hence, remainder = 2.

Question 3

Use the Remainder theorem to find which of the following is a factor of 2x^3 + 3x^2 – 5x – 6.

(i) x + 1

(ii) 2x – 1

(i) To determine if x + 1 is a factor, set x + 1 = 0, which gives x = -1.

Now, substitute x = -1 into the polynomial 2x^3 + 3x^2 - 5x - 6 to find the remainder.

∴ Remainder = 2(-1)^3 + 3(-1)^2 - 5(-1) - 6

= 2(-1) + 3(1) + 5 - 6

= -2 + 3 + 5 - 6

= 8 - 8

The remainder is 0, which means x + 1 is indeed a factor of 2x^3 + 3x^2 - 5x - 6.

(ii) Now, consider 2x - 1 = 0, which implies x = \dfrac{1}{2}.

Substitute x = \dfrac{1}{2} into the polynomial 2x^3 + 3x^2 - 5x - 6 to check the remainder.

Since the remainder is not 0, 2x - 1 is not a factor of 2x^3 + 3x^2 - 5x - 6.

Question 4

If 2x + 1 is a factor of 2x^2 + ax – 3, find the value of a.

Start by setting the factor 2x + 1 to zero: 2x + 1 = 0. This gives us x = -\dfrac{1}{2}.

Given that 2x + 1 is a factor of the polynomial 2x^2 + ax – 3, substituting x = -\dfrac{1}{2} into the polynomial should give a remainder of zero.

Substitute x = -\dfrac{1}{2}:

Calculate each term:

Simplify:

Further simplification gives:

Multiply through by 2 to clear the fraction:

Solve for a:

Hence, a = -5.

Question 5

Find the values of constants a and b when x – 2 and x + 3 both are the factors of expression x^3 + ax^2 + bx – 12.

To determine the values of constants a and b, we start with the factor x - 2, which implies x = 2.

Since x - 2 is a factor of x^3 + ax^2 + bx - 12, substituting x = 2 should result in a remainder of 0.

⇒ (2)^3 + a(2)^2 + b(2) - 12 = 0

⇒ 8 + 4a + 2b - 12 = 0

⇒ 4a + 2b - 4 = 0

⇒ 4a + 2b = 4

⇒ 2(2a + b) = 4

⇒ 2a + b = 2

⇒ b = 2 - 2a ………(i)

Next, consider the factor x + 3, which gives x = -3.

Since x + 3 is also a factor of x^3 + ax^2 + bx - 12, substituting x = -3 should also result in a remainder of 0.

⇒ (-3)^3 + a(-3)^2 + b(-3) - 12 = 0

⇒ -27 + 9a - 3b - 12 = 0

⇒ 9a - 3b - 39 = 0

⇒ 9a - 3b = 39

⇒ 3(3a - b) = 39

⇒ 3a - b = 13

⇒ b = 3a - 13 ………(ii)

Now, equate the expressions for b from (i) and (ii):

⇒ 2 - 2a = 3a - 13

⇒ 3a + 2a = 2 + 13

⇒ 5a = 15

⇒ a = 3.

Substitute a = 3 back into equation (i):

⇒ b = 2 - 2a = 2 - 2(3) = 2 - 6 = -4.

Hence, a = 3 and b = -4.

Question 6

Find the value of k, if 2x + 1 is a factor of (3k + 2)x^3 + (k – 1).

To determine the value of k, start by setting 2x + 1 = 0. Solving for x, we find x = -\dfrac{1}{2}.

Given that 2x + 1 is a factor of (3k + 2)x^3 + (k - 1), it implies that substituting x = -\dfrac{1}{2} into the polynomial should result in a remainder of zero.

Substitute x = -\dfrac{1}{2}:

Simplifying, we have:

Further simplifying the equation:

Combine like terms:

Solving for k, add 10 to both sides:

Divide both sides by 5:

Hence, k = 2.

Question 7

Find the values of m and n so that x – 1 and x + 2 both are factors of

x^3 + (3m + 1)x^2 + nx – 18.

To determine the values of m and n, we begin by considering x - 1 as a factor, which implies x = 1.

Given that x - 1 is a factor of x^3 + (3m + 1)x^2 + nx - 18, substituting x = 1 should yield a remainder of 0:

This simplifies to:

which further reduces to:

Thus, we have the equation:

Next, considering x + 2 as a factor, we have x = -2. Since x + 2 is a factor, substituting x = -2 should also result in a remainder of 0:

This expression simplifies to:

Further reducing gives:

Which simplifies to:

Rearranging gives:

Which simplifies to:

Thus, we have:

Now, solving equations (i) and (ii) together:

From equation (i):
n = 16 - 3m

Substitute n = 16 - 3m into equation (ii):

This simplifies to:

Thus, we have:

Solving for m:

Substitute m = 3 back into equation (i):

Hence, m = 3 and n = 7.

Question 8

When x^3 + 2x^2 – kx + 4 is divided by x – 2, the remainder is k. Find the value of constant k.

We start by setting the divisor equal to zero: x – 2 = 0, which gives us x = 2.

The problem states that the polynomial x^3 + 2x^2 – kx + 4, when divided by x – 2, leaves a remainder of k.

∴ By substituting x = 2 into the polynomial, we equate the result to k.

⇒ (2)^3 + 2(2)^2 – k(2) + 4 = k

⇒ 8 + 8 – 2k + 4 = k

⇒ 20 – 2k = k

Now, solve for k:

⇒ 3k = 20

⇒ k = \dfrac{20}{3} = 6\dfrac{2}{3}.

Thus, k = 6\dfrac{2}{3}.

Question 9

Find the value of a, if the division of ax^3 + 9x^2 + 4x – 10 by x + 3 leaves a remainder 5.

To find the value of a, we begin by setting x + 3 = 0, which implies x = -3.

The polynomial given is ax^3 + 9x^2 + 4x - 10, and we know that dividing it by x + 3 yields a remainder of 5.

∴ By substituting x = -3 into the polynomial, the remainder should be 5:

⇒ a(-3)^3 + 9(-3)^2 + 4(-3) - 10 = 5

⇒ -27a + 81 - 12 - 10 = 5

⇒ -27a + 59 = 5

Rearranging gives:

⇒ 27a = 59 - 5

⇒ 27a = 54

Solving for a:

⇒ a = \dfrac{54}{27} = 2

Hence, a = 2.

Question 10

If x^3 + ax^2 + bx + 6 has x – 2 as a factor and leaves a remainder 3 when divided by x – 3, find the values of a and b.

Given that x - 2 is a factor, we have x = 2.

Since x - 2 is a factor of x^3 + ax^2 + bx + 6, substituting x = 2 should yield a remainder of 0.

∴ (2)^3 + a(2)^2 + b(2) + 6 = 0.

⇒ 8 + 4a + 2b + 6 = 0.

⇒ 4a + 2b + 14 = 0.

This simplifies to 2(2a + b + 7) = 0, giving us 2a + b + 7 = 0.

⇒ b = -(7 + 2a) …….(i)

Now, consider x - 3 = 0, which implies x = 3.

It’s given that dividing x^3 + ax^2 + bx + 6 by x - 3 leaves a remainder of 3.

∴ Substituting x = 3, we have (3)^3 + a(3)^2 + b(3) + 6 = 3.

⇒ 27 + 9a + 3b + 6 = 3.

⇒ 9a + 3b + 33 = 3.

⇒ 9a + 3b = -30.

Thus, 3(3a + b) = -30, leading to 3a + b = -10.

⇒ b = -10 - 3a = -(10 + 3a) ……..(ii)

Equating equations (i) and (ii), we get:

⇒ -(7 + 2a) = -(10 + 3a)

⇒ 7 + 2a = 10 + 3a

⇒ 3a - 2a = 7 - 10

⇒ a = -3.

Substituting a = -3 into equation (i), we find:

⇒ b = -(7 + 2a) = -(7 + 2(-3)) = -(7 - 6) = -1.

Hence, a = -3 and b = -1.

Question 11

What number should be added to 3x^3 – 5x^2 + 6x so that when resulting polynomial is divided by x – 3, the remainder is 8 ?

Suppose we need to add a number, denoted as a, to the polynomial.

∴ The new polynomial becomes 3x^3 - 5x^2 + 6x + a.

Since x - 3 = 0, we have x = 3.

Substituting x = 3 into the polynomial 3x^3 - 5x^2 + 6x + a, the remainder should be 8.

∴ ( 3(3)^3 – 5(3)^2 + 6(3) + a = 8 )

⇒ ( 3(27) – 5(9) + 18 + a = 8 )

⇒ 81 - 45 + 18 + a = 8

⇒ a + 54 = 8

⇒ a = -46

Therefore, the number to be added is -46.

Question 12

What number should be subtracted from x^3 + 3x^2 – 8x + 14 so that on dividing it by x – 2, the remainder is 10 ?

Assume the number that needs to be subtracted is a.

∴ The modified polynomial becomes x^3 + 3x^2 - 8x + 14 - a.

Given x - 2 = 0, we find x = 2.

Substitute x = 2 into the polynomial x^3 + 3x^2 - 8x + 14 - a to ensure the remainder is 10.

∴ (2)^3 + 3(2)^2 - 8(2) + 14 - a = 10.

⇒ 8 + 3(4) - 16 + 14 - a = 10.

⇒ 8 + 12 - 16 + 14 - a = 10.

⇒ 18 - a = 10.

⇒ a = 18 - 10 = 8.

Therefore, the number to be subtracted is 8.

Question 13

The polynomial 2x^3 – 7x^2 + ax – 6 and x^3 – 8x^2 + (2a + 1)x – 16 leave the same remainder when divided by x – 2. Find the value of ‘a’.

We are given two polynomials: 2x^3 - 7x^2 + ax - 6 and x^3 - 8x^2 + (2a + 1)x - 16. Both leave the same remainder when divided by x - 2.

Here, x - 2 = 0 implies x = 2.

∴ By substituting x = 2 into both polynomials, their values should be equal:

∴ 2(2)^3 - 7(2)^2 + a(2) - 6 = (2)^3 - 8(2)^2 + (2a + 1)(2) - 16

⇒ 2(8) - 7(4) + 2a - 6 = 8 - 32 + 4a + 2 - 16

⇒ 16 - 28 + 2a - 6 = 8 - 32 + 4a + 2 - 16

⇒ 2a - 18 = 4a - 38

⇒ 4a - 2a = 38 - 18

⇒ 2a = 20

⇒ a = 10.

Hence, a = 10.

Question 14

If (x – 2) is a factor of the expression 2x^3 + ax^2 + bx – 14 and when the expression is divided by (x – 3), it leaves a remainder 52, find the values of a and b.

Consider the polynomial 2x^3 + ax^2 + bx - 14 where (x - 2) is a factor. This implies that when x = 2, the polynomial evaluates to zero. Substituting x = 2 gives us:

This simplifies to:

Which further reduces to:

Dividing the entire equation by 2, we get:

Rearranging gives us:

Next, we know that dividing the polynomial by (x - 3) leaves a remainder of 52. Thus, substituting x = 3 into the polynomial yields:

This simplifies to:

Which becomes:

Which further reduces to:

Simplifying gives:

Dividing through by 3, we find:

Rearranging gives us:

By equating the expressions for b from equations (i) and (ii), we have:

Simplifying, we get:

Rearranging terms, we find:

Thus, a = 5. Substituting a = 5 back into equation (i):

Hence, a = 5 and b = -11.

Exercise 8(B)

Question 1(a)

For the polynomial x^5 – x^4 + x^3 – 8x^2 + 6x + 15, the maximum number of linear factors is :

• (a) 9
• (b) 6
• (c) 7
• (d) 5

The number of linear factors a polynomial can have is determined by the highest degree of the polynomial. In the given polynomial, x^5 - x^4 + x^3 - 8x^2 + 6x + 15, the highest power of x is 5. ∴, the polynomial can have a maximum of 5 linear factors.

Hence, Option 4 is the correct option.

Question 1(b)

If f(x) = 3x + 8; the value of f(x) + f(-x) is :

• (a) 8
• (b) 16
• (c) -8
• (d) -16

Consider the function given by f(x) = 3x + 8.

To find f(-x), substitute -x into the function:

f(-x) = 3(-x) + 8 = -3x + 8.

Now, calculate f(x) + f(-x):

f(x) + f(-x) = (3x + 8) + (-3x + 8).

Notice that the terms 3x and -3x cancel each other out:

⇒ 3x – 3x + 8 + 8 = 16.

The result is 16.

Hence, Option 2 is the correct option.

Question 1(c)

If x^25 + x^24 is divided by (x + 1), the result is :

• (a) 49
• (b) 1
• (c) 0
• (d) -1

According to the remainder theorem, when a polynomial ( f(x) ) is divided by ( (x – a) ), the remainder is given by ( f(a) ).

Here, we have the polynomial x^{25} + x^{24} and it is being divided by ( (x + 1) ). Notice that x + 1 = 0 implies x = -1.

Now, substitute x = -1 into the polynomial:

Calculating each term, we find:

Adding these results gives:

∴ The remainder is 0.

Therefore, Option 3 is the correct option.

Question 1(d)

Factors of 3x^3 – 2x^2 – 8x are :

• (a) x(3x^2 – 2x – 8)
• (b) x(x – 2)(3x + 4)
• (c) 2x(x – 4)(2x + 1)
• (d) x(x – 4)(2x + 1)

Consider the polynomial:

⇒ 3x^3 – 2x^2 – 8x

First, factor out the common term, which is x:

⇒ x[3x^2 – 2x – 8]

Next, we need to factor the quadratic expression inside the brackets. Start by splitting the middle term:

⇒ x[3x^2 – 6x + 4x – 8]

Group the terms in pairs:

⇒ x[(3x^2 – 6x) + (4x – 8)]

Factor out the common factors from each group:

⇒ x[3x(x – 2) + 4(x – 2)]

Notice that (x – 2) is a common factor:

⇒ x[(3x + 4)(x – 2)]

Thus, the complete factorization is:

⇒ x(3x + 4)(x – 2).

Hence, Option 2 is the correct option.

Question 1(e)

Factors of 4 + 4x – x^2 – x^3 are :

• (a) (2 + x)(2 – x)(1 + x)
• (b) (x – 2)(1 + x)(2 + x)
• (c) (x + 2)(x – 2)(1 – x)
• (d) (2 + x)(x – 1)(2 – x)

Let’s evaluate the polynomial 4 + 4x - x^2 - x^3 by substituting x = 2:

Since the result is 0, x - 2 is indeed a factor of the polynomial.

Now, perform the polynomial division of -x^3 - x^2 + 4x + 4 by x - 2:

The quotient obtained is -x^2 - 3x - 2.

Thus, (-x^3 – x^2 + 4x + 4 = (x – 2)(-x^2 – 3x – 2)).

Breaking it down further:

Simplifying, we have:

Reorganizing the terms gives us:

Hence, Option 1 is the correct option.

Question 2(i)

Using Factor Theorem, show that :

(x – 2) is a factor of x^3 – 2x^2 – 9x + 18. Hence, factorise the expression x^3 – 2x^2 – 9x + 18 completely.

To determine if x - 2 is a factor of the polynomial x^3 - 2x^2 - 9x + 18, we set x - 2 = 0 which gives x = 2.

We substitute x = 2 into the polynomial:

Calculating each term, we have:

Since the remainder is 0, x - 2 is indeed a factor of x^3 - 2x^2 - 9x + 18.

Next, we perform polynomial division of x^3 - 2x^2 - 9x + 18 by x - 2:

The quotient from the division is x^2 - 9.

∴ The polynomial x^3 - 2x^2 - 9x + 18 can be expressed as ((x – 2)(x^2 – 9)).

Now, notice x^2 - 9 can be further factored using the difference of squares:

Thus, the complete factorization of x^3 - 2x^2 - 9x + 18 is:

(x^3 – 2x^2 – 9x + 18 = (x – 2)(x – 3)(x + 3)).

Question 2(ii)

Using Factor Theorem, show that :

(x + 5) is a factor of 2x^3 + 5x^2 – 28x – 15. Hence, factorise the expression 2x^3 + 5x^2 – 28x – 15 completely.

To determine if x + 5 is a factor of 2x^3 + 5x^2 - 28x - 15, let’s set x + 5 = 0 which gives us x = -5.

Now, substitute x = -5 into the polynomial:

Calculate each term:

• (2(-5)^3 = 2(-125) = -250)
• (5(-5)^2 = 5(25) = 125)
• (-28(-5) = 140)
• -15

Adding these results:

Since the remainder is 0, x + 5 is indeed a factor of 2x^3 + 5x^2 - 28x - 15.

Next, divide 2x^3 + 5x^2 - 28x - 15 by x + 5:

The quotient we obtain is 2x^2 - 5x - 3.

Now, let’s factorise 2x^2 - 5x - 3:

Therefore, the complete factorization of 2x^3 + 5x^2 - 28x - 15 is:

(2x^3 + 5x^2 – 28x – 15 = (x + 5)(2x + 1)(x – 3)).

Question 3(i)

Using the remainder theorem, factorise each of the following completely :

3x^3 + 2x^2 – 19x + 6

To determine the factors of the polynomial 3x^3 + 2x^2 - 19x + 6, we apply the remainder theorem. Evaluate the polynomial at x = 2:

Calculate step-by-step:

This shows that (x - 2) is indeed a factor of the polynomial 3x^3 + 2x^2 - 19x + 6.

Next, divide 3x^3 + 2x^2 - 19x + 6 by (x - 2) using polynomial division:

The quotient obtained is 3x^2 + 8x - 3.

Now, factorise 3x^2 + 8x - 3:

Group the terms:

Factor out the common terms:

Thus, 3x^2 + 8x - 3 = (3x - 1)(x + 3).

∴ The complete factorisation of the polynomial 3x^3 + 2x^2 - 19x + 6 is (x - 2)(3x - 1)(x + 3).

Question 3(ii)

Using the remainder theorem, factorise each of the following completely :

2x^3 + x^2 – 13x + 6

To determine the factors of the polynomial 2x^3 + x^2 - 13x + 6, let’s apply the remainder theorem. Plugging x = 2 into the polynomial, we calculate:

This simplifies to:

Since the remainder is 0, (x - 2) is a factor of the polynomial. Now, we divide 2x^3 + x^2 - 13x + 6 by (x - 2):

The quotient obtained is 2x^2 + 5x - 3. Let’s factorize this quadratic expression:

Grouping the terms, we have:

Factoring by grouping gives:

Thus, 2x^2 + 5x - 3 is factored as (2x - 1)(x + 3). Therefore, the complete factorization of the original polynomial is:

Hence, 2x^3 + x^2 - 13x + 6 = (x - 2)(2x - 1)(x + 3).

Question 3(iii)

Using the remainder theorem, factorise each of the following completely :

3x^3 + 2x^2 – 23x – 30

Let’s evaluate the polynomial 3x^3 + 2x^2 - 23x - 30 at x = -2. Substituting, we have:

Calculating each term, we get:

Simplifying further:

Since the remainder is 0, (x + 2) is a factor of the polynomial 3x^3 + 2x^2 - 23x - 30.

Now, let’s perform polynomial division of 3x^3 + 2x^2 - 23x - 30 by x + 2:

The quotient obtained is 3x^2 - 4x - 15.

Now, factorise 3x^2 - 4x - 15:

Grouping terms, we get:

Thus, it factorises to:

∴ 3x^2 - 4x - 15 = (3x + 5)(x - 3).

Hence, 3x^3 + 2x^2 - 23x - 30 = (x + 2)(3x + 5)(x - 3).

Question 3(iv)

Using the remainder theorem, factorise each of the following completely :

4x^3 + 7x^2 – 36x – 63

To determine if ((x + 3)) is a factor of the polynomial 4x^3 + 7x^2 - 36x - 63, we substitute x = -3 into the polynomial:

Calculating step-by-step:

• ((-3)^3 = -27), so (4(-27) = -108)
• ((-3)^2 = 9), so (7(9) = 63)
• (-36(-3) = 108)
• The constant term is -63

Adding these results together:

Since the result is zero, ((x + 3)) is indeed a factor of the polynomial.

Next, we perform polynomial division of 4x^3 + 7x^2 - 36x - 63 by ((x + 3)):

The quotient from this division is 4x^2 - 5x - 21.

Now, let’s factor 4x^2 - 5x - 21:

This can be rewritten as:

Thus, (4x^2 – 5x – 21 = (4x + 7)(x – 3)).

Therefore, the complete factorization of the polynomial is:

(4x^3 + 7x^2 – 36x – 63 = (x + 3)(4x + 7)(x – 3)).

Question 3(v)

Using the remainder theorem, factorise each of the following completely :

x^3 + x^2 – 4x – 4

Let’s evaluate the polynomial x^3 + x^2 - 4x - 4 at x = -1. Substituting, we have:

Since the result is zero, (x + 1) is a factor of the polynomial x^3 + x^2 - 4x - 4.

Now, divide the polynomial x^3 + x^2 - 4x - 4 by (x + 1). The division process is shown as follows:

The quotient obtained is x^2 - 4.

Next, factorise x^2 - 4:

Thus, x^2 - 4 = (x - 2)(x + 2).

Hence, the complete factorisation of x^3 + x^2 - 4x - 4 is (x + 1)(x + 2)(x - 2).

Question 4

Using the Remainder Theorem, factorise the expression 3x^3 + 10x^2 + x – 6. Hence, solve the equation 3x^3 + 10x^2 + x – 6 = 0.

Evaluate the polynomial 3x^3 + 10x^2 + x - 6 at x = -1:

Since the result is zero, (x + 1) is a factor of the polynomial 3x^3 + 10x^2 + x - 6.

Next, perform polynomial division of 3x^3 + 10x^2 + x - 6 by (x + 1):

The quotient from this division is 3x^2 + 7x - 6.

Proceed to factorise 3x^2 + 7x - 6:

Thus, the complete factorisation of 3x^3 + 10x^2 + x - 6 is:

3x^3 + 10x^2 + x - 6 = (x + 1)(3x - 2)(x + 3).

Question 5

Factorise the expression

f(x) = 2x^3 – 7x^2 – 3x + 18.

Hence, find all possible values of x for which f(x) = 0.

Let’s evaluate the polynomial f(x) = 2x^3 - 7x^2 - 3x + 18 at x = 2:

Calculating step-by-step:

1. 2(2)^3 = 16
2. 7(2)^2 = 28
3. 3(2) = 6

Substituting these values, we have:

Since f(2) = 0, it confirms that (x - 2) is a factor of the polynomial.

Now, we will perform polynomial division of 2x^3 - 7x^2 - 3x + 18 by (x - 2):

The quotient obtained is 2x^2 - 3x - 9. Let’s factorise this quadratic expression:

Grouping terms, we have:

Thus, 2x^2 - 3x - 9 factors to:

Therefore, the complete factorisation of f(x) is:

To find the values of x for which f(x) = 0, set each factor to zero:

1. (x - 2) = 0 \Rightarrow x = 2
2. (2x + 3) = 0 \Rightarrow x = -\dfrac{3}{2}
3. (x - 3) = 0 \Rightarrow x = 3

Hence, the values of x that satisfy f(x) = 0 are 2, 3, and -\dfrac{3}{2}.

Question 6

Given that x – 2 and x + 1 are factors of f(x) = x^3 + 3x^2 + ax + b; calculate the values of a and b. Hence, find all the factors of f(x).

The factor x - 2 implies x = 2. Since x - 2 is a factor of f(x) = x^3 + 3x^2 + ax + b, substituting x = 2 into the polynomial gives a remainder of 0.

⇒ 8 + 12 + 2a + b = 0

⇒ 2a + b = -20

⇒ b = -20 - 2a ……..(i)

Similarly, for the factor x + 1, we have x = -1. Substituting x = -1 into the polynomial, the remainder is also 0.

⇒ -1 + 3 - a + b = 0

⇒ 2 - a + b = 0

⇒ b = a - 2 …….(ii)

Using equations (i) and (ii), we equate the two expressions for b:

⇒ -20 - 2a = a - 2

⇒ 3a = -18

⇒ a = -6

Substituting a = -6 back into equation (ii), we find:

⇒ b = a - 2 = -6 - 2 = -8

∴ a = -6 and b = -8.

Next, we perform polynomial division of x^3 + 3x^2 - 6x - 8 by (x - 2):

The quotient is x^2 + 5x + 4.

Factorising x^2 + 5x + 4, we have:

Hence, a = -6, b = -8, and f(x) = (x - 2)(x + 1)(x + 4).

Question 7

The expression 4x^3 – bx^2 + x – c leaves remainders 0 and 30 when divided by x + 1 and 2x – 3 respectively. Calculate the values of b and c. Hence, factorise the expression completely.

We start by setting x + 1 = 0, which gives us x = -1. Since x + 1 is a factor of 4x^3 - bx^2 + x - c, substituting x = -1 in the expression should yield a remainder of 0.

Therefore, 4(-1)^3 - b(-1)^2 + (-1) - c = 0 simplifies to -4 - b - 1 - c = 0. Solving this, we find c = -5 - b …….(i).

Next, we know that dividing 4x^3 - bx^2 + x - c by 2x - 3 leaves a remainder of 30. Thus, substituting x = \dfrac{3}{2} into the expression should result in 30.

Now, equating equations (i) and (ii):

Substituting b = -8 back into equation (i), we get:

Inserting b = -8 and c = 3 into the original polynomial, we have:

Expression = 4x^3 + 8x^2 + x - 3.

Dividing 4x^3 + 8x^2 + x - 3 by x + 1, we perform the division:

The quotient is 4x^2 + 4x - 3.

Factorising 4x^2 + 4x - 3, we have:

= (2x - 1)(2x + 3).

Thus, b = -8, c = 3, and the factorisation of 4x^3 + 8x^2 + x - 3 is (x + 1)(2x - 1)(2x + 3).

Question 8

If x + a is a common factor of expressions f(x) = x^2 + px + q and g(x) = x^2 + mx + n; show that : a = \dfrac{n - q}{m - p}

Given that x + a is a factor of both (f(x)) and (g(x)), it implies that x + a = 0 or x = -a.

Since x + a divides both expressions, we have:

∴ (f(-a) = 0) and (g(-a) = 0).

Equating the expressions:

Simplify both sides:

By subtracting a^2 from both sides, we get:

Rearranging gives:

Factoring out a from the left side:

Dividing both sides by ((m – p)):

Thus, we have shown that a = \dfrac{n - q}{m - p}.

Test Yourself

Question 1(a)

The remainder, when x^3 – x^2 + x – 1 is divided by x + 1, is :

• (a) 0
• (b) -4
• (c) 2
• (d) 4

According to the remainder theorem, when a polynomial ( f(x) ) is divided by ( (x – a) ), the remainder is ( f(a) ). Here, we have the polynomial x^3 - x^2 + x - 1 and it is divided by x + 1. To find the remainder, let’s set x + 1 = 0, which gives us x = -1.

Now, substitute x = -1 into the polynomial:

Calculate each term:
– ((-1)^3 = -1)
– ((-1)^2 = 1)
– -1 = -1
– -1 = -1

Putting it all together:

Notice that the remainder is -4. Hence, Option 2 is the correct option.

Question 1(b)

The number of common factor(s) of two polynomials x^2 – 9 and x^3 – x^2 – 9x + 9 is /are :

• (a) one
• (b) two
• (c) three

To find the common factors of the given polynomials, let’s begin by factorizing each one.

First, consider x^2 - 9. This is a difference of squares, which can be expressed as:

⇒ (x + 3)(x - 3)

Now, look at x^3 - x^2 - 9x + 9. We can factor by grouping:

⇒ x^2(x - 1) - 9(x - 1)

Notice that (x - 1) is a common factor in both terms, so we factor it out:

⇒ (x^2 - 9)(x - 1)

Since x^2 - 9 is already factorized as (x + 3)(x - 3), substitute this back in:

⇒ (x + 3)(x - 3)(x - 1)

Now, compare the factors of both polynomials. The first polynomial, x^2 - 9, has factors (x + 3) and (x - 3). The second polynomial, x^3 - x^2 - 9x + 9, has factors (x + 3), (x - 3), and (x - 1).

The common factors between the two polynomials are (x + 3) and (x - 3).

Thus, there are two common factors. Hence, Option 2 is the correct option.

Question 1(c)

Is (x – 2) a factor of x^3 – 4x^2 – 11x + 30 ?

• (a) yes
• (b) no
• (c) nothing can be said
• (d) none of the above is true

To determine if x - 2 is a factor of x^3 - 4x^2 - 11x + 30, set x - 2 = 0, which gives x = 2.

Now, substitute x = 2 into the polynomial:

Calculating step-by-step:

Substitute these values back:

Simplify the expression:

Since the remainder is 0, it confirms that x - 2 is indeed a factor of x^3 - 4x^2 - 11x + 30.

Hence, option 1 is the correct option.

Question 1(d)

4x^2 – kx + 5 leaves a remainder 2 when divided by x – 1. The value of k is :

• (a) -6
• (b) 6
• (c) 7
• (d) -7

According to the remainder theorem, when a polynomial f(x) is divided by (x - a), the remainder is f(a).

Here, we know that 4x^2 - kx + 5 leaves a remainder of 2 when divided by x - 1.

∴ Substitute x = 1 into the polynomial:

This simplifies to:

Which further simplifies to:

Combining like terms gives:

Solving for k, we get:

Hence, Option 3 is the correct option.

Question 1(e)

If mx^2 – nx + 8 has x – 2 as a factor, then :

• (a) 2m – n = 4
• (b) 2m + n = 4
• (c) 2n + m = 4
• (d) n – 2m = 4

According to the factor theorem, if x - a is a factor of a polynomial ( f(x) ), then substituting x = a into ( f(x) ) should yield a remainder of zero.

In this problem, x - 2 is a factor of the polynomial mx^2 - nx + 8. Thus, we substitute x = 2 in the polynomial and set the result equal to zero:

Simplifying the equation:

We can factor out a 2 from the equation:

Since the equation must hold true, we have:

Rearranging gives:

Hence, Option 4 is the correct option.

Question 1(f)

Two polynomials x^36 – 3x^35 and x – 3.

Assertion (A) : If x – 3 is a factor of x^36 – 3x^35, the remainder is zero.

Reason (R) : The polynomial x – a is factor of polynomial p(x) = x^36 – ax^35, if p(a) = 0

options

• (a) A is true, R is false.
• (b) A is false, R is true.
• (c) Both A and R are true and R is correct reason for A.
• (d) Both A and R are true and R is incorrect reason for A.

Both the assertion (A) and the reason (R) are accurate, and the reason correctly explains the assertion.

According to the factor theorem, a polynomial ((x – a)) is a factor of another polynomial (f(x)) if the remainder when (f(x)) is divided by ((x – a)) is zero, i.e., (f(a) = 0).

Consider (f(x) = x^{36} – 3x^{35}).

If x - 3 is a factor of (f(x)), then by the factor theorem, (f(3) = 0).

∴ Assertion (A) is true.

Now, if we have (p(x) = x^{36} – ax^{35}) and it is divided by x - a, the remainder is given by:

Since (p(a) = 0), it confirms that ((x – a)) is indeed a factor of (p(x)).

∴ Reason (R) is true.

Thus, both assertion (A) and reason (R) are true, and the reason appropriately justifies the assertion.

Hence, option 3 is the correct option.

Question 1(g)

The polynomial 3x^3 + 8x^2 – 15x + k and one of its factors as (x – 1).

Assertion (A) : The value of k = 4.

Reason (R) : x – 1 = 0 ⇒ x = 1.

∴ 3.(1)^3 + 8.(1)^2 – 15 x (1) + k = 0

options

• (a) A is true, R is false.
• (b) A is false, R is true.
• (c) Both A and R are true and R is correct reason for A.
• (d) Both A and R are true and R is incorrect reason for A.

Both the assertion (A) and the reason (R) are valid, and the reason accurately supports the assertion.

Explanation

Consider the polynomial function, ( f(x) = 3x^3 + 8x^2 – 15x + k ).

According to the factor theorem, if ((x – a)) is a factor of ( f(x) ), then ( f(a) = 0 ).

Here, x - 1 = 0 implies x = 1.

It is given that x - 1 is a factor of ( f(x) ).

∴ ( f(1) = 0 ).

Substituting x = 1 in the polynomial,

This simplifies to:

⇒ k = 4.

Therefore, both the assertion and the reason are true, and the reason correctly justifies the assertion.

Hence, Option 3 is the correct option.

Question 1(h)

The polynomial x^2 + x + b has (x + 3) as a factor of it.

Statement 1: The value of b is -4.

Statement 2: (x + 3) is a factor of x^2 + x + b ⇒ (3)^2 + 3 + b = 0.

option

• (a) Both the statements are true.
• (b) Both the statements are false.
• (c) Statement 1 is true, and statement 2 is false.
• (d) Statement 1 is false, and statement 2 is true.

Both the statements are false.

According to the factor theorem, if ((x – a)) is a factor of the polynomial (f(x)), then (f(a) = 0).

⇒ Given x + 3 = 0, this implies x = -3.

Consider the polynomial (f(x) = x^2 + x + b).

Since ((x + 3)) is a factor of x^2 + x + b, we have:

⇒ ((-3)^2 + (-3) + b = 0)

⇒ 9 - 3 + b = 0

⇒ 6 + b = 0

⇒ b = -6.

∴ Statement 1 is incorrect.

Similarly,

⇒ ((-3)^2 + (-3) + b = 0) must hold true.

∴ Statement 2 is incorrect.

Hence, option 2 is the correct option.

Question 2

When x^3 + 3x^2 – mx + 4 is divided by x – 2, the remainder is m + 3. Find the value of m.

We know that when x^3 + 3x^2 - mx + 4 is divided by x - 2, the remainder is m + 3.

∴ Substitute x = 2 into the polynomial:

Calculating each term, we have:

Simplify the left side:

Rearranging the terms gives:

Combine like terms:

Dividing both sides by 3, we find:

Hence, m = 7.

Question 3

What should be subtracted from 3x^3 – 8x^2 + 4x – 3, so that the resulting expression has x + 2 as a factor ?

To determine the number that needs to be subtracted, we must find the remainder when dividing the polynomial 3x^3 - 8x^2 + 4x - 3 by x + 2.

First, set x + 2 = 0, which gives us x = -2.

Next, substitute x = -2 into the polynomial:

Calculating step-by-step:

• (3(-2)^3 = 3(-8) = -24)
• (-8(-2)^2 = -8(4) = -32)
• (4(-2) = -8)
• -3 remains as it is.

Adding these results together:

∴ The number to be subtracted is -67.

Question 4

If (x + 1) and (x – 2) are factors of x^3 + (a + 1)x^2 – (b – 2)x – 6, find the values of a and b. And then, factorise the given expression completely.

To solve this problem, we start with the polynomial:

We know that ((x + 1)) and ((x – 2)) are factors, so let’s use this information.

First, since x + 1 is a factor, set x + 1 = 0 which gives x = -1. Substituting x = -1 into the polynomial should yield a remainder of zero:

Simplifying, we have:

This becomes:

Thus, we find:

From this, we can express b as:

Next, since x - 2 is also a factor, set x - 2 = 0 which gives x = 2. Substituting x = 2 into the polynomial should also yield a remainder of zero:

Simplifying, we have:

This becomes:

So, we find:

Rearranging gives:

Thus, we can express b as:

Now, equate the expressions for b from equations (i) and (ii):

Solving for a, we have:

Substituting a = 1 into equation (i), we find:

Now, substitute a = 1 and b = 7 back into the polynomial:

We divide x^3 + 2x^2 - 5x - 6 by ((x + 1)):

The quotient is x^2 + x - 6.

Now, factorise x^2 + x - 6:

Grouping terms, we have:

This can be written as:

∴ The complete factorization of the polynomial is:

Thus, a = 1, b = 7, and the polynomial x^3 + 2x^2 - 5x - 6 factors to ((x + 1)(x – 2)(x + 3)).

Question 5

If x – 2 is a factor of x^2 + ax + b and a + b = 1, find the values of a and b.

We know that if x - 2 is a factor of x^2 + ax + b, then substituting x = 2 into the polynomial should yield a remainder of 0.

∴ ((2)^2 + a(2) + b = 0)

⇒ 4 + 2a + b = 0

This simplifies to:

⇒ (b = -(2a + 4))

Now, we are also given that a + b = 1. By substituting the expression for b into this equation:

⇒ (a + -(2a + 4) = 1)

⇒ a - 2a - 4 = 1

⇒ -a - 4 = 1

Adding 4 to both sides gives:

⇒ -a = 1 + 4

⇒ a = -5

Substituting a = -5 back into the expression for b:

⇒ (b = -(2(-5) + 4) = -(-10 + 4) = 6)

Hence, a = -5 and b = 6.

Question 6

Find the value of ‘m’, if mx^3 + 2x^2 – 3 and x^2 – mx + 4 leave the same remainder when divided by x – 2.

To find the value of ‘m’, let’s first solve for when the divisor is zero:

x – 2 = 0 ⇒ x = 2.

We are given that the polynomials mx^3 + 2x^2 – 3 and x^2 – mx + 4 have the same remainder when divided by x – 2. This means:

∴ m(2)^3 + 2(2)^2 – 3 = (2)^2 – m(2) + 4.

Calculating each side, we have:

8m + 8 – 3 = 4 – 2m + 4.

Simplifying both sides, we get:

8m + 5 = 8 – 2m.

Rearrange the terms to combine like terms:

8m + 2m = 8 – 5.

Thus, 10m = 3.

Solving for m, we find:

m = \dfrac{3}{10}.

Hence, m = \dfrac{3}{10}.

Question 7

The polynomial px^3 + 4x^2 – 3x + q is completely divisible by x^2 – 1; find the values of p and q. Also for these values of p and q factorize the given polynomial completely.

Given that x^2 - 1 divides the polynomial px^3 + 4x^2 - 3x + q, we know that (x - 1) and (x + 1) are factors.

∴ Substituting x = 1 and x = -1 should yield a remainder of 0.

For x = 1:

⇒ p(1)^3 + 4(1)^2 - 3(1) + q = 0

⇒ p + 4 - 3 + q = 0

⇒ p + q = -1

⇒ p = -1 - q …….(i)

For x = -1:

⇒ p(-1)^3 + 4(-1)^2 - 3(-1) + q = 0

⇒ -p + 4 + 3 + q = 0

⇒ p = 7 + q …….(ii)

By equating equations (i) and (ii):

⇒ -1 - q = 7 + q

⇒ 2q = -1 - 7

⇒ 2q = -8

⇒ q = -4.

Substituting q = -4 into equation (i):

⇒ p = -1 - (-4) = -1 + 4 = 3.

Now, substituting p = 3 and q = -4 into the polynomial px^3 + 4x^2 - 3x + q gives:

= 3x^3 + 4x^2 - 3x - 4.

We divide 3x^3 + 4x^2 - 3x - 4 by x - 1:

The quotient is 3x^2 + 7x + 4.

We factorize 3x^2 + 7x + 4:

= 3x^2 + 3x + 4x + 4

= 3x(x + 1) + 4(x + 1)

= (3x + 4)(x + 1).

Hence, p = 3, q = -4 and 3x^3 + 4x^2 - 3x - 4 = (x - 1)(x + 1)(3x + 4).

Question 8

When the polynomial x^3 + 2x^2 – 5ax – 7 is divided by (x – 1), the remainder is A and when the polynomial x^3 + ax^2 – 12x + 16 is divided by (x + 2), the remainder is B. Find the value of ‘a’ if 2A + B = 0.

Consider the polynomial x^3 + 2x^2 - 5ax - 7 when divided by (x - 1). The remainder is denoted as A.

Since x - 1 = 0, we have x = 1.

Substitute x = 1 into the polynomial:

Simplifying, we find:

Thus, A = -(4 + 5a) ……(i)

Next, consider the polynomial x^3 + ax^2 - 12x + 16 divided by (x + 2), where the remainder is B.

Substitute x = -2 into the polynomial:

Simplifying, we get:

Thus, B = 32 + 4a ……(ii)

We are given that 2A + B = 0.

Substitute the expressions for A and B:

Simplifying:

Solving for a:

Hence, a = 4.

Question 9

(3x + 5) is a factor of the polynomial (a – 1)x^3 + (a + 1)x^2 – (2a + 1)x – 15. Find the value of ‘a’. For this value of ‘a’, factorise the given polynomial completely.

To determine the value of ‘a’, observe that since (3x + 5) is a factor, setting 3x + 5 = 0 gives x = -\dfrac{5}{3}.

Substituting x = -\dfrac{5}{3} into the polynomial (a - 1)x^3 + (a + 1)x^2 - (2a + 1)x - 15, the remainder must be zero:

Simplifying each term:

Combine and simplify:

Clearing the fractions by multiplying through by 27:

Combine like terms:

Solving for ‘a’:

40a = 160
a = 4

Substituting a = 4 back into the polynomial:

This simplifies to:

Factorizing this expression:

Factoring out the common term (3x + 5):

Further factorization gives:

Thus, the value of ‘a’ is 4, and the polynomial 3x^3 + 5x^2 - 9x - 15 can be expressed as (x - \sqrt{3})(x + \sqrt{3})(3x + 5).

Question 10

Using remainder theorem, find the value of k if on dividing 2x^3 + 3x^2 – kx + 5 by x – 2, leaves a remainder 7.

To solve for the value of k, we start by setting x - 2 = 0, which gives x = 2.

We know that dividing the polynomial 2x^3 + 3x^2 - kx + 5 by x - 2 results in a remainder of 7.

∴ Substitute x = 2 into the polynomial:

This simplifies to:

Calculate each term:

Combine the constants:

Rearrange to isolate 2k:

Simplify the right side:

Finally, divide by 2:

k = 13.

Hence, k = 13.

Question 11

Find the value of ‘a’ if (x – a) is a factor of polynomial 3x^3 + x^2 – ax – 81.

According to the factor theorem, if x - a is a factor of a polynomial (f(x)), then the value of (f(a)) must be zero.

Here, the polynomial given is:

Since x - a is a factor, we substitute x = a into the polynomial and set the result equal to zero:

Simplifying the expression, we have:

This simplifies further to:

Solving for a^3, we get:

Dividing both sides by 3:

Taking the cube root of both sides gives:

Hence, a = 3.

Question 12

While factorizing a given polynomial, using remainder and factor theorem, a student finds that x + 3 is a factor of 2x^3 – x^2 – 5x – 2.

(a) Is the student’s, solution correct stating that (x + 3) is a factor of the given polynomial?

(b) Give a valid reason for your answer.

(c) Factorize the given polynomial completely.

Consider the polynomial: 2x^3 - x^2 - 5x - 2.

According to the factor theorem, if x - a is a factor of a polynomial f(x), then f(a) = 0.

Suppose (x + 3) is a factor. Then, by the Factor Theorem, we have:

f(-3) = 0.

Substituting x = -3 into the polynomial, we calculate the remainder:

f(-3) = 2(-3)^3 - (-3)^2 - 5(-3) - 2
= 2(-27) - 9 + 15 - 2
= -54 - 9 + 15 - 2
= -50

Since f(-3) \neq 0, the remainder is not zero.

Therefore, the student’s solution is incorrect.

To factorize the polynomial, let’s try another value. Substituting x = 2 into the polynomial, we find:

f(2) = 2(2)^3 - (2)^2 - 5(2) - 2
= 16 - 4 - 10 - 2
= 16 - 16
= 0

Since f(2) = 0, (x - 2) is indeed a factor of 2x^3 - x^2 - 5x - 2.

Now, dividing 2x^3 - x^2 - 5x - 2 by x - 2:

Thus, 2x^3 - x^2 - 5x - 2 = (x - 2)(2x^2 + 3x + 1).

Continuing with the factorization:

= (x - 2)(2x^2 + 2x + x + 1)
= (x - 2)[2x(x + 1) + 1(x + 1)]
= (x - 2)(x + 1)(2x + 1)

Hence, 2x^3 - x^2 - 5x - 2 = (x - 2)(x + 1)(2x + 1).