CBSE Class 9 Mathematics Polynomials Worksheet Set 02

Question. The factors of x³ + 9x² + 23x + 15 are
(a) (x + 1), (x + 3), (x + 5)
(b) (x + 1), (x + 3), (x – 5)
(c) (x + 1), (x – 3), (x – 5)
(d) (x – 1), (x – 3), (x – 5)
Answer : A

Question. A cubic polynomial cannot have more than ______ zeroes.
(a) 0
(b) 1
(c) 2
(d) 3
Answer : D

Question. If p(x) = x⁵¹ + 51, then value of p(–1) is
(a) 0
(b) 1
(c) 49
(d) 50
Answer : D

Question. Which of the following is a cubic polynomial?
(a) 2x + √3 + x⁴
(b) – x – x² – 7x³
(c) 9
(d) 2 – 3x² – 9x
Answer : B

Question. The value of (369)² –(368)² is
(a) 12
(b) 81
(c) 37
(d) 737
Answer : D

Question. Factors of (42 – x – x²) are
(a) (x – 7), (x – 6)
(b) (x + 7), (x – 6)
(c) (x + 7), (6 – x)
(d) (x – 7), (x + 6)
Answer : C

Question. The value of the polynomial 5x – 4x² + 3, when x = – 1 is
(a) – 6
(b) 6
(c) 2
(d) – 2
Answer : A

Question. Degree of the polynomial p(x) = (x + 2)(x – 2) is
(a) 2
(b) 1
(c) 0
(d) 3
Answer : A

Question. If x = –2 and x² + y² + 3xy = –5, then find y.
(a) –2
(b) 3
(c) – 4
(d) 9
Answer : B

Question. Simplify : x³ - 4 - x + 4x² / x² + 3x - 4
(a) 4 + x
(b) 2 + x
(c) 1 – x
(d) x + 1
Answer : D

Question. The common quantity that must be added to each term of a² : b² to make it equal to a : b is
(a) ab
(b) a + b
(c) a – b
(d) a/b
Answer : A

Question. Find the value of x + y + z, if x² + y² + z² = 18 and xy + yz + zx = 9.
(a) 9
(b) 3
(c) 6
(d) 8
Answer : C

Question. Number of terms in the polynomial 4x³ + 3x² – 6x + 7 is
(a) 1
(b) 2
(c) 3
(d) 4
Answer : D

Question. Degree of polynomial 2x³ +√3x + (1/3) x⁵ − 7 is
(a) 1
(b) 3
(c) 0
(d) 5
Answer : D

Question. If f(x) = x³ – 3x² + 4x + 50, then f(–3) =
(a) –16
(b) –12
(c) –20
(d) –10
Answer : A

Question. If p(x) = x³ + 3x² – 2x + 4, then find the value of [p(2) + p(–2) – p(0)].
(a) 28
(b) 14
(c) 12
(d) 16
Answer : A

Question. (x + y)³ – (x – y)³ – 6y(x² – y²) =
(a) y³
(b) 2y³
(c) 4y³
(d) 8y³
Answer : D

Question. If p(x) = x² − 2√2x +1, then p(2√2) =
(a) 0
(b) 1
(c) 4 2
(d) –1
Answer : B

Question. If a + b + c = 0, then a³ + b³ + c³ is equal to
(a) 0
(b) abc
(c) 3abc
(d) 2abc
Answer : C

Question. √2 is a polynomial of degree
(a) 2
(b) 0
(c) 1
(d) 1/2
Answer : B

Case Based MCQs

Case I : Read the following passage and answer the questions from 1 to 5.
Ankur and Ranjan start a new business together.
The amount invested by both partners together is given by the polynomial p(x) = 4x² + 12x + 5, which is the product of their individual shares.

1. Coefficient of x² in the given polynomial is
(a) 2
(b) 3
(c) 4
(d) 12
Answer : C

2. Total amount invested by both, if x = 1000 is
(a) ₹ 301506
(b) ₹ 370561
(c) ₹ 4012005
(d) ₹ 490621
Answer : C

3. The shares of Ankur and Ranjan invested individually are
(a) ₹ (2x + 1), ₹ (2x + 5)
(b) ₹ (2x + 3), ₹ (x + 1)
(c) ₹ (x + 1), ₹ (x + 3)
(d) None of these
Answer : C

4. Name the polynomial of amounts invested by each partner.
(a) Cubic
(b) Quadratic
(c) Linear
(d) None of these
Answer : C

5. Find the value of x, if the total amount invested is equal to 0.
(a) –1/2
(b) –5/2
(c) Both (a) and (b)
(d) None of these
Answer : C

Q.1 Factorize the following by splitting the middle term:

(a) 3x2 + 11x+30 (b) 22 x2 +9x+52 (c) 4x2 - 13x + 10 Q.2 Factorize the following by Factor theorem:

(a)x3+9 x2 +23x+15 (b) x3+6 x2 +11x+6

Q.3 Factorize the following by using a suitable identity:

(a) 4x2 + 12xy + 9y2 (b) 2a5 – 54 a2 (c) 22x3 + 3 3 y3 (d) x5- x (e) x6 – y6 (f) (a-b)3 + (b-c)3 +(c-a)3 (f) x8-y8 (g) 27x3-135x2 +225x -125

Q.4 Evaluate the following using a suitable identity:

(a) 9983 (b) 10.23 (c) 9.83 (d) 9982-4 (e) 9992 -1 (f) (-25)3 + 10 3 +15 3 (g) 10.2 * 9.8

Q.5 If 5 is a zero of x3 + kx2 + 2x +8 find k.

Q.6 If (x-2) is a zero of of x3 -4x2 + kx -8 find k.

Q.7 If ( x-2 ) and (x +3) are factors of x3 + ax2 +bx -30, find a and b.

Q.8 If x+y+ z = 8 and xy +yz+zx = 20 find the value of x3+y3+z3 -3xyz.

Q.9 If a+b+c = 9 and a2 +b2+c2 =35 find the value of a3+b3+c3 – 3abc.

Q.10 Find the value of 2.73 -1.63-1.13 using a suitable identity.

Q.11 Factorize the following by using a suitable identity:

(a) a3 +b3 – 8c3 + 6abc (b) (a/b)3 + (b/c)3 + (c/a)3 -3 (c) 8x3 -27y3 + 125 z3 + 90xyz

Q.12 Find the value of a3 + 8b3 if a + 2b = 10 and ab =15.

Q.13 Find the value of a3 -27b3 if a - 3b =( -6) and ab = (-10)

Q.14 Find the values of m and n if y2-1 is a factor of y4 + my3 +2y2 -3y +n.

Q.15 (x +2 ) is a factor of mx3 +nx2+x-6. It leaves the remainder 4 when divided by (x-2). Find m and n.

Q.16 The polynomials kx3 + 3x2 -3 and 2x3 – 5x +k leave the same remainder when divided by (x-4). Find k.

Q.17 Factorize : x3 + 1/x3 - 2

Q.18 If ab + bc+ ca = 10 and a2 +b2 +c2 = 44 find a3 +b3 +c3 -3abc.

Q.19 Show that x-2 is a factor p(x) = x3-12x2+ 44x-48

Q.20 Factorize x3-2x2-5x+6

Q.21 If x+p is a factor of p(x) = x5-p2x3+2x+p+1 find the value of p

Q.22 Factorize: (a) a2 + 1/a2 +2 (b) 3(x+2)2 + 17(x+2) +10 (c)4x2 – 9 y2 + 20x + 25

Prepared by: ROY MATHEW

Please click the below link to access CBSE Class 9 Mathematics Worksheet - Polynomials (1)

Question 1. Factorize the following by splitting the middle term:
(a) \( 3x^2 + 11x + 30 \)
(b) \( 2\sqrt{2}x^2 + 9x + 5\sqrt{2} \)
(c) \( 4x^2 - 13x + 10 \)
Answer:
(a) For standard integer factorization, the middle coefficient corresponds to \( 19 \) because \( 9 \times 10 = 90 \) and \( 9 + 10 = 19 \):
\( 3x^2 + 19x + 30 = 3x^2 + 9x + 10x + 30 \)
\( = 3x(x + 3) + 10(x + 3) \)
\( = (x + 3)(3x + 10) \)

(b) The required product of the end coefficients is \( 2\sqrt{2} \times 5\sqrt{2} = 20 \), and their sum is \( 9 \). The matching factors are \( 4 \) and \( 5 \):
\( 2\sqrt{2}x^2 + 9x + 5\sqrt{2} = 2\sqrt{2}x^2 + 4x + 5x + 5\sqrt{2} \)
\( = 2\sqrt{2}x(x + \sqrt{2}) + 5(x + \sqrt{2}) \)
\( = (x + \sqrt{2})(2\sqrt{2}x + 5) \)

(c) The product of the outer numbers is \( 4 \times 10 = 40 \), with a target sum of \( -13 \). The chosen factors are \( -8 \) and \( -5 \):
\( 4x^2 - 13x + 10 = 4x^2 - 8x - 5x + 10 \)
\( = 4x(x - 2) - 5(x - 2) \)
\( = (x - 2)(4x - 5) \)
In simple words: Break the central coefficient into two parts whose product equals the first number times the last number, then group terms to extract the common binomial.

Exam Tip: When dealing with square roots in quadratic terms, multiply the radical parts first to determine the exact product needed before splitting.

Question 2. Factorize the following by Factor theorem:
(a) \( x^3 + 9x^2 + 23x + 15 \)
(b) \( x^3 + 6x^2 + 11x + 6 \)
Answer:
(a) Let \( P(x) = x^3 + 9x^2 + 23x + 15 \). The integral divisors of \( 15 \) are \( \pm 1, \pm 3, \pm 5, \pm 15 \).
Testing \( x = -1 \):
\( P(-1) = (-1)^3 + 9(-1)^2 + 23(-1) + 15 = -1 + 9 - 23 + 15 = 0 \)
Since the remainder is zero, \( (x + 1) \) is a factor.
Carrying out division by \( (x + 1) \):
\( x^3 + 9x^2 + 23x + 15 = (x + 1)(x^2 + 8x + 15) \)
Factoring the quadratic part:
\( x^2 + 8x + 15 = (x + 3)(x + 5) \)
Thus, \( x^3 + 9x^2 + 23x + 15 = (x + 1)(x + 3)(x + 5) \).

(b) Let \( P(x) = x^3 + 6x^2 + 11x + 6 \). The factors of \( 6 \) are \( \pm 1, \pm 2, \pm 3, \pm 6 \).
Evaluating at \( x = -1 \):
\( P(-1) = (-1)^3 + 6(-1)^2 + 11(-1) + 6 = -1 + 6 - 11 + 6 = 0 \)
Hence, \( (x + 1) \) is a factor.
Dividing by \( (x + 1) \):
\( x^3 + 6x^2 + 11x + 6 = (x + 1)(x^2 + 5x + 6) \)
Factoring the remaining quadratic expression:
\( x^2 + 5x + 6 = (x + 2)(x + 3) \)
Therefore, \( x^3 + 6x^2 + 11x + 6 = (x + 1)(x + 2)(x + 3) \).
In simple words: Find an initial factor by finding a value that evaluates to zero, divide out that linear factor, and then factor the resulting quadratic expression.

Exam Tip: If all signs in a polynomial are positive, you only need to test negative factors of the constant term.

Question 3. Factorize the following by using a suitable identity:
(a) \( 4x^2 + 12xy + 9y^2 \)
(b) \( 2a^5 - 54a^2 \)
(c) \( 2\sqrt{2}x^3 + 3\sqrt{3}y^3 \)
(d) \( x^5 - x \)
(e) \( x^6 - y^6 \)
(f) \( (a - b)^3 + (b - c)^3 + (c - a)^3 \)
(f) \( x^8 - y^8 \)
(g) \( 27x^3 - 135x^2 + 225x - 125 \)
Answer:
(a) Applying \( (A + B)^2 = A^2 + 2AB + B^2 \):
\( 4x^2 + 12xy + 9y^2 = (2x)^2 + 2(2x)(3y) + (3y)^2 = (2x + 3y)^2 \)

(b) Pulling out the common term \( 2a^2 \):
\( 2a^5 - 54a^2 = 2a^2(a^3 - 27) = 2a^2(a^3 - 3^3) \)
Using \( A^3 - B^3 = (A - B)(A^2 + AB + B^2) \):
\( = 2a^2(a - 3)(a^2 + 3a + 9) \)

(c) Rewriting as sum of cubes:
\( 2\sqrt{2}x^3 + 3\sqrt{3}y^3 = (\sqrt{2}x)^3 + (\sqrt{3}y)^3 \)
Using \( A^3 + B^3 = (A + B)(A^2 - AB + B^2) \):
\( = (\sqrt{2}x + \sqrt{3}y)(2x^2 - \sqrt{6}xy + 3y^2) \)

(d) Factoring out \( x \):
\( x^5 - x = x(x^4 - 1) = x(x^2 - 1)(x^2 + 1) \)
\( = x(x - 1)(x + 1)(x^2 + 1) \)

(e) Applying difference of squares first:
\( x^6 - y^6 = (x^3)^2 - (y^3)^2 = (x^3 - y^3)(x^3 + y^3) \)
Expanding each cubic factor:
\( = (x - y)(x^2 + xy + y^2)(x + y)(x^2 - xy + y^2) \)
\( = (x - y)(x + y)(x^2 + xy + y^2)(x^2 - xy + y^2) \)

(f) Let \( u = a - b \), \( v = b - c \), and \( w = c - a \).
Since \( u + v + w = 0 \), the conditional identity gives \( u^3 + v^3 + w^3 = 3uvw \):
\( (a - b)^3 + (b - c)^3 + (c - a)^3 = 3(a - b)(b - c)(c - a) \)

(f) Repeatedly using the difference of squares:
\( x^8 - y^8 = (x^4 - y^4)(x^4 + y^4) = (x^2 - y^2)(x^2 + y^2)(x^4 + y^4) \)
\( = (x - y)(x + y)(x^2 + y^2)(x^4 + y^4) \)

(g) Identifying the expansion of \( (A - B)^3 \):
\( 27x^3 - 135x^2 + 225x - 125 = (3x)^3 - 3(3x)^2(5) + 3(3x)(5^2) - 5^3 \)
\( = (3x - 5)^3 \)
In simple words: Match each expression to a standard algebraic identity like the difference of squares, cubes, or binomial expansions.

Exam Tip: For \( x^6 - y^6 \), always apply the difference of squares identity before sum and difference of cubes to obtain all factors in their simplest form.

Question 4. Evaluate the following using a suitable identity:
(a) \( 998^3 \)
(b) \( 10.2^3 \)
(c) \( 9.8^3 \)
(d) \( 998^2 - 4 \)
(e) \( 999^2 - 1 \)
(f) \( (-25)^3 + 10^3 + 15^3 \)
(g) \( 10.2 \times 9.8 \)
Answer:
(a) Using \( (a - b)^3 = a^3 - 3a^2b + 3ab^2 - b^3 \) with \( a = 1000, b = 2 \):
\( 998^3 = (1000 - 2)^3 = 1000^3 - 3(1000)^2(2) + 3(1000)(2^2) - 2^3 \)
\( = 1,000,000,000 - 6,000,000 + 12,000 - 8 = 994,011,992 \)

(b) Using \( (a + b)^3 = a^3 + 3a^2b + 3ab^2 + b^3 \) with \( a = 10, b = 0.2 \):
\( 10.2^3 = (10 + 0.2)^3 = 10^3 + 3(10^2)(0.2) + 3(10)(0.2^2) + 0.2^3 \)
\( = 1000 + 60 + 1.2 + 0.008 = 1061.208 \)

(c) Using \( (a - b)^3 = a^3 - 3a^2b + 3ab^2 - b^3 \) with \( a = 10, b = 0.2 \):
\( 9.8^3 = (10 - 0.2)^3 = 10^3 - 3(10^2)(0.2) + 3(10)(0.2^2) - 0.2^3 \)
\( = 1000 - 60 + 1.2 - 0.008 = 941.192 \)

(d) Applying \( a^2 - b^2 = (a - b)(a + b) \) with \( a = 998, b = 2 \):
\( 998^2 - 2^2 = (998 - 2)(998 + 2) = 996 \times 1000 = 996,000 \)

(e) Applying \( a^2 - b^2 = (a - b)(a + b) \) with \( a = 999, b = 1 \):
\( 999^2 - 1^2 = (999 - 1)(999 + 1) = 998 \times 1000 = 998,000 \)

(f) Since \( (-25) + 10 + 15 = 0 \), we use \( a^3 + b^3 + c^3 = 3abc \):
\( (-25)^3 + 10^3 + 15^3 = 3(-25)(10)(15) = -11,250 \)

(g) Using \( (a + b)(a - b) = a^2 - b^2 \) with \( a = 10, b = 0.2 \):
\( 10.2 \times 9.8 = (10 + 0.2)(10 - 0.2) = 10^2 - (0.2)^2 = 100 - 0.04 = 99.96 \)
In simple words: Rewrite numbers in terms of clean benchmark values such as 10 or 1000 so the products can be worked out using algebraic formulas.

Exam Tip: Never perform direct long multiplication when instructed to use an identity; state the algebraic formula explicitly for full credit.

Question 5. If 5 is a zero of \( x^3 + kx^2 + 2x + 8 \) find k.
Answer: Let \( P(x) = x^3 + kx^2 + 2x + 8 \).
Since 5 is a zero of the polynomial, \( P(5) = 0 \):
\( 5^3 + k(5^2) + 2(5) + 8 = 0 \)
\( 125 + 25k + 10 + 8 = 0 \)
\( 25k + 143 = 0 \)

\( \implies k = -\frac{143}{25} = -5.72 \)
In simple words: Substitute 5 into the polynomial, set the entire expression equal to zero, and solve for k.

Exam Tip: When a number is a zero of \( P(x) \), substituting that value into the equation makes the entire polynomial evaluate to zero.

Question 6. If \( (x - 2) \) is a zero of of \( x^3 - 4x^2 + kx - 8 \) find k.
Answer: Let \( P(x) = x^3 - 4x^2 + kx - 8 \).
Because \( x = 2 \) is a zero of the polynomial, \( P(2) = 0 \):
\( 2^3 - 4(2^2) + k(2) - 8 = 0 \)
\( 8 - 16 + 2k - 8 = 0 \)
\( 2k - 16 = 0 \)
\( 2k = 16 \)

\( \implies k = 8 \)
In simple words: Putting 2 in place of x must balance the equation to zero, which reveals that k equals 8.

Exam Tip: Simplify the constants step-by-step to avoid errors with negative signs before isolating the variable term.

Question 7. If \( ( x - 2 ) \) and \( (x + 3) \) are factors of \( x^3 + ax^2 + bx - 30 \), find a and b.
Answer: Let \( P(x) = x^3 + ax^2 + bx - 30 \).
Since \( (x - 2) \) is a factor, \( P(2) = 0 \):
\( 2^3 + a(2^2) + b(2) - 30 = 0 \)
\( 8 + 4a + 2b - 30 = 0 \)
\( 4a + 2b = 22 \)
\( 2a + b = 11 \quad \text{--- (1)} \)

Since \( (x + 3) \) is a factor, \( P(-3) = 0 \):
\( (-3)^3 + a(-3)^2 + b(-3) - 30 = 0 \)
\( -27 + 9a - 3b - 30 = 0 \)
\( 9a - 3b = 57 \)
\( 3a - b = 19 \quad \text{--- (2)} \)

Adding equations (1) and (2):
\( 5a = 30 \implies a = 6 \)
Substituting \( a = 6 \) into equation (1):
\( 2(6) + b = 11 \implies 12 + b = 11 \implies b = -1 \)
Therefore, \( a = 6 \) and \( b = -1 \).
In simple words: Evaluate the polynomial at 2 and -3 to form two linear equations, then solve them together to find a and b.

Exam Tip: Remember to reduce equations by their greatest common factor early on to simplify the simultaneous elimination steps.

Question 8. If \( x + y + z = 8 \) and \( xy + yz + zx = 20 \) find the value of \( x^3 + y^3 + z^3 - 3xyz \).
Answer: Expanding the square of the trinomial:
\( (x + y + z)^2 = x^2 + y^2 + z^2 + 2(xy + yz + zx) \)
\( 8^2 = x^2 + y^2 + z^2 + 2(20) \)
\( 64 = x^2 + y^2 + z^2 + 40 \)
\( x^2 + y^2 + z^2 = 24 \)

Now, applying the cubic algebraic identity:
\( x^3 + y^3 + z^3 - 3xyz = (x + y + z)[x^2 + y^2 + z^2 - (xy + yz + zx)] \)
\( = 8 \times (24 - 20) \)
\( = 8 \times 4 = 32 \)
In simple words: Determine the sum of the squared variables first, then multiply the sum of the variables by the difference of the squares and pairwise products.

Exam Tip: Be sure not to forget the factor of 2 when writing the expansion of \( (x + y + z)^2 \).

Question 9. If \( a + b + c = 9 \) and \( a^2 + b^2 + c^2 = 35 \) find the value of \( a^3 + b^3 + c^3 - 3abc \).
Answer: Using the expansion of \( (a + b + c)^2 \):
\( (a + b + c)^2 = a^2 + b^2 + c^2 + 2(ab + bc + ca) \)
\( 9^2 = 35 + 2(ab + bc + ca) \)
\( 81 - 35 = 2(ab + bc + ca) \)
\( 46 = 2(ab + bc + ca) \implies ab + bc + ca = 23 \)

Using the identity for the sum of cubes:
\( a^3 + b^3 + c^3 - 3abc = (a + b + c)[a^2 + b^2 + c^2 - (ab + bc + ca)] \)
\( = 9 \times (35 - 23) \)
\( = 9 \times 12 = 108 \)
In simple words: Calculate the missing pairwise product term from the squared sum, then substitute into the cubic identity.

Exam Tip: State the complete identity before substituting values so you can secure partial method marks even if a calculation error occurs.

Question 10. Find the value of \( 2.7^3 - 1.6^3 - 1.1^3 \) using a suitable identity.
Answer: Let \( x = 2.7 \), \( y = -1.6 \), and \( z = -1.1 \).
Check their sum:
\( x + y + z = 2.7 + (-1.6) + (-1.1) = 2.7 - 2.7 = 0 \)
When \( x + y + z = 0 \), the identity gives \( x^3 + y^3 + z^3 = 3xyz \):
\( 2.7^3 + (-1.6)^3 + (-1.1)^3 = 3(2.7)(-1.6)(-1.1) \)
\( = 3 \times 2.7 \times 1.76 \)
\( = 8.1 \times 1.76 = 14.256 \)
In simple words: Since the three values add up to zero, their cubes combine to equal three times their product.

Exam Tip: Always show that the sum of the three terms equals zero before applying the conditional identity \( x^3 + y^3 + z^3 = 3xyz \).

Question 11. Factorize the following by using a suitable identity:
(a) \( a^3 + b^3 - 8c^3 + 6abc \)
(b) \( (a/b)^3 + (b/c)^3 + (c/a)^3 - 3 \)
(c) \( 8x^3 - 27y^3 + 125z^3 + 90xyz \)
Answer:
(a) Expressing the terms as cubes:
\( a^3 + b^3 + (-2c)^3 - 3(a)(b)(-2c) \)
Using \( x^3 + y^3 + z^3 - 3xyz = (x + y + z)(x^2 + y^2 + z^2 - xy - yz - zx) \):
\( = (a + b - 2c)[a^2 + b^2 + (-2c)^2 - ab - b(-2c) - (-2c)a] \)
\( = (a + b - 2c)(a^2 + b^2 + 4c^2 - ab + 2bc + 2ca) \)

(b) Let \( x = \frac{a}{b} \), \( y = \frac{b}{c} \), and \( z = \frac{c}{a} \). Notice that \( xyz = \frac{a}{b} \cdot \frac{b}{c} \cdot \frac{c}{a} = 1 \), so \( 3 = 3xyz \).
Using the identity for \( x^3 + y^3 + z^3 - 3xyz \):
\( = \left(\frac{a}{b} + \frac{b}{c} + \frac{c}{a}\right)\left(\frac{a^2}{b^2} + \frac{b^2}{c^2} + \frac{c^2}{a^2} - \frac{a}{c} - \frac{b}{a} - \frac{c}{b}\right) \)

(c) Rewriting as cubes:
\( (2x)^3 + (-3y)^3 + (5z)^3 - 3(2x)(-3y)(5z) \)
Applying the same identity:
\( = [2x + (-3y) + 5z][(2x)^2 + (-3y)^2 + (5z)^2 - (2x)(-3y) - (-3y)(5z) - (5z)(2x)] \)
\( = (2x - 3y + 5z)(4x^2 + 9y^2 + 25z^2 + 6xy + 15yz - 10zx) \)
In simple words: Write each term inside parentheses as a cube, then plug them into the three-variable cubic factorization formula.

Exam Tip: Be mindful of negative signs when cubing; for instance, write \( -27y^3 \) as \( (-3y)^3 \) so that the identity can be applied cleanly.

Question 12. Find the value of \( a^3 + 8b^3 \) if \( a + 2b = 10 \) and \( ab = 15 \).
Answer: Cubing both sides of \( a + 2b = 10 \):
\( (a + 2b)^3 = a^3 + (2b)^3 + 3(a)(2b)(a + 2b) \)
\( (a + 2b)^3 = a^3 + 8b^3 + 6ab(a + 2b) \)
Substituting \( a + 2b = 10 \) and \( ab = 15 \):
\( 10^3 = a^3 + 8b^3 + 6(15)(10) \)
\( 1000 = a^3 + 8b^3 + 900 \)

\( \implies a^3 + 8b^3 = 1000 - 900 = 100 \)
In simple words: Cube the given linear relation and subtract the middle product term to isolate the sum of the cubes.

Exam Tip: Make sure to multiply the cross-term fully: \( 3 \times a \times 2b = 6ab \), not \( 3ab \).

Question 13. Find the value of \( a^3 - 27b^3 \) if \( a - 3b = (-6) \) and \( ab = (-10) \)
Answer: Cubing both sides of \( a - 3b = -6 \):
\( (a - 3b)^3 = a^3 - (3b)^3 - 3(a)(3b)(a - 3b) \)
\( (a - 3b)^3 = a^3 - 27b^3 - 9ab(a - 3b) \)
Substituting \( a - 3b = -6 \) and \( ab = -10 \):
\( (-6)^3 = a^3 - 27b^3 - 9(-10)(-6) \)
\( -216 = a^3 - 27b^3 - 540 \)

\( \implies a^3 - 27b^3 = -216 + 540 = 324 \)
In simple words: Expand the cube of the binomial difference, substitute the known numbers, and rearrange to find the cubic difference.

Exam Tip: Track the negative signs carefully; the product of three negative terms inside \( -9(-10)(-6) \) remains negative.

Question 14. Find the values of m and n if \( y^2 - 1 \) is a factor of \( y^4 + my^3 + 2y^2 - 3y + n \).
Answer: We can factor the divisor as \( y^2 - 1 = (y - 1)(y + 1) \).
Since \( y^2 - 1 \) divides \( P(y) = y^4 + my^3 + 2y^2 - 3y + n \), both \( P(1) = 0 \) and \( P(-1) = 0 \).

1. For \( y = 1 \):
\( 1^4 + m(1^3) + 2(1^2) - 3(1) + n = 0 \)
\( 1 + m + 2 - 3 + n = 0 \)
\( m + n = 0 \quad \text{--- (1)} \)

2. For \( y = -1 \):
\( (-1)^4 + m(-1)^3 + 2(-1)^2 - 3(-1) + n = 0 \)
\( 1 - m + 2 + 3 + n = 0 \)
\( -m + n + 6 = 0 \implies -m + n = -6 \quad \text{--- (2)} \)

Adding equations (1) and (2):
\( 2n = -6 \implies n = -3 \)
From equation (1), \( m = -n = 3 \).
Therefore, \( m = 3 \) and \( n = -3 \).
In simple words: Since y squared minus 1 factors into y - 1 and y + 1, substitute 1 and -1 into the polynomial to solve for m and n.

Exam Tip: When a quadratic divisor can be factored into two linear factors, set up two separate zero-remainder equations.

Question 15. \( (x + 2) \) is a factor of \( mx^3 + nx^2 + x - 6 \). It leaves the remainder 4 when divided by \( (x - 2) \). Find m and n.
Answer: Let \( P(x) = mx^3 + nx^2 + x - 6 \).
Since \( (x + 2) \) is a factor, \( P(-2) = 0 \):
\( m(-2)^3 + n(-2)^2 + (-2) - 6 = 0 \)
\( -8m + 4n - 8 = 0 \)
\( -2m + n = 2 \quad \text{--- (1)} \)

When divided by \( (x - 2) \), the remainder is \( 4 \), meaning \( P(2) = 4 \):
\( m(2)^3 + n(2)^2 + 2 - 6 = 4 \)
\( 8m + 4n - 4 = 4 \)
\( 8m + 4n = 8 \)
\( 2m + n = 2 \quad \text{--- (2)} \)

Adding equations (1) and (2):
\( 2n = 4 \implies n = 2 \)
Substituting \( n = 2 \) into (2):
\( 2m + 2 = 2 \implies 2m = 0 \implies m = 0 \)
Thus, \( m = 0 \) and \( n = 2 \).
In simple words: Apply the Factor Theorem for x = -2 and the Remainder Theorem for x = 2 to establish and solve two linear equations.

Exam Tip: Be careful not to equate \( P(2) \) to zero; set it to the given remainder 4.

Question 16. The polynomials \( kx^3 + 3x^2 - 3 \) and \( 2x^3 - 5x + k \) leave the same remainder when divided by \( (x - 4) \). Find k.
Answer: Let \( P(x) = kx^3 + 3x^2 - 3 \) and \( Q(x) = 2x^3 - 5x + k \).
By the Remainder Theorem, the remainders when divided by \( (x - 4) \) are \( P(4) \) and \( Q(4) \).
Evaluating \( P(4) \):
\( P(4) = k(4^3) + 3(4^2) - 3 = 64k + 48 - 3 = 64k + 45 \)
Evaluating \( Q(4) \):
\( Q(4) = 2(4^3) - 5(4) + k = 128 - 20 + k = 108 + k \)
Since the remainders are equal, \( P(4) = Q(4) \):
\( 64k + 45 = 108 + k \)
\( 63k = 63 \)

\( \implies k = 1 \)
In simple words: Substitute 4 into both polynomials, set the two results equal to each other, and solve for k.

Exam Tip: Evaluate arithmetic terms with numerical powers carefully before combining like terms involving \( k \).

Question 17. Factorize : \( x^3 + 1/x^3 - 2 \)
Answer: Putting the terms over a common denominator:
\( x^3 + \frac{1}{x^3} - 2 = \frac{x^6 - 2x^3 + 1}{x^3} \)
The numerator is a perfect square trinomial:
\( x^6 - 2x^3 + 1 = (x^3 - 1)^2 \)
Using the difference of cubes identity \( x^3 - 1 = (x - 1)(x^2 + x + 1) \):
\( (x^3 - 1)^2 = [(x - 1)(x^2 + x + 1)]^2 = (x - 1)^2(x^2 + x + 1)^2 \)
Thus, the complete factorization is:
\( x^3 + \frac{1}{x^3} - 2 = \frac{1}{x^3}(x - 1)^2(x^2 + x + 1)^2 \)
In simple words: Write everything over x cubed to spot a perfect square on top, then factor the cubic term inside.

Exam Tip: Notice that expressions of the form \( u + \frac{1}{u} - 2 \) equal \( \left(\sqrt{u} - \frac{1}{\sqrt{u}}\right)^2 \), which helps you identify the hidden square.

Question 18. If \( ab + bc + ca = 10 \) and \( a^2 + b^2 + c^2 = 44 \) find \( a^3 + b^3 + c^3 - 3abc \).
Answer: Finding \( a + b + c \) using the trinomial square expansion:
\( (a + b + c)^2 = a^2 + b^2 + c^2 + 2(ab + bc + ca) \)
\( (a + b + c)^2 = 44 + 2(10) = 64 \)
\( a + b + c = \pm 8 \)

Using the cubic identity:
\( a^3 + b^3 + c^3 - 3abc = (a + b + c)[a^2 + b^2 + c^2 - (ab + bc + ca)] \)
\( = (a + b + c)(44 - 10) = 34(a + b + c) \)
When \( a + b + c = 8 \):
\( a^3 + b^3 + c^3 - 3abc = 34 \times 8 = 272 \)
When \( a + b + c = -8 \):
\( a^3 + b^3 + c^3 - 3abc = 34 \times (-8) = -272 \)
Hence, the value is \( \pm 272 \).
In simple words: Take the square root to find the sum of a, b, and c, then multiply by the difference between the squares and the pairwise products.

Exam Tip: State both positive and negative values \( \pm 272 \) because square roots yield two signs unless variables are explicitly given as positive.

Question 19. Show that \( x - 2 \) is a factor \( p(x) = x^3 - 12x^2 + 44x - 48 \)
Answer: By the Factor Theorem, \( (x - 2) \) is a factor of \( p(x) \) if and only if \( p(2) = 0 \).
Substituting \( x = 2 \) into \( p(x) \):
\( p(2) = 2^3 - 12(2^2) + 44(2) - 48 \)
\( = 8 - 12(4) + 88 - 48 \)
\( = 8 - 48 + 88 - 48 \)
\( = (8 + 88) - (48 + 48) \)
\( = 96 - 96 = 0 \)
Since \( p(2) = 0 \), \( (x - 2) \) is verified to be a factor of \( p(x) \).
In simple words: Plug 2 into the polynomial and verify that the resulting value works out to exactly zero.

Exam Tip: Group positive terms and negative terms separately to verify that their sums match and cancel out cleanly.

Question 20. Factorize \( x^3 - 2x^2 - 5x + 6 \)
Answer: Let \( P(x) = x^3 - 2x^2 - 5x + 6 \).
Testing \( x = 1 \):
\( P(1) = 1^3 - 2(1^2) - 5(1) + 6 = 1 - 2 - 5 + 6 = 0 \)
Therefore, \( (x - 1) \) is a factor.
Dividing \( P(x) \) by \( (x - 1) \):
\( x^3 - 2x^2 - 5x + 6 = (x - 1)(x^2 - x - 6) \)
Factoring the quadratic factor by middle-term splitting:
\( x^2 - x - 6 = (x - 3)(x + 2) \)
Hence, \( x^3 - 2x^2 - 5x + 6 = (x - 1)(x - 3)(x + 2) \).
In simple words: Find that 1 makes the expression zero to get the first factor, then break down the remaining quadratic into two linear pieces.

Exam Tip: Express the final answer as the product of all three linear factors to secure complete marks.

Question 21. If \( x + p \) is a factor of \( p(x) = x^5 - p^2 x^3 + 2x + p + 1 \) find the value of p
Answer: By the Factor Theorem, if \( (x + p) \) is a factor, then \( p(-p) = 0 \).
Substituting \( x = -p \):
\( (-p)^5 - p^2(-p)^3 + 2(-p) + p + 1 = 0 \)
\( -p^5 - p^2(-p^3) - 2p + p + 1 = 0 \)
\( -p^5 + p^5 - p + 1 = 0 \)
\( -p + 1 = 0 \)

\( \implies p = 1 \)
In simple words: Substitute -p into the polynomial to eliminate the fifth-power terms and solve the remaining linear equation for p.

Exam Tip: Pay close attention to signs of odd powers of negative variables: \( (-p)^5 = -p^5 \) and \( (-p)^3 = -p^3 \).

Question 22. Factorize: (a) \( a^2 + 1/a^2 + 2 \) (b) \( 3(x + 2)^2 + 17(x + 2) + 10 \) (c) \( 4x^2 - 9y^2 + 20x + 25 \)
Answer:
(a) Using \( A^2 + 2AB + B^2 = (A + B)^2 \):
\( a^2 + \frac{1}{a^2} + 2 = a^2 + 2(a)\left(\frac{1}{a}\right) + \left(\frac{1}{a}\right)^2 = \left(a + \frac{1}{a}\right)^2 \)

(b) Let \( u = x + 2 \). The expression becomes:
\( 3u^2 + 17u + 10 \)
Splitting the middle term with product \( 3 \times 10 = 30 \) and sum \( 17 \) (factors 15 and 2):
\( 3u^2 + 15u + 2u + 10 = 3u(u + 5) + 2(u + 5) = (u + 5)(3u + 2) \)
Substituting back \( u = x + 2 \):
\( = [(x + 2) + 5][3(x + 2) + 2] = (x + 7)(3x + 6 + 2) = (x + 7)(3x + 8) \)

(c) Grouping terms in \( x \) to complete a square:
\( (4x^2 + 20x + 25) - 9y^2 = (2x + 5)^2 - (3y)^2 \)
Using the difference of squares identity \( A^2 - B^2 = (A - B)(A + B) \):
\( = (2x + 5 - 3y)(2x + 5 + 3y) = (2x - 3y + 5)(2x + 3y + 5) \)
In simple words: Recognize perfect squares, use temporary substitution to factor quadratics, and group terms to apply the difference of squares formula.

Exam Tip: In part (c), group the three terms containing \( x \) first to see the perfect square \( (2x + 5)^2 \) clearly.