How to Use This Page
1. Solve each exercise problem on your own first — attempt every step before looking here. 2. Compare your method with the step-by-step solution shown. 3. If your answer differs, rework from the step where you diverged. 4. Pay special attention to the Common Mistakes section — these are the errors that cost marks in exams. 5. Use the Exam Checklist before your test to ensure you have covered every topic.
Key Concepts (Quick Recap)
Polynomials
- Polynomial: An algebraic expression of the form a_n x^n + a_{n-1} x^{n-1} + ... + a_1 x + a_0 where a_n != 0
- Degree: Highest power of the variable. Example: 5x^3 + 2x - 1 has degree 3
- Types by degree: Constant (0), Linear (1), Quadratic (2), Cubic (3)
- Coefficient: The numerical factor of each term. In 3x^2 + 5x - 7, the coefficient of x^2 is 3, of x is 5
- Zero polynomial: p(x) = 0; degree is undefined (or taken as -1)
- Monomial, Binomial, Trinomial: 1, 2, or 3 terms respectively
Remainder Theorem
- If p(x) is divided by (x - a), then Remainder = p(a)
- No need for long division — just substitute x = a
Factor Theorem
- (x - a) is a factor of p(x) if and only if p(a) = 0
- To check if (x - k) is a factor, evaluate p(k). If result = 0, it is a factor.
Algebraic Identities
- (a + b)^2 = a^2 + 2ab + b^2
- (a - b)^2 = a^2 - 2ab + b^2
- a^2 - b^2 = (a + b)(a - b)
- (a + b)^3 = a^3 + 3a^2b + 3ab^2 + b^3
- (a - b)^3 = a^3 - 3a^2b + 3ab^2 - b^3
- a^3 + b^3 = (a + b)(a^2 - ab + b^2)
- a^3 - b^3 = (a - b)(a^2 + ab + b^2)
- (a + b + c)^2 = a^2 + b^2 + c^2 + 2ab + 2bc + 2ca
Linear Equations in Two Variables
- Standard form: ax + by + c = 0 (where a, b not both zero)
- Solutions are ordered pairs (x, y) that satisfy the equation
- Graph is always a straight line
- To graph: find two points (often intercepts), plot, and join
Exercise 3.1 — Polynomials: Degree, Coefficients, Types
Q1. Which of the following expressions are polynomials? Give reasons.
Q2. Find the degree of each polynomial.
Q3. Write the coefficient of x^2 in each polynomial.
Q4. Classify the following polynomials as monomial, binomial, or trinomial.
Q5. Classify by degree: linear, quadratic, or cubic.
Q6. Find the value of the polynomial p(x) = 2x^2 - 3x + 5 at:
Q7. If p(x) = x^3 - 6x^2 + 11x - 6, find p(1), p(2), and p(3). What do you observe?
p(1) = (1)^3 - 6(1)^2 + 11(1) - 6
= 1 - 6 + 11 - 6
= 0
p(2) = (2)^3 - 6(2)^2 + 11(2) - 6
= 8 - 24 + 22 - 6
= 0
p(3) = (3)^3 - 6(3)^2 + 11(3) - 6
= 27 - 54 + 33 - 6
= 0
Observation: p(1) = p(2) = p(3) = 0
So (x-1), (x-2), and (x-3) are all factors of p(x).
This means p(x) = (x-1)(x-2)(x-3).
Q8. Verify whether the following are zeros of the polynomial p(x) = x^2 - 4x + 3.
Q9. Find the zeros of the polynomial p(x) = x^2 + 5x + 6.
p(x) = x^2 + 5x + 6
Factorise: x^2 + 5x + 6 = x^2 + 2x + 3x + 6
= x(x + 2) + 3(x + 2)
= (x + 2)(x + 3)
For zeros: p(x) = 0
(x + 2)(x + 3) = 0
x + 2 = 0 => x = -2
x + 3 = 0 => x = -3
Zeros are x = -2 and x = -3
Q10. If p(y) = y^2 - 3y + 2, find the value of p(0) + p(1).
p(0) = (0)^2 - 3(0) + 2 = 0 - 0 + 2 = 2
p(1) = (1)^2 - 3(1) + 2 = 1 - 3 + 2 = 0
p(0) + p(1) = 2 + 0 = 2
Answer: 2
Q11. The polynomial p(x) = x^3 - ax^2 + bx + c has zeros 1, 2, and 3. Find a, b, and c.
Since 1, 2, 3 are zeros, the polynomial can be written as:
p(x) = (x - 1)(x - 2)(x - 3)
Expanding step by step:
Step 1: (x - 1)(x - 2) = x^2 - 2x - x + 2 = x^2 - 3x + 2
Step 2: (x^2 - 3x + 2)(x - 3)
= x^2(x - 3) - 3x(x - 3) + 2(x - 3)
= x^3 - 3x^2 - 3x^2 + 9x + 2x - 6
= x^3 - 6x^2 + 11x - 6
Comparing with x^3 - ax^2 + bx + c:
a = 6, b = 11, c = -6
Q12. Give an example of each: (i) a cubic binomial (ii) a quadratic trinomial (iii) a linear monomial
(i) Cubic binomial: x^3 - 5
(degree 3, 2 terms)
(ii) Quadratic trinomial: x^2 + 3x - 7
(degree 2, 3 terms)
(iii) Linear monomial: 4x
(degree 1, 1 term)
Q13. If the polynomial p(x) = kx^2 - 3x + 7 has p(1) = 5, find the value of k.
p(1) = k(1)^2 - 3(1) + 7 = 5
k - 3 + 7 = 5
k + 4 = 5
k = 1
Answer: k = 1
Q14. If p(x) = 2x + 3 and q(x) = x - 1, find p(q(2)).
Step 1: Find q(2)
q(2) = 2 - 1 = 1
Step 2: Find p(q(2)) = p(1)
p(1) = 2(1) + 3 = 2 + 3 = 5
Answer: p(q(2)) = 5
Q15. Write a polynomial with zeros 2 and -1.
If 2 and -1 are zeros, then (x - 2) and (x + 1) are factors.
p(x) = (x - 2)(x + 1)
= x^2 + x - 2x - 2
= x^2 - x - 2
Answer: p(x) = x^2 - x - 2
(Any scalar multiple is also valid.)
Q16. Which of the following are quadratic polynomials? (i) x^2 + 3x - 1 (ii) 2x + 3 (iii) x^3 - x (iv) x^2 + 1
(i) x^2 + 3x - 1: Degree = 2. Yes, it is quadratic.
(ii) 2x + 3: Degree = 1. No, it is linear, not quadratic.
(iii) x^3 - x: Degree = 3. No, it is cubic, not quadratic.
(iv) x^2 + 1: Degree = 2. Yes, it is quadratic.
Q17. Find the number of zeros of the polynomial p(x) = (x - 1)(x + 2)(x - 3)(x + 4).
The polynomial is already in factored form.
Setting p(x) = 0:
(x - 1)(x + 2)(x - 3)(x + 4) = 0
Each factor gives one zero:
x - 1 = 0 => x = 1
x + 2 = 0 => x = -2
x - 3 = 0 => x = 3
x + 4 = 0 => x = -4
Number of zeros = 4
Q18. If p(x) = x^2 + 3x + k and p(-3) = 0, find k.
p(-3) = (-3)^2 + 3(-3) + k = 0
9 - 9 + k = 0
0 + k = 0
k = 0
Answer: k = 0
Q19. Find a quadratic polynomial whose sum of zeros is 5 and product of zeros is 6.
If sum of zeros = S = 5 and product of zeros = P = 6,
then the quadratic polynomial is:
p(x) = x^2 - Sx + P
= x^2 - 5x + 6
Verification: x^2 - 5x + 6 = (x - 2)(x - 3)
Zeros: 2 and 3, Sum = 5, Product = 6. Correct.
Q20. If the zeros of p(x) = x^2 - 8x + k differ by 2, find k.
Let the zeros be alpha and alpha + 2.
Sum of zeros = alpha + (alpha + 2) = 2(alpha) + 2
But sum of zeros = 8 (from polynomial, coefficient of x with opposite sign)
So 2(alpha) + 2 = 8
2(alpha) = 6
alpha = 3
Zeros are 3 and 5.
Product of zeros = 3 * 5 = 15
But product of zeros = k (constant term)
So k = 15
Answer: k = 15
Exercise 3.2 — Remainder Theorem and Factor Theorem
Q1. Find the remainder when p(x) = x^3 + 2x^2 - 3x + 4 is divided by (x - 1).
By Remainder Theorem, Remainder = p(1)
p(1) = (1)^3 + 2(1)^2 - 3(1) + 4
= 1 + 2 - 3 + 4
= 4
Answer: Remainder = 4
Q2. Find the remainder when p(x) = x^3 - 3x^2 + 2x + 5 is divided by (x + 2).
(x + 2) = (x - (-2))
By Remainder Theorem, Remainder = p(-2)
p(-2) = (-2)^3 - 3(-2)^2 + 2(-2) + 5
= -8 - 3(4) - 4 + 5
= -8 - 12 - 4 + 5
= -19
Answer: Remainder = -19
Q3. Find the remainder when 2x^3 - 3x^2 + 4x - 7 is divided by (x - 3).
Remainder = p(3)
p(3) = 2(3)^3 - 3(3)^2 + 4(3) - 7
= 2(27) - 3(9) + 12 - 7
= 54 - 27 + 12 - 7
= 32
Answer: Remainder = 32
Q4. Using the Factor Theorem, determine whether (x - 2) is a factor of p(x) = x^3 - 4x^2 + 4x.
By Factor Theorem, (x - 2) is a factor if p(2) = 0.
p(2) = (2)^3 - 4(2)^2 + 4(2)
= 8 - 4(4) + 8
= 8 - 16 + 8
= 0
Since p(2) = 0, (x - 2) IS a factor of p(x).
Q5. Using the Factor Theorem, determine whether (x + 3) is a factor of p(x) = x^3 + 3x^2 - x - 3.
(x + 3) = (x - (-3))
Check p(-3):
p(-3) = (-3)^3 + 3(-3)^2 - (-3) - 3
= -27 + 3(9) + 3 - 3
= -27 + 27 + 3 - 3
= 0
Since p(-3) = 0, (x + 3) IS a factor of p(x).
Q6. Check whether (x - 1) is a factor of p(x) = x^4 + 3x^3 - 4x + 5.
Check p(1):
p(1) = (1)^4 + 3(1)^3 - 4(1) + 5
= 1 + 3 - 4 + 5
= 5
Since p(1) = 5 != 0, (x - 1) is NOT a factor of p(x).
Q7. Find the remainder when x^4 - 3x^2 + 2x + 1 is divided by (x - 1).
Remainder = p(1)
p(1) = (1)^4 - 3(1)^2 + 2(1) + 1
= 1 - 3 + 2 + 1
= 1
Answer: Remainder = 1
Q8. If (x - 3) is a factor of p(x) = x^3 - 7x + k, find k.
Since (x - 3) is a factor, p(3) = 0.
p(3) = (3)^3 - 7(3) + k = 0
27 - 21 + k = 0
6 + k = 0
k = -6
Answer: k = -6
Q9. If (x + 1) is a factor of p(x) = 2x^3 + kx^2 + 7, find k.
Since (x + 1) = (x - (-1)) is a factor, p(-1) = 0.
p(-1) = 2(-1)^3 + k(-1)^2 + 7 = 0
2(-1) + k(1) + 7 = 0
-2 + k + 7 = 0
k + 5 = 0
k = -5
Answer: k = -5
Q10. Find the remainder when x^3 - 6x^2 + 11x - 6 is divided by (x + 1).
(x + 1) = (x - (-1))
Remainder = p(-1)
p(-1) = (-1)^3 - 6(-1)^2 + 11(-1) - 6
= -1 - 6(1) - 11 - 6
= -1 - 6 - 11 - 6
= -24
Answer: Remainder = -24
Q11. If p(x) = x^3 - 3x^2 + 4x + 50, find p(-3). Also state the remainder when p(x) is divided by (x + 3).
p(-3) = (-3)^3 - 3(-3)^2 + 4(-3) + 50
= -27 - 3(9) - 12 + 50
= -27 - 27 - 12 + 50
= -16
By Remainder Theorem, when p(x) is divided by (x + 3) = (x - (-3)),
the remainder is p(-3) = -16.
Answer: p(-3) = -16, Remainder = -16
Q12. Using the Factor Theorem, show that (x - 1) is a factor of x^3 - 1.
Let p(x) = x^3 - 1.
p(1) = (1)^3 - 1 = 1 - 1 = 0
Since p(1) = 0, by Factor Theorem, (x - 1) is a factor of x^3 - 1.
We can also verify by expansion:
x^3 - 1 = (x - 1)(x^2 + x + 1)
Q13. Find the value of a if (x - 2) is a factor of x^3 - ax + 6.
Since (x - 2) is a factor, p(2) = 0.
p(2) = (2)^3 - a(2) + 6 = 0
8 - 2a + 6 = 0
14 - 2a = 0
2a = 14
a = 7
Answer: a = 7
Q14. Find the value of m if (x + 2) is a factor of x^3 + mx^2 - 4.
Since (x + 2) is a factor, p(-2) = 0.
p(-2) = (-2)^3 + m(-2)^2 - 4 = 0
-8 + 4m - 4 = 0
4m - 12 = 0
4m = 12
m = 3
Answer: m = 3
Q15. Find the remainder when x^2 - 5x + 6 is divided by (x - 2). Verify by long division.
By Remainder Theorem:
p(2) = (2)^2 - 5(2) + 6 = 4 - 10 + 6 = 0
Remainder = 0
Long division verification:
x - 3
------
x-2 ) x^2 - 5x + 6
x^2 - 2x
--------
-3x + 6
-3x + 6
-------
0
Remainder = 0. Confirmed.
Q16. Determine whether (x - 1) and (x + 1) are factors of x^4 - 1.
Let p(x) = x^4 - 1.
For (x - 1):
p(1) = (1)^4 - 1 = 1 - 1 = 0
Since p(1) = 0, (x - 1) IS a factor.
For (x + 1):
p(-1) = (-1)^4 - 1 = 1 - 1 = 0
Since p(-1) = 0, (x + 1) IS a factor.
Both (x - 1) and (x + 1) are factors.
In fact: x^4 - 1 = (x^2 - 1)(x^2 + 1) = (x - 1)(x + 1)(x^2 + 1)
Q17. If (x - 1), (x - 2), and (x - 3) are factors of x^3 + ax^2 + bx + c, find a, b, c.
Since (x-1), (x-2), (x-3) are factors:
p(x) = (x - 1)(x - 2)(x - 3)
Step 1: (x - 1)(x - 2) = x^2 - 3x + 2
Step 2: (x^2 - 3x + 2)(x - 3)
= x^2(x - 3) - 3x(x - 3) + 2(x - 3)
= x^3 - 3x^2 - 3x^2 + 9x + 2x - 6
= x^3 - 6x^2 + 11x - 6
Comparing with x^3 + ax^2 + bx + c:
a = -6, b = 11, c = -6
Q18. Find the remainder when f(x) = x^3 - 2x^2 + 3x - 1 is divided by (2x - 1).
Note: For (2x - 1), we set 2x - 1 = 0, so x = 1/2.
Remainder = f(1/2)
f(1/2) = (1/2)^3 - 2(1/2)^2 + 3(1/2) - 1
= 1/8 - 2(1/4) + 3/2 - 1
= 1/8 - 2/4 + 3/2 - 1
= 1/8 - 1/2 + 3/2 - 1
= 1/8 + (-1/2 + 3/2) - 1
= 1/8 + 1 - 1
= 1/8
Answer: Remainder = 1/8
Q19. If p(x) = x^3 - 3x^2 + 2x + 6, verify that (x + 1) is a factor using Factor Theorem.
Check p(-1):
p(-1) = (-1)^3 - 3(-1)^2 + 2(-1) + 6
= -1 - 3(1) - 2 + 6
= -1 - 3 - 2 + 6
= 0
Since p(-1) = 0, (x + 1) is a factor of p(x). Verified.
Q20. The polynomial p(x) = ax^3 + bx^2 + cx + d has zeros at x = -1, 0, and 1. Find the relationship between a, b, c, and d.
Since zeros are -1, 0, 1:
p(x) = a_0 * x(x - 1)(x + 1) = a_0 * x(x^2 - 1) = a_0(x^3 - x)
So the polynomial is proportional to x^3 - x.
This means the coefficients satisfy:
a : b : c : d = 1 : 0 : (-1) : 0
Therefore: b = 0, d = 0, and c = -a.
The polynomial is p(x) = a(x^3 - x) for any non-zero constant a.
Exercise 3.3 — Factorisation Using Identities
Q1. Factorise: x^2 + 6x + 9
x^2 + 6x + 9
Recognise: this is of the form a^2 + 2ab + b^2 = (a + b)^2
where a = x, b = 3
x^2 + 2(x)(3) + 3^2 = (x + 3)^2
Answer: (x + 3)^2
Q2. Factorise: 4x^2 - 20x + 25
4x^2 - 20x + 25
Recognise: (2x)^2 - 2(2x)(5) + 5^2 = (2x - 5)^2
Answer: (2x - 5)^2
Q3. Factorise: x^2 - 16
x^2 - 16
Using a^2 - b^2 = (a + b)(a - b) where a = x, b = 4:
x^2 - 4^2 = (x + 4)(x - 4)
Answer: (x + 4)(x - 4)
Q4. Factorise: 9a^2 - 25b^2
9a^2 - 25b^2
Using a^2 - b^2 = (a + b)(a - b):
(3a)^2 - (5b)^2 = (3a + 5b)(3a - 5b)
Answer: (3a + 5b)(3a - 5b)
Q5. Factorise: x^2 + 5x + 6
x^2 + 5x + 6
Find two numbers whose sum = 5 and product = 6.
Numbers: 2 and 3 (2 + 3 = 5, 2 * 3 = 6)
x^2 + 5x + 6 = x^2 + 2x + 3x + 6
= x(x + 2) + 3(x + 2)
= (x + 2)(x + 3)
Answer: (x + 2)(x + 3)
Q6. Factorise: x^2 - 7x + 12
x^2 - 7x + 12
Find two numbers whose sum = -7 and product = 12.
Numbers: -3 and -4 (-3 + (-4) = -7, (-3)(-4) = 12)
x^2 - 7x + 12 = x^2 - 3x - 4x + 12
= x(x - 3) - 4(x - 3)
= (x - 3)(x - 4)
Answer: (x - 3)(x - 4)
Q7. Factorise: 2x^2 + 7x + 3
2x^2 + 7x + 3
Find two numbers whose sum = 7 and product = 2 * 3 = 6.
Numbers: 6 and 1 (6 + 1 = 7, 6 * 1 = 6)
2x^2 + 7x + 3 = 2x^2 + 6x + x + 3
= 2x(x + 3) + 1(x + 3)
= (2x + 1)(x + 3)
Answer: (2x + 1)(x + 3)
Q8. Factorise: 6x^2 + 5x - 6
6x^2 + 5x - 6
Find two numbers whose sum = 5 and product = 6 * (-6) = -36.
Numbers: 9 and -4 (9 + (-4) = 5, 9 * (-4) = -36)
6x^2 + 5x - 6 = 6x^2 + 9x - 4x - 6
= 3x(2x + 3) - 2(2x + 3)
= (3x - 2)(2x + 3)
Answer: (3x - 2)(2x + 3)
Q9. Factorise: 27x^3 + 8y^3
27x^3 + 8y^3
Using a^3 + b^3 = (a + b)(a^2 - ab + b^2) where a = 3x, b = 2y:
27x^3 + 8y^3 = (3x + 2y)[(3x)^2 - (3x)(2y) + (2y)^2]
= (3x + 2y)(9x^2 - 6xy + 4y^2)
Answer: (3x + 2y)(9x^2 - 6xy + 4y^2)
Q10. Factorise: 8a^3 - 27b^3
8a^3 - 27b^3
Using a^3 - b^3 = (a - b)(a^2 + ab + b^2) where a = 2a, b = 3b:
8a^3 - 27b^3 = (2a - 3b)[(2a)^2 + (2a)(3b) + (3b)^2]
= (2a - 3b)(4a^2 + 6ab + 9b^2)
Answer: (2a - 3b)(4a^2 + 6ab + 9b^2)
Q11. Factorise: x^2 - 2xy + y^2 - z^2
x^2 - 2xy + y^2 - z^2
Step 1: Group the first three terms.
= (x^2 - 2xy + y^2) - z^2
Step 2: Apply (a - b)^2 = a^2 - 2ab + b^2 on first group.
= (x - y)^2 - z^2
Step 3: Apply a^2 - b^2 = (a + b)(a - b) where a = (x - y), b = z.
= (x - y + z)(x - y - z)
Answer: (x - y + z)(x - y - z)
Q12. Factorise: 4x^2 + 12xy + 9y^2
4x^2 + 12xy + 9y^2
Recognise: (2x)^2 + 2(2x)(3y) + (3y)^2 = (2x + 3y)^2
Answer: (2x + 3y)^2
Q13. Factorise: a^2 - b^2 + 2bc - c^2
a^2 - b^2 + 2bc - c^2
Rearrange and group:
= a^2 - (b^2 - 2bc + c^2)
= a^2 - (b - c)^2
Apply a^2 - b^2 = (a + b)(a - b) where a = a, b = (b - c):
= (a + b - c)(a - b + c)
Answer: (a + b - c)(a - b + c)
Q14. Factorise: 16x^2 - 24x + 9
16x^2 - 24x + 9
Recognise: (4x)^2 - 2(4x)(3) + 3^2 = (4x - 3)^2
Answer: (4x - 3)^2
Q15. Factorise: x^4 - 81
x^4 - 81
Step 1: Apply a^2 - b^2 = (a + b)(a - b) where a = x^2, b = 9:
= (x^2)^2 - 9^2 = (x^2 + 9)(x^2 - 9)
Step 2: (x^2 - 9) can be further factorised:
= (x^2 + 9)(x + 3)(x - 3)
Note: x^2 + 9 cannot be factorised over real numbers (it is always positive).
Answer: (x^2 + 9)(x + 3)(x - 3)
Q16. Factorise: 64a^3 - 125b^3
64a^3 - 125b^3
Using a^3 - b^3 = (a - b)(a^2 + ab + b^2) where a = 4a, b = 5b:
= (4a - 5b)[(4a)^2 + (4a)(5b) + (5b)^2]
= (4a - 5b)(16a^2 + 20ab + 25b^2)
Answer: (4a - 5b)(16a^2 + 20ab + 25b^2)
Q17. Factorise: 3x^2 - 11x + 10
3x^2 - 11x + 10
Find two numbers whose sum = -11 and product = 3 * 10 = 30.
Numbers: -6 and -5 (-6 + (-5) = -11, (-6)(-5) = 30)
3x^2 - 11x + 10 = 3x^2 - 6x - 5x + 10
= 3x(x - 2) - 5(x - 2)
= (3x - 5)(x - 2)
Answer: (3x - 5)(x - 2)
Q18. Factorise: (x + y)^2 - (x - y)^2
Using a^2 - b^2 = (a + b)(a - b):
a = (x + y), b = (x - y)
= [(x + y) + (x - y)][(x + y) - (x - y)]
= [x + y + x - y][x + y - x + y]
= [2x][2y]
= 4xy
Answer: 4xy
Q19. Factorise: 125 + 8x^3
125 + 8x^3
Using a^3 + b^3 = (a + b)(a^2 - ab + b^2) where a = 5, b = 2x:
= (5 + 2x)(5^2 - 5(2x) + (2x)^2)
= (5 + 2x)(25 - 10x + 4x^2)
Answer: (5 + 2x)(4x^2 - 10x + 25)
Q20. Factorise: 49p^2 - 84pq + 36q^2
49p^2 - 84pq + 36q^2
Recognise: (7p)^2 - 2(7p)(6q) + (6q)^2 = (7p - 6q)^2
Answer: (7p - 6q)^2
Exercise 3.4 — Algebraic Identities
Q1. Evaluate using identity: 102^2
102^2 = (100 + 2)^2
Using (a + b)^2 = a^2 + 2ab + b^2:
= 100^2 + 2(100)(2) + 2^2
= 10000 + 400 + 4
= 10404
Answer: 10404
Q2. Evaluate using identity: 98^2
98^2 = (100 - 2)^2
Using (a - b)^2 = a^2 - 2ab + b^2:
= 100^2 - 2(100)(2) + 2^2
= 10000 - 400 + 4
= 9604
Answer: 9604
Q3. Evaluate using identity: 997^2
997^2 = (1000 - 3)^2
Using (a - b)^2 = a^2 - 2ab + b^2:
= 1000^2 - 2(1000)(3) + 3^2
= 1000000 - 6000 + 9
= 994009
Answer: 994009
Q4. Evaluate using identity: 103 * 97
103 * 97 = (100 + 3)(100 - 3)
Using a^2 - b^2 = (a + b)(a - b):
= 100^2 - 3^2
= 10000 - 9
= 9991
Answer: 9991
Q5. Evaluate using identity: 104 * 96
104 * 96 = (100 + 4)(100 - 4)
Using a^2 - b^2 = (a + b)(a - b):
= 100^2 - 4^2
= 10000 - 16
= 9984
Answer: 9984
Q6. Evaluate using identity: 105^3
105^3 = (100 + 5)^3
Using (a + b)^3 = a^3 + 3a^2b + 3ab^2 + b^3:
= 100^3 + 3(100)^2(5) + 3(100)(5)^2 + 5^3
= 1000000 + 3(10000)(5) + 3(100)(25) + 125
= 1000000 + 150000 + 7500 + 125
= 1157625
Answer: 1157625
Q7. Evaluate using identity: 99^3
99^3 = (100 - 1)^3
Using (a - b)^3 = a^3 - 3a^2b + 3ab^2 - b^3:
= 100^3 - 3(100)^2(1) + 3(100)(1)^2 - 1^3
= 1000000 - 30000 + 300 - 1
= 970299
Answer: 970299
Q8. Simplify: (2x + 3y)^2 - (2x - 3y)^2
Using a^2 - b^2 = (a + b)(a - b):
Here a = (2x + 3y), b = (2x - 3y)
= [(2x + 3y) + (2x - 3y)][(2x + 3y) - (2x - 3y)]
= [4x][6y]
= 24xy
Alternative: Using (a+b)^2 - (a-b)^2 = 4ab:
= 4(2x)(3y) = 24xy
Answer: 24xy
Q9. Simplify: (3x + 5)^2 - (3x - 5)^2
Using (a + b)^2 - (a - b)^2 = 4ab where a = 3x, b = 5:
= 4(3x)(5) = 60x
Answer: 60x
Q10. If a + b = 7 and ab = 10, find a^2 + b^2.
Using (a + b)^2 = a^2 + 2ab + b^2:
a^2 + b^2 = (a + b)^2 - 2ab
= 7^2 - 2(10)
= 49 - 20
= 29
Answer: 29
Q11. If a - b = 3 and ab = 10, find a^2 + b^2.
Using (a - b)^2 = a^2 - 2ab + b^2:
a^2 + b^2 = (a - b)^2 + 2ab
= 3^2 + 2(10)
= 9 + 20
= 29
Answer: 29
Q12. If x + 1/x = 5, find x^2 + 1/x^2 and x^3 + 1/x^3.
Given: x + 1/x = 5
For x^2 + 1/x^2:
Square both sides: (x + 1/x)^2 = x^2 + 2(x)(1/x) + 1/x^2
5^2 = x^2 + 2 + 1/x^2
25 = x^2 + 1/x^2 + 2
x^2 + 1/x^2 = 23
For x^3 + 1/x^3:
Using a^3 + b^3 = (a + b)^3 - 3ab(a + b):
x^3 + 1/x^3 = (x + 1/x)^3 - 3(x)(1/x)(x + 1/x)
= 5^3 - 3(1)(5)
= 125 - 15
= 110
Answer: x^2 + 1/x^2 = 23, x^3 + 1/x^3 = 110
Q13. If x - 1/x = 3, find x^2 + 1/x^2 and x^4 + 1/x^4.
Given: x - 1/x = 3
For x^2 + 1/x^2:
Square both sides: (x - 1/x)^2 = x^2 - 2 + 1/x^2
3^2 = x^2 - 2 + 1/x^2
9 = x^2 + 1/x^2 - 2
x^2 + 1/x^2 = 11
For x^4 + 1/x^4:
Square x^2 + 1/x^2 = 11:
(x^2 + 1/x^2)^2 = x^4 + 2 + 1/x^4
11^2 = x^4 + 1/x^4 + 2
121 = x^4 + 1/x^4 + 2
x^4 + 1/x^4 = 119
Answer: x^2 + 1/x^2 = 11, x^4 + 1/x^4 = 119
Q14. Expand using identity: (2a + 3b + 4c)^2
Using (a + b + c)^2 = a^2 + b^2 + c^2 + 2ab + 2bc + 2ca:
(2a + 3b + 4c)^2 = (2a)^2 + (3b)^2 + (4c)^2 + 2(2a)(3b) + 2(3b)(4c) + 2(4c)(2a)
= 4a^2 + 9b^2 + 16c^2 + 12ab + 24bc + 16ca
Answer: 4a^2 + 9b^2 + 16c^2 + 12ab + 24bc + 16ca
Q15. If a^2 + b^2 = 41 and ab = 4, find (a + b)^2 and (a - b)^2.
(a + b)^2 = a^2 + 2ab + b^2 = (a^2 + b^2) + 2ab = 41 + 2(4) = 41 + 8 = 49
(a - b)^2 = a^2 - 2ab + b^2 = (a^2 + b^2) - 2ab = 41 - 2(4) = 41 - 8 = 33
Answer: (a + b)^2 = 49, (a - b)^2 = 33
Q16. Evaluate: 5.3^2 - 4.7^2
Using a^2 - b^2 = (a + b)(a - b):
a = 5.3, b = 4.7
= (5.3 + 4.7)(5.3 - 4.7)
= 10.0 * 0.6
= 6.0
Answer: 6.0
Q17. Evaluate using identity: 10.5 * 9.5
10.5 * 9.5 = (10 + 0.5)(10 - 0.5)
Using a^2 - b^2 = (a + b)(a - b):
= 10^2 - (0.5)^2
= 100 - 0.25
= 99.75
Answer: 99.75
Q18. Simplify: (3x + 2y)^2 + (3x - 2y)^2
Using (a + b)^2 + (a - b)^2 = 2(a^2 + b^2):
where a = 3x, b = 2y
= 2[(3x)^2 + (2y)^2]
= 2[9x^2 + 4y^2]
= 18x^2 + 8y^2
Answer: 18x^2 + 8y^2
Q19. Simplify: (3x + 2y)^2 - (3x - 2y)^2
Using (a + b)^2 - (a - b)^2 = 4ab:
where a = 3x, b = 2y
= 4(3x)(2y) = 24xy
Answer: 24xy
Q20. If a + b + c = 9 and ab + bc + ca = 26, find a^2 + b^2 + c^2.
Using (a + b + c)^2 = a^2 + b^2 + c^2 + 2(ab + bc + ca):
9^2 = a^2 + b^2 + c^2 + 2(26)
81 = a^2 + b^2 + c^2 + 52
a^2 + b^2 + c^2 = 81 - 52
a^2 + b^2 + c^2 = 29
Answer: 29
Q21. Evaluate: 999^2
999^2 = (1000 - 1)^2
Using (a - b)^2 = a^2 - 2ab + b^2:
= 1000^2 - 2(1000)(1) + 1^2
= 1000000 - 2000 + 1
= 998001
Answer: 998001
Q22. If a^2 + b^2 + c^2 = 50 and ab + bc + ca = 14, find (a + b + c)^2.
Using (a + b + c)^2 = a^2 + b^2 + c^2 + 2(ab + bc + ca):
(a + b + c)^2 = 50 + 2(14) = 50 + 28 = 78
Answer: 78
Exercise 3.5 — Linear Equations in Two Variables
Q1. Express y = 2x + 3 in standard form (ax + by + c = 0).
y = 2x + 3
=> 2x - y + 3 = 0
Standard form: 2x - y + 3 = 0
Here a = 2, b = -1, c = 3
Q2. Find three solutions of the equation 2x + y = 7.
2x + y = 7 => y = 7 - 2x
Substituting different values of x:
When x = 0: y = 7 - 0 = 7. Solution: (0, 7)
When x = 1: y = 7 - 2 = 5. Solution: (1, 5)
When x = 2: y = 7 - 4 = 3. Solution: (2, 3)
Verification:
(0, 7): 2(0) + 7 = 7 ✓
(1, 5): 2(1) + 5 = 7 ✓
(2, 3): 2(2) + 3 = 7 ✓
Answer: (0, 7), (1, 5), (2, 3)
Q3. Find three solutions of the equation x - 2y = 4.
x - 2y = 4 => x = 4 + 2y
When y = 0: x = 4 + 0 = 4. Solution: (4, 0)
When y = -1: x = 4 + 2(-1) = 2. Solution: (2, -1)
When y = 1: x = 4 + 2(1) = 6. Solution: (6, 1)
Answer: (4, 0), (2, -1), (6, 1)
Q4. Check whether (2, 3) is a solution of 3x - 2y = 0.
LHS = 3(2) - 2(3) = 6 - 6 = 0
RHS = 0
Since LHS = RHS, (2, 3) IS a solution of 3x - 2y = 0.
Q5. Check whether (1, 2) is a solution of x + 2y = 5.
LHS = 1 + 2(2) = 1 + 4 = 5
RHS = 5
Since LHS = RHS, (1, 2) IS a solution of x + 2y = 5.
Q6. If x = 3 is a solution of 5x - 2y = 7, find y.
Substitute x = 3 in 5x - 2y = 7:
5(3) - 2y = 7
15 - 2y = 7
2y = 8
y = 4
Answer: y = 4. The solution is (3, 4).
Q7. Draw the graph of y = x + 2.
y = x + 2
Table of values:
x | y = x + 2
---|----------
-1 | 1
0 | 2
1 | 3
2 | 4
Plot the points (-1, 1), (0, 2), (1, 3), (2, 4) and draw a straight line through them.
The line cuts the y-axis at (0, 2) [y-intercept = 2]
The line cuts the x-axis at (-2, 0) [x-intercept: set y = 0, x = -2]
Q8. Draw the graph of 2x + y = 6. Find where it crosses the x-axis and y-axis.
2x + y = 6 => y = 6 - 2x
Table of values:
x | y = 6 - 2x
---|----------
0 | 6
1 | 4
2 | 2
3 | 0
x-intercept (where y = 0): 2x + 0 = 6 => x = 3. Point: (3, 0)
y-intercept (where x = 0): 0 + y = 6 => y = 6. Point: (0, 6)
Plot (0, 6), (1, 4), (2, 2), (3, 0) and draw a straight line.
The line crosses the x-axis at (3, 0) and y-axis at (0, 6).
Q9. Draw the graph of x - y = 1.
x - y = 1 => y = x - 1
Table of values:
x | y = x - 1
---|----------
-1 | -2
0 | -1
1 | 0
2 | 1
3 | 2
Plot the points and draw a straight line.
x-intercept: (1, 0), y-intercept: (0, -1)
Q10. Solve graphically: x + y = 5 and x - y = 1.
For x + y = 5:
x | y
---|----
0 | 5
5 | 0
For x - y = 1:
x | y
---|----
0 | -1
1 | 0
Plot both lines on the same graph.
The intersection point satisfies both equations:
x + y = 5 ... (1)
x - y = 1 ... (2)
Adding (1) and (2): 2x = 6 => x = 3
Substituting in (1): 3 + y = 5 => y = 2
Intersection point: (3, 2)
Answer: x = 3, y = 2
Q11. Find the value of k if (2, k) is a solution of 3x - y = 5.
Substitute x = 2 in 3x - y = 5:
3(2) - k = 5
6 - k = 5
k = 1
Answer: k = 1
Q12. How many solutions does the equation 2x + 3y = 12 have?
A linear equation in two variables has infinitely many solutions.
Each pair (x, y) that satisfies the equation is a solution.
For example:
(0, 4), (3, 2), (6, 0), (-3, 6), etc. — infinitely many.
Answer: Infinitely many solutions.
Q13. Write the equation of a line passing through (0, 5) with slope 2.
Using the point-slope form: y - y_1 = m(x - x_1)
where (x_1, y_1) = (0, 5) and m = 2:
y - 5 = 2(x - 0)
y - 5 = 2x
2x - y + 5 = 0
Answer: 2x - y + 5 = 0 (or y = 2x + 5)
Q14. Find four solutions of the equation pi * x + y = 12.
pi * x + y = 12 => y = 12 - pi * x
When x = 0: y = 12. Solution: (0, 12)
When x = 1: y = 12 - pi. Solution: (1, 12 - pi)
When x = 2: y = 12 - 2pi. Solution: (2, 12 - 2pi)
When x = -1: y = 12 + pi. Solution: (-1, 12 + pi)
Answer: (0, 12), (1, 12 - pi), (2, 12 - 2pi), (-1, 12 + pi)
Q15. The cost of 3 pens and 4 pencils is Rs 50. Express this as a linear equation in two variables and find two solutions.
Let cost of a pen = x, cost of a pencil = y.
Equation: 3x + 4y = 50
Find two solutions:
When x = 10: 3(10) + 4y = 50 => 4y = 20 => y = 5. Solution: (10, 5)
When x = 2: 3(2) + 4y = 50 => 4y = 44 => y = 11. Solution: (2, 11)
Answer: 3x + 4y = 50; two solutions: (10, 5) and (2, 11)
Q16. Draw the graph of the equation y = 3x. What is special about this line?
y = 3x
Table of values:
x | y = 3x
---|--------
-1 | -3
0 | 0
1 | 3
2 | 6
This line passes through the origin (0, 0).
Any equation of the form y = mx (no constant term) passes through the origin.
Answer: The line y = 3x passes through the origin.
Q17. Solve: x + y = 7 and 2x - y = 2 graphically.
For x + y = 7:
x | y
---|----
0 | 7
7 | 0
For 2x - y = 2:
x | y
---|----
0 | -2
1 | 0
Algebraic verification:
x + y = 7 ... (1)
2x - y = 2 ... (2)
Adding: 3x = 9 => x = 3
From (1): 3 + y = 7 => y = 4
Intersection point: (3, 4)
Answer: x = 3, y = 4
Q18. The taxi fare in a city is Rs 10 for the first km and Rs 6 for each additional km. Express this as a linear equation.
Total fare for d km:
Fare = 10 + 6(d - 1) for d > 1
= 10 + 6d - 6
= 6d + 4
Let y = total fare, x = distance in km.
y = 6x + 4
Standard form: 6x - y + 4 = 0
Verification:
For 2 km: y = 6(2) + 4 = 16. Actual: 10 + 6(1) = 16 ✓
For 5 km: y = 6(5) + 4 = 34. Actual: 10 + 6(4) = 34 ✓
Answer: y = 6x + 4 (for x >= 1)
Q19. If the point (3, k) lies on the line 2x + 3y = 15, find k.
Substitute x = 3:
2(3) + 3k = 15
6 + 3k = 15
3k = 9
k = 3
Answer: k = 3
Q20. Find the coordinates where the line 4x - 3y = 12 intersects the axes.
x-intercept (y = 0): 4x - 3(0) = 12 => 4x = 12 => x = 3. Point: (3, 0)
y-intercept (x = 0): 4(0) - 3y = 12 => -3y = 12 => y = -4. Point: (0, -4)
Answer: x-intercept: (3, 0), y-intercept: (0, -4)
Exercise 3.6 — Word Problems and Applications
Q1. The sum of two numbers is 25 and their difference is 7. Find the numbers.
Let the two numbers be x and y.
x + y = 25 ... (1)
x - y = 7 ... (2)
Adding (1) and (2): 2x = 32 => x = 16
Subtracting (2) from (1): 2y = 18 => y = 9
Verification: 16 + 9 = 25 ✓, 16 - 9 = 7 ✓
Answer: The numbers are 16 and 9.
Q2. The perimeter of a rectangle is 44 cm and its length is 2 cm more than its breadth. Find the dimensions.
Let breadth = b cm, then length = (b + 2) cm.
Perimeter = 2(length + breadth) = 44
2(b + 2 + b) = 44
2(2b + 2) = 44
4b + 4 = 44
4b = 40
b = 10
Length = 10 + 2 = 12 cm
Verification: 2(12 + 10) = 2(22) = 44 ✓
Answer: Length = 12 cm, Breadth = 10 cm
Q3. Five years ago, a father was 3 times as old as his son. Now their total age is 60 years. Find their present ages.
Let father's present age = F, son's present age = S.
Five years ago: F - 5 = 3(S - 5) ... (1)
Present total: F + S = 60 ... (2)
From (1): F - 5 = 3S - 15 => F = 3S - 10
Substituting in (2):
3S - 10 + S = 60
4S = 70
S = 17.5
F = 60 - 17.5 = 42.5
Verification: 5 years ago, father was 37.5, son was 12.5.
37.5 / 12.5 = 3 ✓
Answer: Father = 42.5 years, Son = 17.5 years
Q4. A number is 4 times another. Their sum is 45. Find the numbers.
Let the smaller number = x, then the larger number = 4x.
x + 4x = 45
5x = 45
x = 9
Larger number = 4 * 9 = 36
Verification: 36 + 9 = 45 ✓, 36 = 4(9) ✓
Answer: The numbers are 36 and 9.
Q5. The sum of digits of a two-digit number is 9. If 27 is added, the digits are reversed. Find the number.
Let the tens digit = x and the units digit = y.
Number = 10x + y
Sum of digits: x + y = 9 ... (1)
After adding 27, digits reverse: 10x + y + 27 = 10y + x ... (2)
From (2): 10x + y + 27 = 10y + x
9x - 9y = -27
x - y = -3 ... (2')
From (1): x + y = 9
From (2'): x - y = -3
Adding: 2x = 6 => x = 3
From (1): 3 + y = 9 => y = 6
Number = 10(3) + 6 = 36
Verification: 36 + 27 = 63 (digits reversed) ✓
Answer: The number is 36.
Q6. The area of a rectangle is 120 sq cm. If the length is 2 cm more than the breadth, find the dimensions.
Let breadth = b cm, then length = (b + 2) cm.
Area = length * breadth = 120
(b + 2)(b) = 120
b^2 + 2b - 120 = 0
Factorise: Find two numbers whose sum = 2 and product = -120.
Numbers: 12 and -10.
b^2 + 12b - 10b - 120 = 0
b(b + 12) - 10(b + 12) = 0
(b + 12)(b - 10) = 0
b = -12 (rejected, breadth cannot be negative)
b = 10
Length = 10 + 2 = 12 cm
Verification: 12 * 10 = 120 sq cm ✓
Answer: Length = 12 cm, Breadth = 10 cm
Q7. Divide Rs 2000 between two people such that one gets Rs 200 more than the other.
Let the smaller share = x, then the larger share = x + 200.
x + (x + 200) = 2000
2x + 200 = 2000
2x = 1800
x = 900
Larger share = 900 + 200 = 1100
Verification: 900 + 1100 = 2000 ✓, 1100 - 900 = 200 ✓
Answer: Rs 1100 and Rs 900.
Q8. A train travels 300 km at a uniform speed. If the speed had been 5 km/hr more, the journey would have taken 2 hours less. Find the speed of the train.
Let the speed of the train = s km/hr.
Time taken = 300/s hours.
If speed = s + 5: Time = 300/(s + 5)
300/s - 300/(s + 5) = 2
300(s + 5) - 300s / [s(s + 5)] = 2
300s + 1500 - 300s = 2s(s + 5)
1500 = 2s^2 + 10s
2s^2 + 10s - 1500 = 0
s^2 + 5s - 750 = 0
Factorise: Find two numbers whose sum = 5 and product = -750.
Numbers: 30 and -25.
(s + 30)(s - 25) = 0
s = -30 (rejected) or s = 25
Answer: Speed = 25 km/hr.
Q9. The numerator of a fraction is 3 less than the denominator. If the numerator and denominator are both increased by 2, the fraction becomes 3/4. Find the fraction.
Let denominator = d, then numerator = d - 3.
Original fraction = (d - 3)/d.
After increasing both by 2:
(d - 3 + 2)/(d + 2) = 3/4
(d - 1)/(d + 2) = 3/4
Cross multiply:
4(d - 1) = 3(d + 2)
4d - 4 = 3d + 6
d = 10
Numerator = 10 - 3 = 7
Original fraction = 7/10
Verification: (7 + 2)/(10 + 2) = 9/12 = 3/4 ✓
Answer: The fraction is 7/10.
Q10. Factorise: x^2 + 14x + 49 and hence find the value of (3p)^2 + 14(3p) + 49 when p = 2.
Factorise: x^2 + 14x + 49
= (x + 7)^2 [since 49 = 7^2 and 14 = 2 * 7]
Now substitute x = 3p and p = 2:
(3p + 7)^2 = (3(2) + 7)^2 = (6 + 7)^2 = 13^2 = 169
Answer: (x + 7)^2, and the value when p = 2 is 169.
Q11. Using suitable identity, find the value of (2.3)^2 + (0.7)^2 + 2(2.3)(0.7).
Recognise: a^2 + b^2 + 2ab = (a + b)^2 where a = 2.3, b = 0.7
= (2.3 + 0.7)^2
= 3.0^2
= 9
Answer: 9
Q12. If the difference of squares of two consecutive integers is 17, find the integers.
Let the two consecutive integers be n and n + 1.
Difference of squares: (n + 1)^2 - n^2 = 17
Using a^2 - b^2 = (a + b)(a - b):
= [(n + 1) + n][(n + 1) - n] = 17
= (2n + 1)(1) = 17
2n + 1 = 17
2n = 16
n = 8
The integers are 8 and 9.
Verification: 9^2 - 8^2 = 81 - 64 = 17 ✓
Answer: 8 and 9
Q13. Simplify: (a^2 - b^2)(a^2 + b^2)(a^4 + b^4)
Step 1: (a^2 - b^2)(a^2 + b^2) = a^4 - b^4 [using a^2 - b^2]
Step 2: (a^4 - b^4)(a^4 + b^4) = a^8 - b^8 [using a^2 - b^2 again]
Answer: a^8 - b^8
Q14. A farmer has 60 animals (hens and cows). The total number of legs is 180. How many hens and cows does he have?
Let number of hens = h, number of cows = c.
h + c = 60 ... (1) [total animals]
2h + 4c = 180 ... (2) [total legs: hens have 2 legs, cows have 4]
From (1): h = 60 - c
Substituting in (2):
2(60 - c) + 4c = 180
120 - 2c + 4c = 180
2c = 60
c = 30
h = 60 - 30 = 30
Verification: 30 + 30 = 60 ✓, 2(30) + 4(30) = 60 + 120 = 180 ✓
Answer: 30 hens and 30 cows.
Q15. The cost of 2 pencils and 3 pens is Rs 34. The cost of 4 pencils and 5 pens is Rs 62. Find the cost of each.
Let cost of a pencil = p, cost of a pen = q.
2p + 3q = 34 ... (1)
4p + 5q = 62 ... (2)
Multiply (1) by 2: 4p + 6q = 68 ... (3)
Subtract (2) from (3): q = 6
Substituting q = 6 in (1):
2p + 3(6) = 34
2p + 18 = 34
2p = 16
p = 8
Verification: 2(8) + 3(6) = 16 + 18 = 34 ✓, 4(8) + 5(6) = 32 + 30 = 62 ✓
Answer: Pencil = Rs 8, Pen = Rs 6.
Q16. Factorise: x^3 - 23x^2 + 142x - 120 given that (x - 1) is a factor.
Since (x - 1) is a factor, divide p(x) by (x - 1).
Using synthetic division with root 1:
1 | 1 -23 142 -120
| 1 -22 120
|-------------------------
1 -22 120 0
So p(x) = (x - 1)(x^2 - 22x + 120)
Factorise x^2 - 22x + 120:
Find two numbers: sum = 22, product = 120.
Numbers: 10 and 12.
x^2 - 22x + 120 = x^2 - 10x - 12x + 120
= x(x - 10) - 12(x - 10)
= (x - 10)(x - 12)
So p(x) = (x - 1)(x - 10)(x - 12)
Answer: (x - 1)(x - 10)(x - 12)
Q17. Evaluate using a suitable identity: 10.2 * 9.8
10.2 * 9.8 = (10 + 0.2)(10 - 0.2)
Using a^2 - b^2 = (a + b)(a - b):
= 10^2 - (0.2)^2
= 100 - 0.04
= 99.96
Answer: 99.96
Q18. If (x + y) = 12 and xy = 32, find x^2 + y^2.
x^2 + y^2 = (x + y)^2 - 2xy
= 12^2 - 2(32)
= 144 - 64
= 80
Answer: x^2 + y^2 = 80
Q19. The length of a rectangular garden is 3 m more than its breadth. The area is 180 sq m. Find the dimensions.
Let breadth = b m, then length = (b + 3) m.
Area = length * breadth
(b + 3)(b) = 180
b^2 + 3b - 180 = 0
Factorise: Find two numbers whose sum = 3 and product = -180.
Numbers: 15 and -12.
b^2 + 15b - 12b - 180 = 0
b(b + 15) - 12(b + 15) = 0
(b + 15)(b - 12) = 0
b = -15 (rejected) or b = 12
Length = 12 + 3 = 15 m
Verification: 15 * 12 = 180 ✓
Answer: Length = 15 m, Breadth = 12 m
Q20. Find two numbers such that their sum is 15 and product is 56.
Let the numbers be x and y.
x + y = 15 ... (1)
xy = 56 ... (2)
From (1): y = 15 - x
Substituting in (2): x(15 - x) = 56
15x - x^2 = 56
x^2 - 15x + 56 = 0
Factorise: Find two numbers whose sum = 15 and product = 56.
Numbers: 7 and 8.
(x - 7)(x - 8) = 0
x = 7 or x = 8
If x = 7, y = 8. If x = 8, y = 7.
Verification: 7 + 8 = 15 ✓, 7 * 8 = 56 ✓
Answer: The numbers are 7 and 8.
Common Mistakes
1. Sign errors in Remainder/Factor Theorem: When evaluating p(-a), students often make sign errors. For p(x) = x^3 - 2x^2 + x - 5, p(-2) = (-2)^3 - 2(-2)^2 + (-2) - 5 = -8 - 8 - 2 - 5 = -23. Always compute (-a)^n carefully: (-a)^2 = a^2 but (-a)^3 = -a^3.
2. Confusing degree of zero polynomial: The zero polynomial p(x) = 0 does not have degree 0. Its degree is undefined (sometimes taken as -1). A constant polynomial like p(x) = 7 has degree 0. Do not mix these up.
3. Wrong identity selection: Students write (a + b)^3 = a^3 + b^3 (missing the middle terms). Remember the full expansion: (a + b)^3 = a^3 + 3a^2b + 3ab^2 + b^3. Only a^3 + b^3 = (a + b)(a^2 - ab + b^2) is a separate identity.
4. Splitting the middle term incorrectly: For 2x^2 + 7x + 3, the two numbers must have sum = 7 AND product = 2 3 = 6. Students often forget to multiply the leading coefficient by the constant when finding the product condition. The numbers 6 and 1 work (6 + 1 = 7, 6 1 = 6).
5. Forgetting the coefficient in a^2 - b^2: 9x^2 - 16 = (3x)^2 - 4^2 = (3x + 4)(3x - 4), NOT (9x + 16)(9x - 16). Always take the square root of each coefficient.
6. Missing solutions in two-variable equations: A linear equation in two variables has infinitely many solutions, not just one. Every point on the line is a solution. For 2x + y = 6, (0, 6), (1, 4), (3, 0), (-1, 8) are all valid solutions.
7. Graph plotting with insufficient points: Always find at least 2 points (preferably 3) before drawing a line. Using only 1 point or the origin without a second point leads to incorrect lines. Also verify that all computed points satisfy the original equation.
Exam Checklist
- [ ] I can identify whether an expression is a polynomial (no negative or fractional exponents)
- [ ] I can state the degree, leading coefficient, and type (linear/quadratic/cubic) of any polynomial
- [ ] I can evaluate a polynomial at a given value of x by substitution
- [ ] I can find zeros of a quadratic polynomial by factorisation
- [ ] I can apply the Remainder Theorem to find remainders without long division
- [ ] I can use the Factor Theorem to check if (x - a) is a factor of p(x)
- [ ] I can use the Factor Theorem to find unknown constants (like k in p(x) = x^3 + kx^2 + ...)
- [ ] I can factorise quadratic expressions by splitting the middle term
- [ ] I can factorise expressions using all standard identities: (a+b)^2, (a-b)^2, a^2-b^2, (a+b)^3, (a-b)^3, a^3+b^3, a^3-b^3
- [ ] I can evaluate numerical expressions like 102^2, 99^3, 103*97 using appropriate identities
- [ ] I can find multiple solutions of a linear equation in two variables
- [ ] I can solve pairs of linear equations graphically and algebraically
- [ ] I can set up and solve word problems involving linear equations
Internal Links
- [Class 9 Maths Home](/class/9/maths) — All chapters and resources
- [Class 9 Maths Algebra Chapter Guide](/class/9/maths/algebra) — Concept explanations and examples
- [Class 9 Maths Geometry Solutions](/class/9/maths/solutions/geometry-exercise-solutions) — Next chapter
- [Class 8 Maths Algebra Solutions](/class/8/maths/solutions/algebra-exercise-solutions) — Previous year foundation
- [TN State Board Class 9 Home](/class/9?board=tamil-nadu-state-board) — All subjects
- [KalviGavani Home](/) — Classes 6-12 resources
Frequently Asked Questions
If a polynomial p(x) is divided by (x - a), the remainder is p(a). This allows us to find the remainder without performing long division. For example, if p(x) = x^2 + 3x + 2 is divided by (x - 1), the remainder is p(1) = 1 + 3 + 2 = 6.