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Class 9 Mathematics Notes – Chapter 4: Factorization and Algebraic Manipulation

Punjab Board | 2026

Exercise 4.1

Q1. Factorize by identifying common factors.

(i) \(6x+12\)

Answer:

\[ 6x+12=6(x+2) \]

(ii) \(15y^2+20y\)

Answer:

\[ 15y^2+20y=5y(3y+4) \]

(iii) \(-12x^2-3x\)

Answer:

\[ -12x^2-3x=-3x(4x+1) \]

(iv) \(4a^2b+8ab^2\)

Answer:

\[ 4a^2b+8ab^2=4ab(a+2b) \]

(v) \(xy-3x^2+2x\)

Answer:

\[ xy-3x^2+2x=x(y-3x+2) \]

(vi) \(3a^2b-9ab^2+15ab\)

Answer:

\[ 3a^2b-9ab^2+15ab=3ab(a-3b+5) \]

Q2. Factorize and represent pictorially.

(i) \(5x+15\)

Answer:

\[ 5x+15=5(x+3) \]

(ii) \(x^2+4x+3\)

Answer:

\[ x^2+4x+3=(x+1)(x+3) \]

(iii) \(x^2+6x+8\)

Answer:

\[ x^2+6x+8=(x+2)(x+4) \]

(iv) \(x^2+4x+4\)

Answer:

\[ x^2+4x+4=(x+2)^2 \]

Q3. Factorize.

(i) \(x^2+x-12\)

Solution:

\[ \begin{aligned} &=x^2+4x-3x-12\\ &=x(x+4)-3(x+4)\\ &=\mathbf{(x+4)(x-3)} \end{aligned} \]

(ii) \(x^2+7x+10\)

Answer:

\[ x^2+7x+10=(x+5)(x+2) \]

(iii) \(x^2-6x+8\)

Answer:

\[ x^2-6x+8=(x-4)(x-2) \]

(iv) \(x^2-x-56\)

Answer:

\[ x^2-x-56=(x-8)(x+7) \]

(v) \(x^2-10x-24\)

Answer:

\[ x^2-10x-24=(x-12)(x+2) \]

(vi) \(y^2+4y-12\)

Answer:

\[ y^2+4y-12=(y+6)(y-2) \]

(vii) \(y^2+13y+36\)

Answer:

\[ y^2+13y+36=(y+9)(y+4) \]

(viii) \(x^2-x-2\)

Answer:

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

Q4. Factorize.

(i) \(2x^2+7x+3\)

Solution:

\[ \begin{aligned} &=2x^2+6x+x+3\\ &=2x(x+3)+(x+3)\\ &=\mathbf{(x+3)(2x+1)} \end{aligned} \]

(ii) \(2x^2+11x+15\)

Solution:

\[ \begin{aligned} &=2x^2+6x+5x+15\\ &=2x(x+3)+5(x+3)\\ &=\mathbf{(x+3)(2x+5)} \end{aligned} \]

(iii) \(4x^2+13x+3\)

Solution:

\[ \begin{aligned} &=4x^2+12x+x+3\\ &=4x(x+3)+(x+3)\\ &=\mathbf{(x+3)(4x+1)} \end{aligned} \]

(iv) \(3x^2+5x+2\)

Answer:

\[ 3x^2+5x+2=(3x+2)(x+1) \]

(v) \(3y^2-11y+6\)

Answer:

\[ 3y^2-11y+6=(3y-2)(y-3) \]

(vi) \(2y^2-5y+2\)

Answer:

\[ 2y^2-5y+2=(2y-1)(y-2) \]

(vii) \(4z^2-11z+6\)

Answer:

\[ 4z^2-11z+6=(4z-3)(z-2) \]

(viii) \(6+7x-3x^2\)

Answer:

\[ 6+7x-3x^2=(x+3)(2-3x) \]

Exercise 4.2

Q1. Factorize each of the following expressions.

(i) \(4x^4+81y^4\)

Solution:

\[ \begin{aligned} &=(2x^2+9y^2)^2-(6xy)^2\\ &=\mathbf{(2x^2-6xy+9y^2)(2x^2+6xy+9y^2)} \end{aligned} \]

(ii) \(a^4+64b^4\)

Solution:

\[ \begin{aligned} &=(a^2+8b^2)^2-(4ab)^2\\ &=\mathbf{(a^2-4ab+8b^2)(a^2+4ab+8b^2)} \end{aligned} \]

(iii) \(x^4+4x^2+16\)

Solution:

\[ \begin{aligned} &=(x^2+4)^2-(2x)^2\\ &=\mathbf{(x^2-2x+4)(x^2+2x+4)} \end{aligned} \]

(iv) \(x^4-14x^2+1\)

Answer as supplied:

\[ (x^2-2\sqrt{3}x-1)(x^2+2\sqrt{3}x-1) \]

Note: The supplied factorization does not expand to the given polynomial. The correct factorization over the real numbers is:

\[ (x^2-2\sqrt{3}x+1)(x^2+2\sqrt{3}x+1) \]

because the product is \(x^4-10x^2+1\), so this also does not match the original expression. The expression \(x^4-14x^2+1\) requires a different factorization; please verify the textbook question or supplied answer.

(v) \(x^4-30x^2y^2+9y^4\)

Answer as supplied:

\[ (x^2-3xy-3y^2)(x^2+3xy-3y^2) \]

(vi) \(x^4+11x^2y^2+y^4\)

Answer as supplied:

\[ (x^2-3xy+y^2)(x^2+3xy+y^2) \]

Q2. Factorize each of the following expressions.

(i) \((x+1)(x+2)(x+3)(x+4)+1\)

Solution:

\[ \begin{aligned} &=(x^2+5x+4)(x^2+5x+6)+1\\ &=\mathbf{(x^2+5x+5)^2} \end{aligned} \]

(ii) \((x+2)(x-7)(x-4)(x-1)+17\)

Answer as supplied:

\[ (x^2-5x-5)(x^2-5x-3) \]

(iii) \((2x^2+7x+3)(2x^2+7x+5)+1\)

Answer:

\[ (2x^2+7x+4)^2 \]

(iv) \((3x^2+5x+3)(3x^2+5x+5)-3\)

Answer:

\[ (3x^2+5x+6)(3x^2+5x+2) \]

(v) \((x+1)(x+2)(x+3)(x+6)-3x^2\)

Answer:

\[ (x^2+4x+6)(x^2+4x+3) \]

(vi) \((x+1)(x-1)(x+2)(x-2)+13x^2\)

Answer:

\[ (x^2+3x+2)(x^2-3x+2) \]

Q3. Factorize.

(i) \(8x^3+12x^2+6x+1\)

Answer:

\[ (2x+1)^3 \]

(ii) \(27a^3+108a^2b+144ab^2+64b^3\)

Answer:

\[ (3a+4b)^3 \]

(iii) \(x^3+48x^2y+108xy^2+216y^3\)

Answer as supplied:

\[ (x+6y)^3 \]

Note: Expanding \((x+6y)^3\) gives \(x^3+18x^2y+108xy^2+216y^3\), so the coefficient of \(x^2y\) does not match the supplied expression.

(iv) \(8x^3-125y^3+150xy^2-60x^2y\)

Answer:

\[ (2x-5y)^3 \]

Q4. Factorize.

(i) \(125a^3-1\)

Answer:

\[ (5a-1)(25a^2+5a+1) \]

(ii) \(64x^3+125\)

Answer:

\[ (4x+5)(16x^2-20x+25) \]

(iii) \(x^6-27\)

Answer:

\[ (x^2-3)(x^4+3x^2+9) \]

(iv) \(1000a^3+1\)

Answer:

\[ (10a+1)(100a^2-10a+1) \]

(v) \(343x^3+216\)

Answer:

\[ (7x+6)(49x^2-42x+36) \]

(vi) \(27-512y^3\)

Answer:

\[ (3-8y)(9+24y+64y^2) \]

Exercise 4.3

Q1. Find the HCF by the factorization method.

(i) \(21x^2y,\ 35xy^2\)

Answer:

\[ \boxed{7xy} \]

(ii) \(4x^2-9y^2,\ 2x^2-3xy\)

Answer as supplied:

\[ 2x-3y \]

(iii) \(x^3-1,\ x^2+x+1\)

Answer:

\[ x^2+x+1 \]

(iv) \(a^3+2a^2-3a,\ 2a^3+5a^2-3a\)

Answer:

\[ a(a+3) \]

(v) \(t^2+3t-4,\ t^2+5t+4,\ t^2-1\)

Answer:

\[ t+1 \]

(vi) \(x^2+15x+56,\ x^2+5x-24,\ x^2+8x\)

Answer:

\[ x+8 \]

Q2. Find the HCF of the following expressions by using the division method.

(i) \(27x^3+9x^2-3x-9,\ 3x-2\)

Answer as supplied:

\[ 3x-2 \]

(ii) \(x^3-9x^2+21x-15,\ x^2-4x+3\)

Answer:

\[ x-3 \]

(iii) \(2x^3+2x^2+2x+2,\ 6x^3+12x^2+6x+12\)

Answer as supplied:

\[ x^2+1 \]

(iv) \(2x^3-4x^2+6x,\ x^3-2x,\ 3x^2-6x\)

Answer as supplied:

\[ x(x-2) \]

Q3. Find the LCM of the following expressions by using the prime factorization method.

(i) \(2a^2b,\ 4ab^2,\ 6ab\)

Answer:

\[ \boxed{12a^2b^2} \]

(ii) \(x^2+x,\ x^3+x^2\)

Answer:

\[ \boxed{x^2(x+1)} \]

(iii) \(a^2-4a+4,\ a^2-2a\)

Answer:

\[ \boxed{a(a-2)^2} \]

(iv) \(x^4-16,\ x^3-4x\)

Answer:

\[ \boxed{x(x-2)(x+2)(x^2+4)} \]

(v) \(16-4x^2,\ x^2+x-6,\ 4-x^2\)

Answer:

\[ \boxed{4(2-x)(2+x)(x+3)} \]

Q4. The HCF of two polynomials is \(y-7\) and their LCM is \(y^3-10y^2+11y+70\). If one of the polynomials is \(y^2-5y-14\), find the other.

Solution:

The relation between two polynomials is:

\[ \text{Product of polynomials}=\text{HCF}\times\text{LCM} \]

Therefore,

\[ \text{Other polynomial} =\frac{\text{LCM}\times\text{HCF}}{\text{Given polynomial}} \]

\[ =\frac{(y^3-10y^2+11y+70)(y-7)} {(y-7)(y+2)} \]

\[ =\frac{y^3-10y^2+11y+70}{y+2} \]

Answer as supplied:

\[ y^2-12y+35 \]

Note: The division shown does not yield the supplied answer. The polynomial \(y^3-10y^2+11y+70\) is not divisible by \(y+2\) without a remainder, so the question or its given LCM should be checked.

Q5. The LCM and HCF of two polynomials \(p(x)\) and \(q(x)\) are \(36x^3(x+a)(x^3-a^3)\) and \(x^2(x-a)\), respectively. If \(p(x)=4x^2(x^2-a^2)\), find \(q(x)\).

Solution:

\[ q(x)=\frac{\text{LCM}\times\text{HCF}}{p(x)} \]

Answer as supplied:

\[ \boxed{9x^3(x^2+ax+a^2)} \]

Q6. The HCF and LCM of two polynomials are \(x+a\) and \(12x^2(x+a)(x^2-a^2)\), respectively. Find the product of the two polynomials.

Solution:

\[ \text{Product}=\text{LCM}\times\text{HCF} \]

\[ =(x+a)\times12x^2(x+a)(x^2-a^2) \]

Answer:

\[ \boxed{12x^2(x+a)^2(x^2-a^2)} \]

Exercise 4.4

Q1. Find the square root of the following polynomials by the factorization method.

(i) \(x^2-8x+16\)

Answer:

\[ \pm(x-4) \]

(ii) \(9x^2+12x+4\)

Answer:

\[ \pm(3x+2) \]

(iii) \(36a^2+84a+49\)

Answer:

\[ \pm(6a+7) \]

(iv) \(64y^2-32y+4\)

Answer:

\[ \pm2(4y-1) \]

(v) \(200t^2-120t+18\)

Answer:

\[ \pm2(10t-3) \]

(vi) \(40x^2+120x+90\)

Answer as supplied:

\[ \pm\sqrt{10}(2x+3) \]

Q2. Find the square root of the following polynomials by the division method.

(i) \(4x^4-28x^3+37x^2+42x+9\)

Answer:

\[ \pm(2x^2-7x-3) \]

(ii) \(121x^4-198x^3-183x^2+216x+144\)

Answer:

\[ \pm(11x^2-9x-12) \]

(iii) \(x^4-10x^3+27x^2y^2-10xy^3+y^4\)

Answer as supplied:

\[ \pm(x^2-5xy+y^2) \]

(iv) \(4x^4-12x^3+37x^2-42x+49\)

Answer:

\[ \pm(2x^2-3x+7) \]

Q3. An investor’s return \(R(x)\) in rupees after investing \(x\) thousand rupees is given by \(R(x)=-x^2+6x-8\). Find the investment levels that result in zero return.

Solution:

\[ -x^2+6x-8=0 \]

\[ -(x^2-6x+8)=0 \]

\[ -(x-4)(x-2)=0 \]

Therefore,

\[ x=2\quad\text{or}\quad x=4 \]

Answer: Rs. 2,000 or Rs. 4,000.

Q4. A company’s profit \(P(x)\) in rupees from selling \(x\) units is \(P(x)=x^3-15x^2+75x-125\). Find the break-even point(s).

Solution:

\[ x^3-15x^2+75x-125=0 \]

\[ (x-5)^3=0 \]

\[ x=5 \]

Answer: 5 units.

Q5. The potential energy \(V(x)\) is given by \(V(x)=2x^3-6x^2+4x\). Determine where potential energy is zero.

Solution:

\[ 2x^3-6x^2+4x=0 \]

\[ 2x(x^2-3x+2)=0 \]

\[ 2x(x-1)(x-2)=0 \]

Answer:

\[ \boxed{x=0,\ 1,\ 2} \]

Q6. The deflection \(Y(x)\) of a beam is \(Y(x)=2x^2-8x+6\). Find the points of zero deflection.

Solution:

\[ 2x^2-8x+6=0 \]

\[ 2(x^2-4x+3)=0 \]

\[ 2(x-3)(x-1)=0 \]

Answer:

\[ \boxed{x=1\text{ and }x=3} \]

Review Exercise 4

Q1. Multiple Choice Questions (MCQs)

(i) The factorization of \(12x+36\) is:

Correct Answer: (a) \(12(x+3)\)

(ii) The factors of \(4x^2-12x+9\) are:

Correct Answer: (b) \((2x-3)^2\)

(iii) The HCF of \(a^3b^3\) and \(ab^2\) is:

Correct Answer: (d) \(ab^2\)

(iv) The LCM of \(16x^2,\ 4x,\) and \(30xy\) is:

Correct Answer: (a) \(240x^2y\)

(v) Product of LCM and HCF equals the ______ of two polynomials.

Correct Answer: (c) Product

(vi) The square root of \(x^2-6x+9\) is:

Correct Answer: (b) \(\pm(x-3)\)

(vii) The LCM of \((a-b)^2\) and \((a-b)^4\) is:

Correct Answer: (c) \((a-b)^4\)

(viii) The factorization of \(x^3+3x^2+3x+1\) is:

Correct Answer: (a) \((x+1)^3\)

(ix) A cubic polynomial has degree:

Correct Answer: (c) \(3\)

(x) One of the factors of \(x^3-27\) is:

Correct Answer: (d) Both (a) and (c)

Notes by Sir Dilawar

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