9th Class Math Chapter 4 Notes PDF 2026 – Factorization — Solved Exercises

Q1(i) — Factorize: $6 x + 12$

Factorize by taking out the common factor.

Both terms share a factor of $6$ .

Answer: $6 (x + 2)$

Q1(ii) — Factorize: $15 y^{2} + 20 y$

Factorize by taking out the common factor.

Both terms share a factor of $5 y$ .

Answer: $5 y (3 y + 4)$

Q1(iii) — Factorize: $- 12 x^{2} - 3 x$

Factorize by taking out the common factor.

Both terms share a factor of $- 3 x$ .

Answer: $- 3 x (4 x + 1)$

Q1(iv) — Factorize: $4 a^{2} b + 8 a b^{2}$

Factorize by taking out the common factor.

Both terms share a factor of $4 ab$ .

Answer: $4 ab (a + 2 b)$

Q1(v) — Factorize: $x y - 3 x^{2} + 2 x$

Factorize by taking out the common factor.

All terms share a factor of $x$ .

Answer: $x (y - 3 x + 2)$

Q1(vi) — Factorize: $3 a^{2} b - 9 a b^{2} + 15 ab$

Factorize by taking out the common factor.

All terms share a factor of $3 ab$ .

Answer: $3 ab (a - 3 b + 5)$

Q2(i) — Factorize: $5 x + 15$

Factorize.

Common factor $5$ .

Answer: $5 (x + 3)$

Q2(ii) — Factorize: $x^{2} + 4 x + 3$

Factorize.

Need two numbers multiplying to $3$ , summing to $4$ : $1, 3$ .

Answer: $(x + 1) (x + 3)$

Q2(iii) — Factorize: $x^{2} + 6 x + 8$

Factorize.

Two numbers multiplying to $8$ , summing to $6$ : $2, 4$ .

Answer: $(x + 2) (x + 4)$

Q2(iv) — Factorize: $x^{2} + 4 x + 4$

Factorize.

Perfect square: $2^{2} = 4$ , $2 (2) = 4$ .

Answer: $(x + 2)^{2}$

Q2(v) — Factorize: $x^{2} + x - 12$

Factorize.

Two numbers multiplying to $- 12$ , summing to $1$ : $4, - 3$ .

Answer: $(x + 4) (x - 3)$

Q2(vi) — Factorize: $x^{2} + 7 x + 10$

Factorize.

Two numbers multiplying to $10$ , summing to $7$ : $2, 5$ .

Answer: $(x + 2) (x + 5)$

Q2(vii) — Factorize: $x^{2} - 6 x + 8$

Factorize.

Two numbers multiplying to $8$ , summing to $- 6$ : $- 2, - 4$ .

Answer: $(x - 2) (x - 4)$

Q2(viii) — Factorize: $x^{2} - x - 56$

Factorize.

Two numbers multiplying to $- 56$ , summing to $- 1$ : $- 8, 7$ .

Answer: $(x - 8) (x + 7)$

Q2(ix) — Factorize: $x^{2} - 10 x - 24$

Factorize.

Two numbers multiplying to $- 24$ , summing to $- 10$ : $- 12, 2$ .

Answer: $(x - 12) (x + 2)$

Q2(x) — Factorize: $y^{2} + 4 y - 12$

Factorize.

Two numbers multiplying to $- 12$ , summing to $4$ : $6, - 2$ .

Answer: $(y + 6) (y - 2)$

Q2(xi) — Factorize: $y^{2} + 13 y + 36$

Factorize.

Two numbers multiplying to $36$ , summing to $13$ : $4, 9$ .

Answer: $(y + 4) (y + 9)$

Q2(xii) — Factorize: $x^{2} - x - 2$

Factorize.

Two numbers multiplying to $- 2$ , summing to $- 1$ : $- 2, 1$ .

Answer: $(x - 2) (x + 1)$

Q2(xiii) — Factorize: $2 x^{2} + 7 x + 3$

Factorize.

Split middle term: $2 x^{2} + 6 x + x + 3 = 2 x (x + 3) + 1 (x + 3)$ .

Answer: $(2 x + 1) (x + 3)$

Q2(xiv) — Factorize: $2 x^{2} + 11 x + 15$

Factorize.

Split: $2 x^{2} + 6 x + 5 x + 15 = 2 x (x + 3) + 5 (x + 3)$ .

Answer: $(2 x + 5) (x + 3)$

Q2(xv) — Factorize: $4 x^{2} + 13 x + 3$

Factorize.

Split: $4 x^{2} + 12 x + x + 3 = 4 x (x + 3) + 1 (x + 3)$ .

Answer: $(4 x + 1) (x + 3)$

Q2(xvi) — Factorize: $3 x^{2} + 5 x + 2$

Factorize.

Split: $3 x^{2} + 3 x + 2 x + 2 = 3 x (x + 1) + 2 (x + 1)$ .

Answer: $(3 x + 2) (x + 1)$

Q2(xvii) — Factorize: $3 y^{2} - 11 y + 6$

Factorize.

Split: $3 y^{2} - 9 y - 2 y + 6 = 3 y (y - 3) - 2 (y - 3)$ .

Answer: $(3 y - 2) (y - 3)$

Q2(xviii) — Factorize: $2 y^{2} - 5 y + 2$

Factorize.

Split: $2 y^{2} - 4 y - y + 2 = 2 y (y - 2) - 1 (y - 2)$ .

Answer: $(2 y - 1) (y - 2)$

Q2(xix) — Factorize: $4 z^{2} - 11 z + 6$

Factorize.

Split: $4 z^{2} - 8 z - 3 z + 6 = 4 z (z - 2) - 3 (z - 2)$ .

Answer: $(4 z - 3) (z - 2)$

Q2(xx) — Factorize: $6 + 7 x - 3 x^{2}$

Factorize.

Rewrite as $- 3 x^{2} + 7 x + 6 = - (3 x^{2} - 7 x - 6)$ .
$3 x^{2} - 7 x - 6$ splits as $3 x^{2} - 9 x + 2 x - 6 = 3 x (x - 3) + 2 (x - 3) = (3 x + 2) (x - 3)$ .
So $6 + 7 x - 3 x^{2} = - (3 x + 2) (x - 3) = (3 x + 2) (3 - x)$ .

Answer: $(3 x + 2) (3 - x)$

Q3(i) — Factorize: $4 x^{4} + 81 y^{4}$

Factorize.

Write as $(2 x^{2})^{2} + (9 y^{2})^{2}$ ; add and subtract $2 (2 x^{2}) (9 y^{2}) = 36 x^{2} y^{2}$ .
$= (2 x^{2} + 9 y^{2})^{2} - (6 x y)^{2}$ .

Answer: $(2 x^{2} + 6 x y + 9 y^{2}) (2 x^{2} - 6 x y + 9 y^{2})$

Q3(ii) — Factorize: $a^{4} + 64 b^{4}$

Factorize.

Write as $(a^{2})^{2} + (8 b^{2})^{2}$ ; add/subtract $2 (a^{2}) (8 b^{2}) = 16 a^{2} b^{2}$ .
$= (a^{2} + 8 b^{2})^{2} - (4 ab)^{2}$ .

Answer: $(a^{2} + 4 ab + 8 b^{2}) (a^{2} - 4 ab + 8 b^{2})$

Q3(iii) — Factorize: $x^{4} + 4 x^{2} + 16$

Factorize.

$(x^{2} + 4)^{2} = x^{4} + 8 x^{2} + 16$ ; difference from target is $4 x^{2}$ .
$= (x^{2} + 4)^{2} - 4 x^{2}$ .

Answer: $(x^{2} - 2 x + 4) (x^{2} + 2 x + 4)$

Q3(iv) — Factorize: $x^{4} - 14 x^{2} + 1$

Factorize.

$(x^{2} + 1)^{2} = x^{4} + 2 x^{2} + 1$ ; difference from target is $- 16 x^{2}$ .
$= (x^{2} + 1)^{2} - 16 x^{2}$ .

Answer: $(x^{2} - 4 x + 1) (x^{2} + 4 x + 1)$

Q3(v) — Factorize: $x^{4} - 30 x^{2} y^{2} + 9 y^{4}$

Factorize.

$(x^{2} + 3 y^{2})^{2} = x^{4} + 6 x^{2} y^{2} + 9 y^{4}$ ; difference is $- 36 x^{2} y^{2}$ .
$= (x^{2} + 3 y^{2})^{2} - 36 x^{2} y^{2}$ .

Answer: $(x^{2} - 6 x y + 3 y^{2}) (x^{2} + 6 x y + 3 y^{2})$

Q3(vi) — Factorize: $x^{4} - 7 x^{2} y^{2} + y^{4}$

Factorize.

$(x^{2} + y^{2})^{2} = x^{4} + 2 x^{2} y^{2} + y^{4}$ ; difference is $- 9 x^{2} y^{2}$ .
$= (x^{2} + y^{2})^{2} - 9 x^{2} y^{2}$ .

Answer: $(x^{2} - 3 x y + y^{2}) (x^{2} + 3 x y + y^{2})$

Q4(i) — Factorize: $(x + 1) (x + 2) (x + 3) (x + 4) + 1$

Factorize.

Pair factors with equal sum of roots: $(x + 1) (x + 4) = x^{2} + 5 x + 4$ , $(x + 2) (x + 3) = x^{2} + 5 x + 6$ .
Let $y = x^{2} + 5 x + 4$ : expression $= y (y + 2) + 1 = (y + 1)^{2}$ .

Answer: $(x^{2} + 5 x + 5)^{2}$

Q4(ii) — Factorize: $(x + 2) (x - 7) (x - 4) (x - 1) + 17$

Factorize.

Roots $- 2, 7, 4, 1$ ; pair with equal sums: $- 2 + 7 = 5 = 4 + 1$ , so pair $(x + 2) (x - 7)$ and $(x - 4) (x - 1)$ .
$(x + 2) (x - 7) = x^{2} - 5 x - 14$ , $(x - 4) (x - 1) = x^{2} - 5 x + 4$ .
Let $y = x^{2} - 5 x$ : expression $= (y - 14) (y + 4) + 17 = y^{2} - 10 y - 39 = (y - 13) (y + 3)$ .

Answer: $(x^{2} - 5 x - 13) (x^{2} - 5 x + 3)$

Q4(iii) — Factorize: $(2 x^{2} + 7 x + 3) (2 x^{2} + 7 x + 5) + 1$

Factorize.

Let $y = 2 x^{2} + 7 x + 3$ : expression $= y (y + 2) + 1 = (y + 1)^{2}$ .

Answer: $(2 x^{2} + 7 x + 4)^{2}$

Q4(iv) — Factorize: $(3 x^{2} + 5 x + 3) (3 x^{2} + 5 x + 5) - 3$

Factorize.

Let $y = 3 x^{2} + 5 x + 3$ : expression $= y (y + 2) - 3 = y^{2} + 2 y - 3 = (y + 3) (y - 1)$ .
$(y + 3) = 3 x^{2} + 5 x + 6$ ; $(y - 1) = 3 x^{2} + 5 x + 2 = (3 x + 2) (x + 1)$ .

Answer: $(3 x^{2} + 5 x + 6) (3 x + 2) (x + 1)$

Q4(v) — Factorize: $(x + 1) (x + 2) (x + 3) (x + 6) - 3 x^{2}$

Factorize.

$(x + 1) (x + 6) = x^{2} + 7 x + 6$ , $(x + 2) (x + 3) = x^{2} + 5 x + 6$ ; full expansion gives $x^{4} + 12 x^{3} + 47 x^{2} + 72 x + 36$ , minus $3 x^{2}$ : $x^{4} + 12 x^{3} + 44 x^{2} + 72 x + 36$ .
This factors as $(x^{2} + a x + 6) (x^{2} + c x + 6)$ with $a + c = 12$ , $a c = 32$ : solving gives $a = 8, c = 4$ .

Answer: $(x^{2} + 8 x + 6) (x^{2} + 4 x + 6)$

Q4(vi) — Factorize: $(x + 1) (x - 1) (x + 2) (x - 2) - 16 x^{2}$

Factorize.

$(x + 1) (x - 1) = x^{2} - 1$ , $(x + 2) (x - 2) = x^{2} - 4$ ; product $= x^{4} - 5 x^{2} + 4$ , minus $16 x^{2}$ : $x^{4} - 21 x^{2} + 4$ .
$(x^{2} + 2)^{2} = x^{4} + 4 x^{2} + 4$ ; difference from target is $- 25 x^{2}$ : $= (x^{2} + 2)^{2} - 25 x^{2}$ .

Answer: $(x^{2} - 5 x + 2) (x^{2} + 5 x + 2)$

Q5(i) — Factorize: $8 x^{3} + 12 x^{2} + 6 x + 1$

Factorize.

Check against $(2 x + 1)^{3} = 8 x^{3} + 12 x^{2} + 6 x + 1$ — matches exactly.

Answer: $(2 x + 1)^{3}$

Q5(ii) — Factorize: $27 a^{3} + 108 a^{2} b + 144 a b^{2} + 64 b^{3}$

Factorize.

Check against $(3 a + 4 b)^{3} = 27 a^{3} + 108 a^{2} b + 144 a b^{2} + 64 b^{3}$ — matches exactly.

Answer: $(3 a + 4 b)^{3}$

Q5(iii) — Factorize: $x^{3} + 18 x^{2} y + 108 x y^{2} + 216 y^{3}$

Factorize.

Check against $(x + 6 y)^{3} = x^{3} + 18 x^{2} y + 108 x y^{2} + 216 y^{3}$ — matches exactly.

Answer: $(x + 6 y)^{3}$

Q5(iv) — Factorize: $8 x^{3} - 125 y^{3} + 150 x y^{2} - 60 x^{2} y$

Factorize.

Rearrange: $8 x^{3} - 60 x^{2} y + 150 x y^{2} - 125 y^{3}$ .
Check against $(2 x - 5 y)^{3} = 8 x^{3} - 60 x^{2} y + 150 x y^{2} - 125 y^{3}$ — matches exactly.

Answer: $(2 x - 5 y)^{3}$

Q6(i) — Factorize: $125 a^{3} - 1$

Factorize.

$= (5 a)^{3} - 1^{3}$ , using $a^{3} - b^{3} = (a - b) (a^{2} + ab + b^{2})$ .

Answer: $(5 a - 1) (25 a^{2} + 5 a + 1)$

Q6(ii) — Factorize: $64 x^{3} + 125$

Factorize.

$= (4 x)^{3} + 5^{3}$ , using $a^{3} + b^{3} = (a + b) (a^{2} - ab + b^{2})$ .

Answer: $(4 x + 5) (16 x^{2} - 20 x + 25)$

Q6(iii) — Factorize: $x^{6} - 27$

Factorize.

$= (x^{2})^{3} - 3^{3}$ .

Answer: $(x^{2} - 3) (x^{4} + 3 x^{2} + 9)$

Q6(iv) — Factorize: $1000 a^{3} + 1$

Factorize.

$= (10 a)^{3} + 1^{3}$ .

Answer: $(10 a + 1) (100 a^{2} - 10 a + 1)$

Q6(v) — Factorize: $343 x^{3} + 216$

Factorize.

$= (7 x)^{3} + 6^{3}$ .

Answer: $(7 x + 6) (49 x^{2} - 42 x + 36)$

Q6(vi) — Factorize: $27 - 512 y^{3}$

Factorize.

$= 3^{3} - (8 y)^{3}$ .

Answer: $(3 - 8 y) (9 + 24 y + 64 y^{2})$

Q7(i) — HCF: $21 x^{2} y, 35 x y^{2}$

Find the HCF by factorization.

$21 x^{2} y = 3 \cdot 7 \cdot x^{2} y$ ; $35 x y^{2} = 5 \cdot 7 \cdot x y^{2}$ .
Common factor: $7 x y$ .

Answer: $7 x y$

Q7(ii) — HCF: $4 x^{2} - 9 y^{2}, 2 x^{2} - 3 x y$

Find the HCF by factorization.

$4 x^{2} - 9 y^{2} = (2 x - 3 y) (2 x + 3 y)$ .
$2 x^{2} - 3 x y = x (2 x - 3 y)$ .
Common factor: $(2 x - 3 y)$ .

Answer: $2 x - 3 y$

Q7(iii) — HCF: $x^{3} - 1, x^{2} + x + 1$

Find the HCF by factorization.

$x^{3} - 1 = (x - 1) (x^{2} + x + 1)$ .
Common factor: $(x^{2} + x + 1)$ .

Answer: $x^{2} + x + 1$

Q7(iv) — HCF: $a^{3} + 2 a^{2} - 3 a, 2 a^{3} + 5 a^{2} - 3 a$

Find the HCF by factorization.

$a^{3} + 2 a^{2} - 3 a = a (a + 3) (a - 1)$ .
$2 a^{3} + 5 a^{2} - 3 a = a (2 a - 1) (a + 3)$ .
Common factor: $a (a + 3)$ .

Answer: $a (a + 3)$

Q7(v) — HCF: $t^{2} - 3 t - 4, t^{2} + 5 t + 4, t^{2} - 1$

Find the HCF by factorization.

$t^{2} - 3 t - 4 = (t - 4) (t + 1)$ .
$t^{2} + 5 t + 4 = (t + 4) (t + 1)$ .
$t^{2} - 1 = (t - 1) (t + 1)$ .
Common factor: $(t + 1)$ .

Answer: $t + 1$

Q7(vi) — HCF: $x^{2} + 15 x + 56, x^{2} + 5 x - 24, x^{2} + 8 x$

Find the HCF by factorization.

$x^{2} + 15 x + 56 = (x + 7) (x + 8)$ .
$x^{2} + 5 x - 24 = (x + 8) (x - 3)$ .
$x^{2} + 8 x = x (x + 8)$ .
Common factor: $(x + 8)$ .

Answer: $x + 8$

Q8(i) — HCF: $27 x^{3} + 9 x^{2} - 3 x - 10, 3 x - 2$

Find the HCF by the division method.

Divide $27 x^{3} + 9 x^{2} - 3 x - 10$ by $3 x - 2$ : quotient $9 x^{2} + 9 x + 5$ , remainder $0$ .
Since the remainder is $0$ , the divisor itself is the HCF.

Answer: $3 x - 2$

Q8(ii) — HCF: $x^{3} - 9 x^{2} + 23 x - 15, x^{2} - 4 x + 3$

Find the HCF by the division method.

$x = 1$ and $x = 3$ are roots of the cubic, giving $x^{3} - 9 x^{2} + 23 x - 15 = (x - 1) (x - 3) (x - 5)$ .
$x^{2} - 4 x + 3 = (x - 1) (x - 3)$ .
Common factor: $(x - 1) (x - 3) = x^{2} - 4 x + 3$ .

Answer: $x^{2} - 4 x + 3$

Q8(iii) — HCF: $2 x^{3} + 2 x^{2} + 2 x + 2, 6 x^{3} + 12 x^{2} + 6 x + 12$

Find the HCF by the division method.

$2 x^{3} + 2 x^{2} + 2 x + 2 = 2 (x + 1) (x^{2} + 1)$ .
$6 x^{3} + 12 x^{2} + 6 x + 12 = 6 (x + 2) (x^{2} + 1)$ .
Common factors: numeric $g cd (2, 6) = 2$ , and $(x^{2} + 1)$ .

Answer: $2 (x^{2} + 1)$

Q8(iv) — HCF: $2 x^{3} - 4 x^{2} - 16 x, x^{3} - 4 x, 3 x^{2} - 6 x$

Find the HCF by the division method.

$2 x^{3} - 4 x^{2} - 16 x = 2 x (x - 4) (x + 2)$ .
$x^{3} - 4 x = x (x - 2) (x + 2)$ .
$3 x^{2} - 6 x = 3 x (x - 2)$ .
Only $x$ is common to all three.

Answer: $x$

Q9(i) — LCM: $2 a^{2} b, 4 a b^{2}, 6 ab$

Find the LCM.

$2 a^{2} b = 2 \cdot a^{2} \cdot b$ ; $4 a b^{2} = 2^{2} \cdot a \cdot b^{2}$ ; $6 ab = 2 \cdot 3 \cdot a \cdot b$ .
Take the highest power of each factor: $2^{2} \cdot 3 \cdot a^{2} \cdot b^{2}$ .

Answer: $12 a^{2} b^{2}$

Q9(ii) — LCM: $x^{2} + x, x^{3} + x^{2}$

Find the LCM.

$x^{2} + x = x (x + 1)$ ; $x^{3} + x^{2} = x^{2} (x + 1)$ .
Highest powers: $x^{2} (x + 1)$ .

Answer: $x^{2} (x + 1)$

Q9(iii) — LCM: $a^{2} - 4 a + 4, a^{2} - 2 a$

Find the LCM.

$a^{2} - 4 a + 4 = (a - 2)^{2}$ ; $a^{2} - 2 a = a (a - 2)$ .
Highest powers: $a (a - 2)^{2}$ .

Answer: $a (a - 2)^{2}$

Q9(iv) — LCM: $x^{4} - 16, x^{3} - 4 x$

Find the LCM.

$x^{4} - 16 = (x - 2) (x + 2) (x^{2} + 4)$ ; $x^{3} - 4 x = x (x - 2) (x + 2)$ .
Combine all distinct factors at highest power: $x (x - 2) (x + 2) (x^{2} + 4)$ .

Answer: $x (x - 2) (x + 2) (x^{2} + 4)$

Q9(v) — LCM: $16 - 4 x^{2}, x^{2} + x - 6, 4 - x^{2}$

Find the LCM.

$16 - 4 x^{2} = 4 (2 - x) (2 + x)$ ; $x^{2} + x - 6 = (x + 3) (x - 2)$ ; $4 - x^{2} = (2 - x) (2 + x)$ .
Combine the distinct factors (coefficient $4$ , $(x - 2)$ , $(x + 2)$ , $(x + 3)$ ).

Answer: $4 (x - 2) (x + 2) (x + 3)$

Q10(i) — Find the other polynomial

Two polynomials have HCF $= (y - 7)$ and LCM $= y^{3} - 10 y^{2} + 11 y + 70$ . One polynomial is $y^{2} - 5 y - 14$ . Find the other.

$y^{2} - 5 y - 14 = (y - 7) (y + 2)$ .
Using HCF $\times$ LCM $=$ product of the two polynomials: other $= \frac{HCF \times LCM}{given polynomial} = \frac{( y - 7 ) ( y ^{3} - 10 y ^{2} + 11 y + 70 )}{( y - 7 ) ( y + 2 )} = \frac{y ^{3} - 10 y ^{2} + 11 y + 70}{y + 2}$ .
Dividing: $y^{3} - 10 y^{2} + 11 y + 70 \div (y + 2) = y^{2} - 12 y + 35 = (y - 5) (y - 7)$ .

Answer: $(y - 5) (y - 7) = y^{2} - 12 y + 35$

Q10(ii) — Find $q (x)$

$p (x)$ and $q (x)$ have LCM $= 36 x^{3} (x + a) (x^{3} - a^{3})$ and HCF $= x^{2} (x - a)$ . If $p (x) = 4 x^{2} (x^{2} - a^{2})$ , find $q (x)$ .

$p (x) \times q (x) = HCF \times LCM$ .
$p (x) = 4 x^{2} (x - a) (x + a)$ ; HCF $\times$ LCM $= x^{2} (x - a) \times 36 x^{3} (x + a) (x^{3} - a^{3}) = 36 x^{5} (x - a)^{2} (x + a) (x^{2} + a x + a^{2})$ (using $x^{3} - a^{3} = (x - a) (x^{2} + a x + a^{2})$ ).
$q (x) = \frac{36 x ^{5} ( x - a ) ^{2} ( x + a ) ( x ^{2} + a x + a ^{2} )}{4 x ^{2} ( x - a ) ( x + a )} = 9 x^{3} (x - a) (x^{2} + a x + a^{2}) = 9 x^{3} (x^{3} - a^{3})$ .

Answer: $q (x) = 9 x^{3} (x^{3} - a^{3})$

Q10(iii) — Product of two polynomials

HCF $= (x + a)$ , LCM $= 12 x^{2} (x + a) (x^{2} - a^{2})$ . Find the product of the two polynomials.

Product $=$ HCF $\times$ LCM $= (x + a) \times 12 x^{2} (x + a) (x^{2} - a^{2})$ .
$= 12 x^{2} (x + a)^{2} (x - a) (x + a) = 12 x^{2} (x + a)^{3} (x - a)$ .

Answer: $12 x^{2} (x + a)^{3} (x - a)$

Q11(i) — Square root: $x^{2} - 8 x + 16$

Find the square root by factorization.

$x^{2} - 8 x + 16 = (x - 4)^{2}$ .

Answer: $x - 4$

Q11(ii) — Square root: $9 x^{2} + 12 x + 4$

Find the square root by factorization.

$9 x^{2} + 12 x + 4 = (3 x + 2)^{2}$ .

Answer: $3 x + 2$

Q11(iii) — Square root: $36 a^{2} + 84 a + 49$

Find the square root by factorization.

$(6 a)^{2} = 36 a^{2}$ , $2 (6 a) (7) = 84 a$ , $7^{2} = 49$ : $36 a^{2} + 84 a + 49 = (6 a + 7)^{2}$ .

Answer: $6 a + 7$

Q11(iv) — Square root: $64 y^{2} - 32 y + 4$

Find the square root by factorization.

$64 y^{2} - 32 y + 4 = (8 y - 2)^{2}$ .

Answer: $8 y - 2$

Q11(v) — Square root: $200 t^{2} - 120 t + 18$

Find the square root by factorization.

Factor out $2$ : $2 (100 t^{2} - 60 t + 9) = 2 (10 t - 3)^{2}$ .
This is $2 \times$ a perfect square, so its square root carries a $2$ factor.

Answer: $2 (10 t - 3)$

Q11(vi) — Square root: $40 x^{2} + 120 x + 90$

Find the square root by factorization.

Factor out $10$ : $10 (4 x^{2} + 12 x + 9) = 10 (2 x + 3)^{2}$ .

Answer: $10 (2 x + 3)$

Q12(i) — Square root: $4 x^{4} - 28 x^{3} + 37 x^{2} + 42 x + 9$

Find the square root by the division method.

First term: $4 x^{4} = 2 x^{2}$ .
Next term: $- 28 x^{3} \div (2 \times 2 x^{2}) = - 7 x$ ; root so far $2 x^{2} - 7 x$ ; subtracting leaves $- 12 x^{2} + 42 x$ .
Next term: $- 12 x^{2} \div (2 \times 2 x^{2}) = - 3$ ; root $2 x^{2} - 7 x - 3$ ; check gives remainder $0$ .

Answer: $2 x^{2} - 7 x - 3$

Q12(ii) — Square root: $121 x^{4} - 198 x^{3} - 183 x^{2} + 216 x + 144$

Find the square root by the division method.

First term: $121 x^{4} = 11 x^{2}$ .
Next: $- 198 x^{3} \div 22 x^{2} = - 9 x$ ; root $11 x^{2} - 9 x$ ; remainder $- 264 x^{2} + 216 x$ .
Next: $- 264 x^{2} \div 22 x^{2} = - 12$ ; root $11 x^{2} - 9 x - 12$ ; check gives remainder $0$ .

Answer: $11 x^{2} - 9 x - 12$

Q12(iii) — Square root: $x^{4} - 10 x^{3} y + 27 x^{2} y^{2} - 10 x y^{3} + y^{4}$

Find the square root by the division method.

First term: $x^{4} = x^{2}$ .
Next: $- 10 x^{3} y \div 2 x^{2} = - 5 x y$ ; root $x^{2} - 5 x y$ ; remainder $2 x^{2} y^{2} - 10 x y^{3}$ .
Next: $2 x^{2} y^{2} \div 2 x^{2} = y^{2}$ ; root $x^{2} - 5 x y + y^{2}$ ; check gives remainder $0$ .

Answer: $x^{2} - 5 x y + y^{2}$

Q12(iv) — Square root: $4 x^{4} - 12 x^{3} + 37 x^{2} - 42 x + 49$

Find the square root by the division method.

First term: $4 x^{4} = 2 x^{2}$ .
Next: $- 12 x^{3} \div 4 x^{2} = - 3 x$ ; root $2 x^{2} - 3 x$ ; remainder $28 x^{2} - 42 x$ .
Next: $28 x^{2} \div 4 x^{2} = 7$ ; root $2 x^{2} - 3 x + 7$ ; check gives remainder $0$ .

Answer: $2 x^{2} - 3 x + 7$

WP1 — Investment return: $R (x) = - x^{2} + 6 x - 8$

Factorize $R (x)$ and find the investment levels giving zero return.

$R (x) = - x^{2} + 6 x - 8 = - (x^{2} - 6 x + 8) = - (x - 2) (x - 4)$ .
$R (x) = 0 \Rightarrow x = 2$ or $x = 4$ .

Answer: Zero return at $x = 2$ and $x = 4$ (thousand rupees)

WP2 — Break-even: $P (x) = x^{3} - 15 x^{2} + 75 x - 125$

Find the break-even point(s).

Recognize the perfect cube: $(x - 5)^{3} = x^{3} - 15 x^{2} + 75 x - 125$ .
$P (x) = 0 \Rightarrow x = 5$ (a triple root).

Answer: Break-even at $x = 5$ units

WP3 — Potential energy: $V (x) = 2 x^{3} - 6 x^{2} + 4 x$

Determine where $V (x) = 0$ .

$V (x) = 2 x (x^{2} - 3 x + 2) = 2 x (x - 1) (x - 2)$ .
$V (x) = 0 \Rightarrow x = 0, 1, 2$ .

Answer: $x = 0, 1, 2$

WP4 — Beam deflection: $Y (x) = 2 x^{2} - 8 x + 6$

Find the points of zero deflection.

$Y (x) = 2 (x^{2} - 4 x + 3) = 2 (x - 1) (x - 3)$ .
$Y (x) = 0 \Rightarrow x = 1$ or $x = 3$ .

Answer: $x = 1$ and $x = 3$

Q13(i) — Factorize: $4 x^{3} + 18 x^{2} - 12 x$

Factorize.

Common factor $2 x$ : $2 x (2 x^{2} + 9 x - 6)$ .
The quadratic $2 x^{2} + 9 x - 6$ has discriminant $81 + 48 = 129$ (not a perfect square), so it doesn't factor further over rationals.

Answer: $2 x (2 x^{2} + 9 x - 6)$

Q13(ii) — Factorize: $x^{3} + 64 y^{3}$

Factorize.

$= (x)^{3} + (4 y)^{3} = (x + 4 y) (x^{2} - 4 x y + 16 y^{2})$ .

Answer: $(x + 4 y) (x^{2} - 4 x y + 16 y^{2})$

Q13(iii) — Factorize: $x^{3} y^{3} - 8$

Factorize.

$= (x y)^{3} - 2^{3} = (x y - 2) (x^{2} y^{2} + 2 x y + 4)$ .

Answer: $(x y - 2) (x^{2} y^{2} + 2 x y + 4)$

Q13(iv) — Factorize: $- x^{2} - 23 x - 60$

Factorize.

$= - (x^{2} + 23 x + 60)$ ; factors of $60$ summing to $23$ : $3, 20$ .

Answer: $- (x + 3) (x + 20)$

Q13(v) — Factorize: $2 x^{2} + 7 x + 3$

Factorize.

Split: $2 x^{2} + 6 x + x + 3 = 2 x (x + 3) + 1 (x + 3)$ .

Answer: $(2 x + 1) (x + 3)$

Q13(vi) — Factorize: $x^{4} + 64$

Factorize using the Sophie Germain identity.

$x^{4} + 4 (2)^{4}$ form: $(x^{2} + 2 (2)^{2})^{2} - (2 \cdot x \cdot 2)^{2} = (x^{2} + 8)^{2} - (4 x)^{2}$ .

Answer: $(x^{2} - 4 x + 8) (x^{2} + 4 x + 8)$

Q13(vii) — Factorize: $x^{4} + 2 x^{2} + 9$

Factorize.

$(x^{2} + 3)^{2} = x^{4} + 6 x^{2} + 9$ ; difference from target is $- 4 x^{2}$ .
$= (x^{2} + 3)^{2} - 4 x^{2}$ .

Answer: $(x^{2} - 2 x + 3) (x^{2} + 2 x + 3)$

Q13(viii) — Factorize: $(x + 3) (x + 4) (x + 5) (x + 6) - 360$

Factorize.

Pair $(x + 3) (x + 6) = x^{2} + 9 x + 18$ , $(x + 4) (x + 5) = x^{2} + 9 x + 20$ ; let $y = x^{2} + 9 x + 18$ .
Expression $= y (y + 2) - 360 = y^{2} + 2 y - 360 = (y + 20) (y - 18)$ .
$y + 20 = x^{2} + 9 x + 38$ ; $y - 18 = x^{2} + 9 x = x (x + 9)$ .

Answer: $x (x + 9) (x^{2} + 9 x + 38)$

Q13(ix) — Factorize: $(x^{2} + 6 x + 3) (x^{2} + 6 x - 9) + 36$

Factorize.

Let $y = x^{2} + 6 x$ : expression $= (y + 3) (y - 9) + 36 = y^{2} - 6 y + 9 = (y - 3)^{2}$ .

Answer: $(x^{2} + 6 x - 3)^{2}$

Q14(i) — LCM & HCF: $4 x^{3} + 12 x^{2}, 8 x^{2} + 16 x$

Find both the LCM and HCF.

$4 x^{3} + 12 x^{2} = 4 x^{2} (x + 3)$ ; $8 x^{2} + 16 x = 8 x (x + 2)$ .
HCF: common numeric factor $4$ , common variable $x$ → $4 x$ .
LCM: numeric $lcm (4, 8) = 8$ , highest power $x^{2}$ , both binomials needed → $8 x^{2} (x + 3) (x + 2)$ .

Answer: HCF $= 4 x$ ; LCM $= 8 x^{2} (x + 2) (x + 3)$

Q14(ii) — LCM & HCF: $x^{3} + 3 x^{2} - 4 x, x^{2} - 4 x + 3$

Find both the LCM and HCF.

$x^{3} + 3 x^{2} - 4 x = x (x + 4) (x - 1)$ ; $x^{2} - 4 x + 3 = (x - 1) (x - 3)$ .
HCF $= (x - 1)$ ; LCM $= x (x + 4) (x - 1) (x - 3)$ .

Answer: HCF $= x - 1$ ; LCM $= x (x + 4) (x - 1) (x - 3)$

Q14(iii) — LCM & HCF: $x^{2} + 8 x + 16, x^{2} - 16$

Find both the LCM and HCF.

$x^{2} + 8 x + 16 = (x + 4)^{2}$ ; $x^{2} - 16 = (x - 4) (x + 4)$ .
HCF $= (x + 4)$ ; LCM $= (x + 4)^{2} (x - 4)$ .

Answer: HCF $= x + 4$ ; LCM $= (x + 4)^{2} (x - 4)$

Q14(iv) — LCM & HCF: $x^{3} - 9 x, x^{2} - x - 6$

Find both the LCM and HCF.

$x^{3} - 9 x = x (x - 3) (x + 3)$ ; $x^{2} - x - 6 = (x - 3) (x + 2)$ .
HCF $= (x - 3)$ ; LCM $= x (x - 3) (x + 3) (x + 2)$ .

Answer: HCF $= x - 3$ ; LCM $= x (x - 3) (x + 2) (x + 3)$

Q15 — Square root of $16 x^{4} + 8 x^{2} + 1$ (factorization and division)

Find the square root by both methods.

Factorization: recognize $16 x^{4} + 8 x^{2} + 1 = (4 x^{2} + 1)^{2}$ directly.
Division method: first term $16 x^{4} = 4 x^{2}$ ; since there's no $x^{3}$ term, the next quotient term is $0$ ; bringing down gives $8 x^{2}$ ; dividing by $2 (4 x^{2}) = 8 x^{2}$ gives $1$ ; final check confirms remainder $0$ .

Answer: $4 x^{2} + 1$ (both methods agree)

WP5 — Loan cost: $C (x) = x^{2} - 8 x + 15$

Factorize $C (x)$ and find the repayment period(s) where the modeled cost is zero.

$C (x) = x^{2} - 8 x + 15 = (x - 3) (x - 5)$ .
$C (x) = 0 \Rightarrow x = 3$ or $x = 5$ .

Answer: $x = 3$ or $x = 5$ years

Answer Key Highlights

●Common-factor and trinomial factorizations (Parts A–B): see full working above for all 26 items.
●Perfect cubes (Part E): $(2 x + 1)^{3}$ , $(3 a + 4 b)^{3}$ , $(x + 6 y)^{3}$ , $(2 x - 5 y)^{3}$ .
●Sum/difference of cubes (Part F): $(5 a - 1) (25 a^{2} + 5 a + 1)$ ; $(4 x + 5) (16 x^{2} - 20 x + 25)$ ; $(x^{2} - 3) (x^{4} + 3 x^{2} + 9)$ ; $(10 a + 1) (100 a^{2} - 10 a + 1)$ ; $(7 x + 6) (49 x^{2} - 42 x + 36)$ ; $(3 - 8 y) (9 + 24 y + 64 y^{2})$ .
●HCF/LCM (Parts G–I, O): worked example-by-example above.
●HCF-LCM relationship (Part J): $(y - 5) (y - 7)$ ; $q (x) = 9 x^{3} (x^{3} - a^{3})$ ; product $= 12 x^{2} (x + a)^{3} (x - a)$ .
●Square roots (Parts K, L, P): $x - 4$ ; $3 x + 2$ ; $6 a + 7$ ; $8 y - 2$ ; $2 (10 t - 3)$ ; $10 (2 x + 3)$ ; $2 x^{2} - 7 x - 3$ ; $11 x^{2} - 9 x - 12$ ; $x^{2} - 5 x y + y^{2}$ ; $2 x^{2} - 3 x + 7$ ; $4 x^{2} + 1$ .
●Word problems (Parts M, Q): zero return at $x = 2, 4$ ; break-even at $x = 5$ ; $V = 0$ at $x = 0, 1, 2$ ; zero deflection at $x = 1, 3$ ; loan cost zero at $x = 3, 5$ years.

9th Class Math Chapter 4 Notes

Exercise 4.1 — Common Factors

Exercise 4.1 — Trinomial Factorization

Exercise 4.2 — Biquadratics via the Difference-of-Squares Trick

Exercise 4.2 — Product-of-Four-Factors-Plus-Constant Trick

Exercise 4.2 — Perfect Cube Recognition

Exercise 4.2 — Sum and Difference of Cubes

Exercise 4.3 — HCF by Factorization Method

Exercise 4.3 — HCF by Division Method

Exercise 4.3 — LCM by Prime Factorization

Exercise 4.3 — HCF / LCM Relationship Problems

Exercise 4.4 — Square Root by Factorization

Exercise 4.4 — Square Root by Division Method

Exercise 4.4 — Word Problems: Factorizing to Find Zeros

Review Exercise 4 — More Factorization

Review Exercise 4 — LCM and HCF Together

Review Exercise 4 — Square Root by Both Methods

Review Exercise 4 — Word Problem: Loan Repayment

More 9th Class Math notes

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