CBSE Class 8 Maths Factorization HOTs

HOTS

Question. Find all the factors of 14xy².
Answer : 2 × 7 × x × y × y

Question. Factorize : x² – 30x – 216.
Answer : (x – 36) (x + 6)

Question. Complete the brackets –
(x + a) (x + b) = x² + ( ) x + ( )
Answer : a + b, ab

Question. Complete it : x² – x – 156 = (_______) × (x + 12).
Answer : (x – 13)

Question. Give the factors which are common in 6xyz and 9yz²
Answer : 3yz

Question. Find dividend, when
Divisor = x + 3, Quotient = x + 2, Remainder = 0.
Answer : x² + 5x + 6

Question. Find two numbers ‘a’ and ‘b’ so that ab = 36 and (a – b) = 5
Answer : a = 9, b = 4

Question. Factorize : (x + y)² – (x – y)².
Answer : 4xy

Question. Simplify : x² – 10x – 96.
Answer : (x – 16) (x + 6)

Question. Simplify : x² - 12x - 45 / x² + 4x + 3
Answer : x - 15 / x + 1

Question. Find x² – y² if x = 5, y = 7.
Answer : –24

Question. Which is the factor common in all terms? 3xy + 15x²y + 9xy² + 21y³
Answer : 3y

Question. Simplify

Answer: (x–2)

Question. Find the value of ‘a’ and ‘b’ so that x⁴+x³+8x2+ax+b is divisible by (x²+1).
Answer: a=1, b=7

Question. If both (p+1) and (p–1) are factors of ap³+p²–2p+b; find the value of ‘a’ and ‘b’.
Answer: a=2, b=–1

Question. Factorize : 
a. 5 √5x 2 + 30x + 8 √5  b. 9a³b+41a²b²+20ab³   c. x⁸+x⁴+1
Answer: a. √5(√5 x +4) (√5x + 2)
b. ab(9a+5b)(a+4b)
c. (x²+1+x)(x²+1–x)(x⁴–x²+1)

CHALLENGES

1. Factorise the following :
a. x²+6x+9 b. 1–8x+16x² c. 4x2–81y²
d. 4a²+4ab+b² e. a²b²+c²d²–a²c²–b²d²

2. Factorise the following:
a. x²+7x+12  b. x²+x–12    c. x²–3x–18
d. x²+4x–21  e. x²–4x–192  f. x4–5x²+4
g. x⁴–13x²y²+36y4

3. Factorise the following:
a. 2x²+7x+6  b. 3x²–17x+20 c. 6x²–5x–14
d. 4x²+12xy+5y² e. 4x⁴–5x²+1

4. Factorise the following:
a. x⁸–y⁸ b. a¹²x⁴–a⁴x¹²
c. x⁴+x²+1 d. x⁴+5x²+9

5. Factorise x⁴+4y⁴. Use this to prove that 2011⁴+64 is a composite number.

6. Prove the identity (x+y+ z)²= x²+y²+z²+2xy+2yz+2zx. Use this to factorise the expression x8+4x2+4.

7. Prove that 41×61 can be written as the sum of two perfect squares.

8. Factorise: b²–12ac–4a²–9c².

9. Factorise: 4a–3+16a²+64a³.

10. Factorise : a⁴–5a³–12a²–5a+1.

SUMMARY

1. The process of writing an expression as the product of two or more expressions is called factorization.

2. There are four ways to factorize
a. Common factor method
b. Grouping method
c. Using identities
d. Splitting the middle term

3. Basic identities are
(a+b)²=a²+2ab+b²
(a–b)²=a²–2ab+b²
(a+b)(a–b)=a²–b²
(x+a)(x+b)=x²+(a+b)x+ab

4. Factorization is the inverse of multiplication. Division is the inverse of multiplication. So, factorization makes division easy.

5. For dividing an algebraic expression by a monomial, there are two ways
a. Term by term
b. Common factor method

6. For dividing an algebraic expression by a polynomial, there are two ways
a. Factorization and cancelling the common factors
b. Long division method.

ERROR ANALYSIS

1. Students make mistakes while copying down an expression.

2. Students fail to identify which identity is to be applied in a particular question.

3. While performing long division, they miss on arranging the dividend in the standard form.

ACTIVITY I

To verify the identity (a+b)²=a²+2ab+b²
Material required
a. White chart paper    b. Cardboard
c. Geometry box         d. Pair of scissors
e. Fevistick                 f. Colour box

Steps
1 On a white chart paper, draw and cut a square of side 8cm (= a), another square of side 3cm (=b) and two rectangles each of length 8 cm and breadth 3 cm (as shown in fig. (i)).
2. Colour the bigger square red, the smaller square light red and each rectangle grey.

3. Paste the red square on the card board.
4. Arrange the other cut outs on the cardboard (as shown in fig. (ii))•
5. Name the figure (as shown in fig. (ii)).

Demonstration
AC = AB + BC = a + b
CE = CD + DE = a + b
∴ ACEG is a square of side a + b.
Now, the area of square ACEG = area of square ABPH + area of rectangle BCDP
+ area of rectangle PFGH + area of square DEFP
⇒ (a + b)(a + b) = (a × a) + a × b + a x b + (b × b)
⇒ (a + b)² = a² + ab + ab + b²
⇒ (a + b)² = a² + 2ab + b².

ACTIVITY II

To verify the identity (a - b)² = a² – 2ab + b²
Material required
a. White chart paper          b. Cardboard
c. Geometry box               d. Pair of scissors
e. Fevistick                       f. Colour box

Steps
1. On a white chart paper, draw and cut a square of side 5 cm, another square of side 3 cm and two rectangles of dimensions 5 cm x 3 cm and 8 cm x 3 cm (as shown in fig- (i)).
2. Colour the bigger square red, smaller square as light red, bigger rectangle grey and smaller rectangle dark grey. 

3. Paste the red square on the cardboard.
4. Arrange the other cut outs on the cardboard (as shown in fig. (ii)).
5. Name the figure (as shown in fig. (ii)).

Demonstration
Suppose AB = 8 cm = a, DE = 3 cm = b
∴ PQ = PC – QC = 8 cm – 3 cm = 5 cm = a –b
and BD = BC + CD = 3 cm + 5 cm = 8 cm = a
∴ PQRS is a square of side a-b and ABDS is a square of side a.
CEFQ is a rectangle whose length = a – b + b = a and breadth = b.
Now, the area of square PQRS = area of square ABDS + area of square DEFR
– area of rectangle ABCP – area of rectangle CEFQ
⇒ (a – b)(a – b) = a × a + b × b – a × b – a × b
⇒ (a – b)² = a² + b² – ab – ab
⇒ (a – b)² = a² – 2ab + b².