WBBSE Solutions For Class 8 Maths Algebra Chapter 4 Factorization Of Algebraic Expression

Algebra Chapter 4 Factorization Of Algebraic Expression

⇒ Suppose area of a rectangle is (9x² – 16y²) sq. unit.

⇒ To find out the length of each side of a rectangle it is required to factorisation of the algebraic expression first.

⇒ Area = 9x² – 16y² = (3x)²– (4y)² = (3x + 4y) (3x – 4y)

⇒ So the length and breadth of the rectangle are (3x + 4y) unit and (3x – 4y) unit respectively.

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Necessary identities which help to factorise.

1. a² – b² = (a + b) (a – b)
2. a³ + b³ = (a + b) (a² – ab + b²)
3. a³ – b³ = (a – b) (a² + ab + b²)
4. x² +(a + b)x + ab = (x + a) (x + b)

Algebra Chapter 4 Factorization Of Algebraic Expression Examples

Example 1. Comparing the following algebraic expression with the identity. x² + (a + b)x + ab = (x + a) (x + b), find the value of a and b and factorise

1. x² – 5x + 6
2. x² – 5x – 6
3. (x + y)² – 7(x + y) + 12

Solution:

1. x² – 5x + 6

Given

f(x) = x² – 5x + 6

= x² + {(-3)+(-2)}x + (-3) x (-2) = (x – 3)(x – 2)

Here a = -3, b = -2

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2. x² – 5x – 6

Given

f(x) = x² – 5x – 6

= x² + {(-6) + 1}x + (-6) x 1

= (x – 6) (x + 1)

Here a = -6, b = 1

3. (x + y)² -7(x + y) + 12

Given

f(x) = (x + y)2 -7(x + y) + 12

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

= (x + y = 4)(x + y − 3)

Here a = -4, b = -3

Example 2. Resolve into factors by middle-term break. x² + 3x – 40

Solution: The factors of 40 are (4, 10), (2, 20), (5, 8), (1, 40)

Again, 3 = 8 – 5.

x² + 3x – 40

= x² + (8 – 5)x – 40

= x² + 8x – 5x – 40

= x(x + 8) – 5 (x + 8)

= (x + 8)(x – 5)

x² + 3x – 40 = (x + 8)(x – 5)

Example 3. x² – 13x + 40

Solution: x² – (8 + 5)x + 40

= x²– 8x – 5x + 40

40 = 5 x 8
13 = 5 + 8

= x(x – 8) – 5(x – 8)

= (x – 8)(x – 5)

x² – 13x + 40 = (x – 8)(x – 5)

Example 4. (a + 1)(a + 3)(a – 4)(a – 6) + 24

Solution: (a + 1)(a + 3)(a – 4)(a – 6) + 24

= (a + 1)(a – 4)(a + 3)(a − 6) + 24

= (a² – 4a + a – 4)(a² – 6a + 3a – 18) + 24

= (a² – 3a -4)(a² – 3a – 18) + 24

= (x – 4)(x – 18)+ 24

= x² – 18x – 4x + 72 + 24

= x² – 22x + 96

= x² – (16 + 6)x + 96

= x – 16x – 6x + 96

96 = 6 x 16

22 = 6 + 16

= x(x – 16) – 6(x – 16)

= (x – 16)(x – 6)(a² – 3a – 16)(a² – 3a – 6)

(a + 1)(a + 3)(a – 4)(a – 6) + 24 = (x – 16)(x – 6)(a² – 3a – 16)(a² – 3a – 6)

Example 5. x² + 4pqx – (p² – q²)²

Solution:

Given

x² + 4pqx – (p + q)² (p − q)²

= x² + {(p + q)² – (p − q)² x – (p + q)²(p − q)²

= x2 + (p + q)² x – (p − q)² x – (p + q)²(p − q)²

= x{x + (p + q)²} – (p − q)² {x + (p + q)²}

= {x + (p + q)²} {x – (p − q)²}

x² + 4pqx – (p² – q²)² = {x + (p + q)²} {x – (p − q)²}

Example 6. To factorize ax² + bx + c, find two quantities such that their algebraic sum is b (co-efficient of x) and their product is ac. [The product of coefficient of x² and the term free from x] Resolve 6y² – y – 15

Solution: Here, a = 6, b = -1, c = -15

a x c = 6x – 15 = 90

90= 1 x 90 = 2 x 45 = 3 x 30 = 5 x 18 = 6 x 15 = 9 x 10

10 – 9 = 1

∴ 6y² – y – 15

= 6y² – (10 – 9)y – 15

= 6y² – 10y+ 9y – 15

= 2y(3y – 5) + 3(3y – 5)

= (3y – 5)(2y + 3)

10 × 9 = 90

10 – 9 = 1

Example 7. Resolve into factors: 6(x + y)² + 5 (x² – y²) – 6 (x − y)²

Solution:

Given

6 (x + y)² + 5(x + y)(x – y) – 6(x – y)²

[Let x + y = a;x – y = b]

[9 x 4 = 6 x 6 = 36; 9 – 4 = 5]

= 6a² + 5ab – 6b²

= 6a² + (9 – 4)ab – 6b²

= 6a² + 9ab – 4ab – 6b²

= 3a(2a+3b) – 2b (2a + 3b)

= (2a + 3b)(3a – 2b)

[Substituting the value of a and b]

= {2(x + y) + 3 (x – y)} {3(x + y) – 2(x – y)}

= (2x + 2y + 3x – 3y)(3x + 3y – 2x + 2y) = (5x – y) (x + 5y)

6(x + y)² + 5 (x² – y²) – 6 (x − y)² = (2x + 2y + 3x – 3y)(3x + 3y – 2x + 2y) = (5x – y) (x + 5y)

Example 8. Resolve into factors: x² – 2abx + (a² – b²)

Solution:

Given

x² -2abx + (a² – b²)

= x² – 2ax + (a + b)(a – b)

= x² – {(a + b) + (a – b)}x + (a + b) (a – b)

= x2 – (a + b)x – (a – b)x + (a + b)(a – b)

= x(x – a – b) – (a – b) (x – a – b) = (x – a – b) (x – a + b)

x² – 2abx + (a² – b²) = x(x – a – b) – (a – b) (x – a – b) = (x – a – b) (x – a + b)

Example 9. Resolve into factors: a² + 1 – \(\frac{6}{a^2}\)

Solution: \(a^2+\frac{3 a}{a}-\frac{2 a}{a}-\frac{6}{a^2}\)

= \(a\left(a+\frac{3}{a}\right)-\frac{2}{a}\left(a+\frac{3}{a}\right)=\left(a+\frac{3}{a}\right)\left(a-\frac{2}{a}\right)\)

Example 10. Resolve into factors: \(2\left(a^2+\frac{1}{a^2}\right)-\left(a-\frac{1}{a}\right)-7\)

Solution:

Given

\(2\left(a^2+\frac{1}{a^2}\right)-\left(a-\frac{1}{a}\right)-7\)

= \(2\left\{\left(a-\frac{1}{a}\right)^2+2 \cdot a \cdot \frac{1}{a}\right\}-\left(a-\frac{1}{a}\right)-7=2\left(a-\frac{1}{a}\right)^2+4-\left(a-\frac{1}{a}\right)-7\)

= \(2\left(a-\frac{1}{a}\right)^2-\left(a-\frac{1}{a}\right)-3=2\left(a-\frac{1}{a}\right)^2-\{3-2\}\left(a-\frac{1}{a}\right)-3\)

= \(2\left(a-\frac{1}{a}\right)^2-3\left(a-\frac{1}{a}\right)+2\left(a-\frac{1}{a}\right)-3\)

= \(\left(a-\frac{1}{a}\right)\left\{2\left(a-\frac{1}{a}\right)-3\right\}+1\left\{2\left(a-\frac{1}{a}\right)-3\right\}\)

= \(\left\{2\left(a-\frac{1}{a}\right)-3\right\}\left(a-\frac{1}{a}+1\right)=\left\{2 a-\frac{2}{a}-3\right\}\left(a-\frac{1}{a}+1\right)\)

= \(\left(2 a-3-\frac{2}{a}\right)\left(a-\frac{1}{a}+1\right)=\left(2 a-(4-1)-\frac{2}{a}\right)\left(a-\frac{1}{a}+1\right)\)

= \(\left(2 a-4+1-\frac{2}{a}\right)\left(a-\frac{1}{a}+1\right)\)

= \(\left\{2(a-2)+\frac{1}{a}(a-2)\right\}\left(a-\frac{1}{a}+1\right)=(a-2)\left(2+\frac{1}{a}\right)\left(a-\frac{1}{a}+1\right)\)

Example 11. Resolve into factors: a (a + 1) x² – x – a (a− 1)

Solution:

Given

a (a + 1)x² – x – a(a – 1)

= a (a + 1)x² – {a² – (a² – 1)}x – a(a− 1)

= a (a + 1)x² – a² x + (a² – 1)x – a(a – 1)

= ax {(a + 1)x – a} + (a – 1) {(a + 1)x – a}

= {(a + 1)x – a} (ax + a – 1) = (ax + x – a)(ax + a – 1)

a (a + 1) x² – x – a (a− 1) = {(a + 1)x – a} (ax + a – 1) = (ax + x – a)(ax + a – 1)

Example 12. Factorising the expressions into factors by expressing them as the difference of two sqaures

1. x² + px + q

Solution: \(x^2+2 \cdot x \cdot \frac{p}{2}+\left(\frac{p}{2}\right)^2+q-\left(\frac{p}{2}\right)^2\)

= \(\left(x+\frac{p}{2}\right)^2-\left(\frac{p^2}{4}-q\right)^2=\left(x+\frac{p}{2}\right)^2-\left(\sqrt{\frac{p^2}{4}-q}\right)^2\)

= \(\left(x+\frac{p}{2}+\sqrt{\frac{p^2}{4}-q}\right)\left(x+\frac{p}{2}-\sqrt{\frac{p^2}{4}-q}\right)\)

2. x² + 5x + 6

= \(\left(x+\frac{5}{2}\right)^2-\left(\frac{1}{2}\right)^2=\left(x+\frac{5}{2}+\frac{1}{2}\right)\left(x+\frac{5}{2}-\frac{1}{2}\right) \)

= \(\left(x+\frac{6}{2}\right)\left(x+\frac{4}{2}\right)=(x+3)(x+2)\)

3. \(7 x^2-30 x+8\)

= \(7\left(x^2-\frac{30}{7} x+\frac{8}{7}\right)\)

= \(7\left\{x^2-2 \cdot x \cdot \frac{15}{7}+\left(\frac{15}{7}\right)^2-\left(\frac{15}{7}\right)^2+\frac{8}{7}\right\}\)

= \(7\left\{\left(x-\frac{15}{7}\right)^2-\left(\frac{225}{49}-\frac{8}{9}\right)\right\}=7\left\{\left(x-\frac{15}{7}\right)^2-\left(\frac{225-56}{49}\right)\right\}\)

= \(7\left\{\left(x-\frac{15}{7}\right)^2-\frac{169}{49}\right\}=7\left\{\left(x-\frac{15}{7}\right)^2-\left(\frac{13}{7}\right)^2\right\}\)

= \(7\left(x-\frac{15}{7}+\frac{13}{7}\right)\left(x-\frac{15}{7}-\frac{13}{7}\right)=7\left(x-\frac{2}{7}\right)\left(x-\frac{28}{7}\right)=(7 x-2)(x-4)\)

4. \(12 a^2+13 a b-4 b^2\)

= \(12\left(a^2+\frac{13}{12} a b-\frac{4}{12} b^2\right)\)

= \(12\left\{a^2+2 \cdot a \cdot \frac{13 b}{24}+\left(\frac{13 b}{24}\right)^2-\left(\frac{13 b}{24}\right)^2-\frac{b^2}{3}\right\}\)

= \(12\left\{\left(a+\frac{13 b}{24}\right)^2-\left(\frac{169 b^2}{576}+\frac{b^2}{3}\right)\right\}=12\left\{\left(a+\frac{13 b}{24}\right)^2-\left(\frac{169 b^2+192 b^2}{576}\right)\right\}\)

= \(12\left\{\left(a+\frac{13 b}{24}\right)^2-\left(\frac{361 b^2}{576}\right)\right\}=12\left\{\left(a+\frac{13 b}{24}\right)^2-\left(\frac{19 b}{24}\right)^2\right\}\)

= \(12\left(a+\frac{13 b}{24}+\frac{19 b}{24}\right)\left(a+\frac{13 b}{24}-\frac{19 b}{24}\right)=12\left(a+\frac{32 b}{24}\right)\left(a-\frac{6 b}{24}\right)\)

= \(12\left(a+\frac{4 b}{3}\right)\left(a-\frac{b}{4}\right)=3\left(a+\frac{4 b}{3}\right) 4\left(a-\frac{b}{4}\right)=(3 a+4 b)(4 a-b)\)

Example 13. Resolve into factors: ax² – (a² – 2)x – 2a

Solution:

Given

ax² – (a² – 2)x – 2a

= ax² – a²x + 2x – 2a = ax(x – a) + 2(x − a)

= (x – a)(ax + 2)

ax² – (a² – 2)x – 2a = (x – a)(ax + 2)

Example 14. Choose the correct answer:

1. (a – b) is equal to

1. \(\left(a^{\frac{1}{2}}+b^{\frac{1}{2}}\right)\left(a^{\frac{1}{2}}-b^{\frac{1}{2}}\right)\)
2. \((\sqrt{a}-\sqrt{b})^2\)
3. \(\left(a^{\frac{1}{3}}+b^{\frac{1}{3}}\right)\left(a^{\frac{1}{3}}-b^{\frac{1}{3}}\right)\)
4. None of these

Solution: \(a-b=\left(a^{\frac{1}{2}}\right)^2-\left(b^{\frac{1}{2}}\right)^2=\left(a^{\frac{1}{2}}+b^{\frac{1}{2}}\right)\left(a^{\frac{1}{2}}-b^{\frac{1}{2}}\right)\)

∴ So the correct answer is 1. \(\left(a^{\frac{1}{2}}+b^{\frac{1}{2}}\right)\left(a^{\frac{1}{2}}-b^{\frac{1}{2}}\right)\)

2. Factorised form of (p² – 17p – 38) is

1. (p – 19)(p – 2)
2. (p – 19)(p + 2)
3. (p + 19)(p – 2)
4. (p + 19)(p + 2)

Solution: p² – 17p – 38

= p² – (19 – 2)p – 38 = p² – 19p + 2p – 38

= p(p – 19)+ 2(p – 19)= (p – 19)(p + 2)

∴ The correct answer is 2. (p – 19)(p + 2)

3. If (5x² – 4x – 9) = (x + 1) (5x + p), then the value of p is

1. 9
2. 5
3. -9
4. None Of these

Solution: 5x² – 4x – 9

= 5x² – 9x + 5x – 9 = x(5x – 9) + 1 (5x – 9)

= (5x – 9)(x + 1)

(x + 1)(5x + p) = (x + 1)(5x – 9)

5x + p = 5x – 9 ⇒ p = -9

∴ The correct answer is 3. -9

Example 15. Write ‘True’ or ‘False’:

1. The factors of (x² – xy – 30y²) is (x + 5y) (x − 6y).

Solution: x² – xy – 30y² = x² – 6xy + 5xy – 30y²

= x(x – 6y) + 5y(x – 6y) = (x – 6y)(x + 5y)

∴ The statement is true.

2. Difference of twp square of the expression x² – x -12 is \(\left(\dot{x}-\frac{1}{2}\right)^2-\left(\frac{7}{2}\right)^2\)

Solution: \(x^2-x-12\)

= \(x^2-2 \cdot x \cdot \frac{1}{2}+\left(\frac{1}{2}\right)^2-\left(\frac{1}{2}\right)^2-12\)

= \(\left(x-\frac{1}{2}\right)^2-\left(\frac{1}{4}+12\right)=\left(x-\frac{1}{2}\right)^2-\frac{49}{4}\)

= \(\left(x-\frac{1}{2}\right)^2-\left(\frac{7}{2}\right)^2\)

∴ The statement is true.

3. Factorised form of (a + b)³ – c³ is (a + b+ c)(a² + b² + c² + 2ab + ac + bc)

Solution: (a + b)³ – c³

= (a + b – c) {(a + b)² + (a + b) c + c²}

= (a + b – c) (a² + 2ab + b² + ac + bc + c²)

∴ The statement is false.

Example 16. Fill in the blanks:

1. (x + a) (x + b) = x² + (a + b) x + ________

Solution: (x + a) (x + b) = x(x + b) + a(x + b)

= x² + bx + ax + ab = x2 + (a + b)x + ab.

(x + a) (x + b) = x² + (a + b) x + = x² + bx + ax + ab = x2 + (a + b)x + ab.

2. 67² – 37² = 30 x _________

Solution: 67² – 37² = 30 x ________

= (67 – 37)(67 +37) = 30 x 104

67² – 37² = 30 x = (67 – 37)(67 +37) = 30 x 104

3. (a + b + c)³ = a³ + b³ + c³ + ________

Solution: 3 (a + b)(b + c)(c + a).

(a + b + c)³ = a³ + b³ + c³ + 3 (a + b)(b + c)(c + a).