Algebra Chapter 5 Factorisation
⇒ By factorisation of a polynomial, it means that the polynomial should be represented as the product of two or more polynomials.
⇒ Each of the polynomials obtained is called a factor of the original one.
⇒ If f(x) = ax3 + bx + c, then we first find the zero of the polynomial.
⇒ Let f(α) = 0 then (x – α) is a factor of fx).
⇒ This method is called vanishing or trial or zero method.
⇒ We are going to use these formulae to factorise.
- a2 – b2 = (a + b) (a – b)
- a3 + b3 = (a + b) (a2 – ab + b2)
- a3 – b3 = (a – b) (a2 + ab + b2)
- a3 + b3 + c3 – 3abc = (a + b + c) (a2 + b2 +c – ab – bc – ca) = \(\frac{1}{2}\)(a + b + c ) {(a – b)2 + (b − c)2 + (c − a)2}
- If a + b + c = 0 then a3 + b3 + c3 = 3abc
- If a3 + b3 + c3 = 3 abc then either a + bc = 0 or a = b = c
- For quadratic expressions, we break the middle term of factorise like x2 + (a + b)x + ab = x2 + ax + bx+ ab = x (x + a) + b (x + a) = (x + a) (x + b)
Read and Learn More WBBSE Solutions For Class 9 Maths
Algebra Chapter 5 Factorisation True Or False
Example 1. (2a + 1) is a factor of 8a3+ 8a – 5.
Solution: The correct answer is False.
Example 2. x + \(\frac{3}{2}\) is a factor of 2x3 + x2 – 9x – 9.
Solution: The correct answer is True.
Example 3. If f(x) = x3 + 3x – 4 then (x – 1) is a factor of x3 + 3x – 4
Solution: The correct answer is True.
Example 4. If a + b + c ≠ 0 and a3 +b3 + c3 =3abc then a ≠ b ≠ c
Solution: The correct answer is False.
Example 5. There are 6 factors of 1 – x12
Solution: The correct answer is True.
Example 6. Value of a3 +b3 + c3 – 3abc when a = 999, b = 998, c = 997 is 8982.
Solution: The correct answer is True.
Example 7. (x – a) (x – b) = x2 + (a + b) x + ab
Solution: The correct answer is False.
Example 8. If x + y + z = 0 then x3 + y3 + z3 = -3xyz
Solution: The correct answer is False.
Example 9. x and (x + 9) are two factors of x2 + 9x.
Solution: The correct answer is True.
Example 10. (x – 1) and (x2 + x + 4) are two factors of x3 + 3x – 4.
Solution: The correct answer is True.
Algebra Chapter 5 Factorisation Fill In The Blanks
Example 1. 2x3 + x2 – 9x – 9 = (2x + 3) x (______)
Solution: x2
Example 2. x2 – 2ax + (a + b)(a – b) = (x – a – b) x (________)
Solution: x – a + b.
Example 3. 4x2 – 12xy + 9y2 + 2x – 3y =(2x – 3y) x (__________)
Solution: 2x – 3y + 1.
Example 4. x2 + 5x + 6 = x2 + x (_______) + 6
Solution: 3 + 2
Example 5. x2 – 5x + 6 = x2 – x (________) + 6
Solution: 3 + 2.
Example 6. a6 + 5a3 + 8 = (a2)3 + (_______)3+ 23 -3a2(_______).2
Solution: -a, – a.
Example 7. (a – b)3 + (b – c)3 + (c – a)3 = 3 ______
Solution: (a – b)(b – c)(c – a).
Example 8. (80)3 – (20)3 – (60)3 = _______
Solution: 96000.
Example 9. If a – b = -1 then a3 – b3+ 3ab + 1 = _______
Solution: 0
Example 10. If a + b + c = 0 then a3 + b3 – 3abc + c3 = _______
Solution: 0
Algebra Chapter 5 Factorisation Short Answer Type Questions
Example 1. Simplify: \(\frac{\left(b^2-c^2\right)^3+\left(c^2-a^2\right)^3+\left(a^2-b^2\right)^3}{(b-c)^3+(c-a)^3+(a-b)^3}\)
Solution: \(b^2-c^2+c^2-a^2+a^2-b^2=0\)
∴ \(\left(b^2-c^2\right)^3+\left(c^2-a^2\right)^3+\left(a^2-b^2\right)^3\)
Similarly, \(3\left(b^2-c^2\right)\left(c^2-a^2\right)\left(a^2-b^2\right)\)
∴ \(\frac{\left(b^2-c^2\right)^3+\left(c^2-a^2\right)^3+\left(a^2-b^2\right)^3}{(b-c)^3+(c-a)^3+(a-b)^3}=\frac{3\left(b^2-c^2\right)\left(c^2-a^2\right)\left(a^2-b^2\right)}{3(b-c)(c-a)(a-b)}\)
= (b+c)(c+a)(a+b)
\(\frac{\left(b^2-c^2\right)^3+\left(c^2-a^2\right)^3+\left(a^2-b^2\right)^3}{(b-c)^3+(c-a)^3+(a-b)^3}\) = (b+c)(c+a)(a+b)
Example 2. Let us write the relation of a, b, c if a3 +b3 + c3 – 3abc = 0 and a + b + c ≠ 0.
Solution: a3 + b3 + c3 – 3abc
= \(\frac{1}{2}\) (a + b + c) {(a – b)2 + (b − c)2 + (c − a)2} = 0
⇒ Now, a + b + c ≠ 0
∴ (a – b)2 + (b – c)2 + (c – a)2 = 0
⇒ If sum of the squares are zero, then each term is zero
∴ a – b = 0, a = b and b – c = 0, b = c
∴ a = b = c.
Example 3. If a2 – b2 = 224 or b are negative integers (a < b) then find the values of a and b.
Solution: a2 – b2 = 225 – 1 = (15)2– (1)2
⇒ or, a2 – b2 = (-15)2 – (-1)2
⇒ a and b are negative integers and a < b
∴ a = -15, b = -1.
The values of a and b -15 and -1.
Example 4. Let us write the value of (x – a)3 + (x − b)3 + (x − c)3 – 3(x -a)(x – b)(x – c) if 3x = a + b + c.
Solution: Now x – a + x – b + x – c
= 3x – (a + b + c)
= 3x – 3x = 0
∴ (x – a)3 + (x – b)3 + (x−c)3 = 3(x − a)(x – b)(x – c)
⇒ or, (x – a)3 + (x – b)3 + (x – c)3 – 3 (x – a)(x – b)(x – c) = 0
The value of (x – a)3 + (x − b)3 + (x − c)3 – 3(x -a)(x – b)(x – c) is 0.
Example 5. Let us write the values of a and p if 2x2 + px + 6 = (2x – a)(x-2)
Solution: 2x2 + px + 6 = 2x2 – 4x – ax + 2a
= 2x2 + x (-4 – a) + 2a
⇒ Comparing both sides by coefficients of x and constant terms, we get 2a = 6, a = 3 and p = – 4- a = -4 – 3 = -7.
The values of a and p are 3 and -7 .
Example 6. Find the values of (10)3 – 53 – 53.
Solution: 10 + (-5) + (-5) = 0
∴ (10)3 + (-5)3 + (-5)3 =3.10(-5)(-5) = 750
The values of (10)3 – 53 – 53 = 750
Example 7. Find the value of \(\frac{10^3+5^3}{10^2-25}\)
Solution: \(\frac{10^3+5^3}{10^2-25}=\frac{(10+5)\left(10^2-10 \times 5+5^2\right)}{10^2-5 \times 10+5^2}\)
= 10 + 5 = 15
The value of \(\frac{10^3+5^3}{10^2-25}\) = 15
Example 8. If a3 – 0.216 = (a – 0.6) (a2 + 0.6a + k), then find the value of k.
Solution: a3 – 0.216 = (a – 0.6) (a2 + 0.6a + (0.6)2}
∴ k = (0.6)2 = 0.36
The value of k = 0.36
Example 9. Factorise 8x3 – y3 – 12xy + 6xy2.
Solution: (2x)3 + (-y)3 – 6xy (2x – y)
= (2x)3 + (-y)3 + 3 x 2x (-y) (2x + (-y)}
= (2x – y)3 = (2x – y) (2x – y)(2x – y)
8x3 – y3 – 12xy + 6xy2 = (2x – y)3 = (2x – y) (2x – y)(2x – y)
Example 10. If x + y = – 4 then find the value of x3 + y3 – 12xy + 64.
Solution: x3 + y3 – 12xy + 64
= x3 + y3 + 43 – 3xy x 4
= (x + y + 4) (x2 + y2 + 16 – xy – 4x – 4y)
= 0 x (x2 + y2 + 16 – xy – 4x – 4y) = 0
The value of x3 + y3 – 12xy + 64 = 0