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Chapter 7: Factorization Exercise – 7.8 Question: 1 Resolve of the following quadratic equation trinomials into factors: 2x^{2 }+ 5x + 3 Solution: The given expression is 2x^{2 }+ 5x + 3. (Co-efficient of x^{2 }= 2, co-efficient of x = 5 and the constant term = 3) We will split the co-efficient of x into two parts such that their sum is 5 and their product equals to the product of the co-efficient of x^{2} and the constant term, i.e., 2 × 3 = 6. Now, 2 + 3 = 5 And 2 × 3 = 6 Replacing the middle term 5x by 2x + 3x, we have: 2x^{2 }+ 5x + 3 = 2x^{2 }+ 2x + 3x + 3 = (2x^{2 }+ 2x) + (3x + 3) = 2x (x + 1) + 3 (x + 1) = (2x + 3) (x + 1) Question: 2 Resolve of the following quadratic equation trinomials into factors: 2x^{2 }- 3x - 2 Solution: The given expression is 2x^{2 }– 3x – 2. (Co-efficient of x^{2 }= 2, co-efficient of x = -3 and the constant term = -2) We will split the co-efficient of x into two parts such that their sum is -3 and their product equals to the product of the co-efficient of x^{2} and the constant term, i.e., 2 × (-2) = - 4 Now, (- 4) + 1 = - 3 And (- 4) × 1 = – 4 Replacing the middle term 3x by – 4x + x, we have: 2x^{2 }- 3x - 2 = 2x^{2 }- 4x + x -2 = (2x^{2 }– 4x) + (x - 2) = 2x(x - 2) + 1(x – 2) = (x – 2)(2x + 1) Question: 3 Resolve of the following quadratic equation trinomials into factors: 3x^{2 }+ 10x + 3 Solution: The given expression is 3x^{2 }+ 10x + 3. (Co-efficient of x^{2 }= 3, co-efficient of x = 10 and the constant term = 3) We will split the co-efficient of x into two parts such that their sum is 10 and their product equals to the product of the co-efficient of x^{2} and the constant term, i.e., 3 × 3 = 9 Now, 9 + 1 = 10 And 9 × 1 = 9 Replacing the middle term 10x by 9x + x, we have: 3x^{2 }+ 10x + 3 = 3x^{2 }+ 9x + x + 3 = (3x^{2 }+ 9x) + (x + 3) = 3x(x + 3) + 1(x + 3) = (x + 3)(3x + 1) Question: 4 Resolve of the following quadratic equation trinomials into factors: 7x – 6 – 2x^{2} Solution: The given expression is 7x – 6 – 2x^{2}. (Co-efficient of x^{2 }= -2, co-efficient of x = 7 and the constant term = – 6) We will split the co-efficient of x into two parts such that their sum is 7 and their product equals to the product of the co-efficient of x^{2} and the constant term, i.e., (-2) × (- 6) = 12 Now, 4 + 3 = 7 And 4 × 3 = 12 Replacing the middle term 7x by 4x + 3x, we have: 7x - 6 - 2x^{2 }= -2x^{2 }+ 4x + 3x - 6 = (-2x^{2 }+ 4x) + (3x – 6) = 2x (2 - x) – 3(2 - x) = (2x – 3) (2 – x) Question: 5 Resolve of the following quadratic equation trinomials into factors: 7x^{2 }– 19x – 6 Solution: The given expression is 7x^{2 }– 19x – 6. (Co-efficient of x^{2 }= 7, co-efficient of x = -19 and the constant term = – 6) We will split the co-efficient of x into two parts such that their sum is -19 and their product equals to the product of the co-efficient of x^{2} and the constant term, i.e., 7 × (– 6) = 9 Now, (-21) + 2 = -19 And (-21) × 2 = - 42 Replacing the middle term -19x by -21x + 2x, we have: 7x^{2 }– 19x – 6 = 7x^{2 }– 21x + 2x -6 = (7x^{2 }– 21x) + (2x – 6) = 7x(x – 3) + 2(x – 3) = (x – 3) (7x + 2) Question: 6 Resolve of the following quadratic equation trinomials into factors: 28 – 31x – 5x^{2} Solution: The given expression is 28 – 31x – 5x^{2}. (Co-efficient of x^{2 }= -5, co-efficient of x = -31 and the constant term = 28) We will split the co-efficient of x into two parts such that their sum is -31 and their product equals to the product of the co-efficient of x^{2} and the constant term, i.e., ( -5 ) × ( 28 ) = -140 Now, (-35) + 4 = -31 And (-35) × 4 = -140 Replacing the middle term -31x by -35x + 4x, we have: 28 – 31x – 5x^{2 }= -5x^{2 }– 35x + 4x + 28 = (-5x^{2 }– 35x) + (4x + 28) = -5x(x + 7) + 4(x + 7) = (4 - 5x)(x + 7) Question: 7 Resolve of the following quadratic equation trinomials into factors: 3 + 23y – 8y^{2} Solution: The given expression is 3 + 23y – 8y^{2}. (Co-efficient of y^{2 }= -8, co-efficient of y = 23 and the constant term = 3) We will split the co-efficient of x into two parts such that their sum is 23 and their product equals to the product of the co-efficient of x^{2} and the constant term, i.e., ( -8 ) × 3 = -24 Now, (-1) + 24 = 23 And (-1) × 24 = -24 Replacing the middle term 23y by -y + 24y, we have: 3 + 23y – 8y^{2 }= -8y^{2 }– y + 24y + 3 = (-8y^{2 }– y) + (24y + 3) = – y(8y + 1) + 3(8y + 1) = (8y + 1)( y + z) Question: 8 Resolve of the following quadratic equation trinomials into factors: 11x^{2 }– 54x + 63 Solution: The given expression is 11x^{2 }– 54x + 63. (Co-efficient of x^{2 }= 11, co-efficient of x = – 54 and the constant term = 63) We will split the co-efficient of x into two parts such that their sum is -19 and their product equals to the product of the co-efficient of x^{2} and the constant term, i.e., 11 × 63 = 693 Now, (-33) + (-21) =- 54 And (-33) × (-21) = 693 Replacing the middle term -54x by -33x – 21x, we have: 11x^{2 }– 54x + 63 = 11x^{2 }– 33x – 21x + 63 = (11x^{2 }– 33x) + (-21x + 63) = 11x(x – 3) - 21(x – 3) = (x - 3)( 11x - 21) Question: 9 Resolve of the following quadratic equation trinomials into factors: 7x – 6x^{2 }+ 20 Solution: The given expression is 7x – 6x^{2 }+ 20. (Co-efficient of x^{2 }= -6, co-efficient of x = 7 and the constant term = 20) We will split the co-efficient of x into two parts such that their sum is -19 and their product equals to the product of the co-efficient of x^{2} and the constant term, i.e., (– 6) × 20 = – 120 Now, (15) + (- 8) = 7 And (15) × (-8) = -120 Replacing the middle term 7x by 15x - 8x, we have: 7x – 6x^{2 }+ 20 = – 6x^{2 }+ 15x - 8x + 20 = (-6x^{2 }+ 15x) + (- 8x + 20) = 3x (-2x + 5) + 4(- 2x + 5) = (-2x + 5)(3x + 4) Question: 10 Resolve of the following quadratic equation trinomials into factors: 3x^{2 }+ 22x + 35 Solution: The given expression is 3x^{2 }+ 22x + 35. (Co-efficient of x^{2 }= 3, co-efficient of x = 22 and the constant term = 35) We will split the co-efficient of x into two parts such that their sum is -19 and their product equals to the product of the co-efficient of x^{2} and the constant term, i.e., 3 × 35 = 105 Now, (15) + (7) = 22 And (15) × (7) = 105 Replacing the middle term 22x by 15x + 7x, we have: 3x^{2 }+ 22x + 35 = 3x^{2 }+ 15x + 7x + 35 = (3x^{2 }+ 15x) + (7x + 35) = 3x(x + 5) + 7(x + 5) = (x + 5)( 3x + 7) Question: 19 Resolve of the following quadratic equation trinomials into factors: 12x^{2 }– 17xy + 6y^{2} Solution: The given expression is 12x^{2 }– 17xy + 6y^{2}. (Co-efficient of x^{2 }= 12, co-efficient of x = -17y and the constant term = 6y^{2}) We will split the co-efficient of x into two parts such that their sum is -17y and their product equals to the product of the co-efficient of x^{2} and the constant term, i.e., 12 × 6y^{2} = 72y^{2} Now, (-9y) + (-8y) = -17y And (-9y) × (-8y) = 72y^{2} Replacing the middle term -17xy by -9xy – 8xy, we have: 12x^{2 }– 17xy + 6y^{2 }= 12x^{2 }– 9xy – 8xy + 6y^{2} = (12x^{2 }– 9xy) – (8xy + 6y^{2}) = 3x(4x – 3y) -2y(4x – 3y) = (4x – 3y)( 3x – 2y) Question: 20 Resolve of the following quadratic equation trinomials into factors: 6x^{2 }– 5xy – 6y^{2} Solution: The given expression is 6x^{2 }– 5xy – 6y^{2}. (Co-efficient of x^{2 }= 6, co-efficient of x = -5y and the constant term = – 6y^{2}) We will split the co-efficient of x into two parts such that their sum is -17y and their product equals to the product of the co-efficient of x^{2} and the constant term, i.e., 6 × (– 6y^{2}) = –36y^{2} Now, (-9y) + (4y) = - 5y And (-9y) × (4y) = -36y^{2} Replacing the middle term -5xy by -9xy + 4xy, we have: 6x^{2 }– 5xy – 6y^{2 }= 6x^{2 }– 9xy + 4xy – 6y^{2} = (6x^{2 }– 9xy) + (4xy – 6y^{2}) = 3x(2x – 3y) + 2y(2x – 3y) = (2x – 3y)(3x + 2y) Question: 13 Resolve of the following quadratic equation trinomials into factors: 6x^{2 }– 13xy + 2y^{2} Solution: The given expression is 6x^{2 }– 13xy + 2y^{2}. (Co-efficient of x^{2 }= 6, co-efficient of x = -13y and the constant term = 2y^{2}) We will split the co-efficient of x into two parts such that their sum is -13y and their product equals to the product of the co-efficient of x^{2} and the constant term, i.e., 6 × ( 2y^{2} ) = 12y^{2} Now, (-12y) + (-y) = -13y And (-12y) × (-y) = 12y^{2} Replacing the middle term -13xy by -12xy – xy, we have: 6x^{2 }– 13xy + 2y^{2 }= 6x^{2 }– 12xy – xy + 2y^{2} = (6x^{2 }– 12xy) – (xy – 2y^{2}) = 6x(x – 2y) – y(x – 2y) = (x – 2y)(6x – y) Question: 14 Resolve of the following quadratic equation trinomials into factors: 14x^{2 }+ 11xy – 15y^{2} Solution: The given expression is 14x^{2 }+ 11xy – 15y^{2}. (Co-efficient of x^{2 }= 14, co-efficient of x = 11y and the constant term = -15y^{2}) We will split the co-efficient of x into two parts such that their sum is 11y and their product equals to the product of the co-efficient of x^{2} and the constant term, i.e., 14 × (-15y^{2}) = -210y^{2} Now, (21y) + (-10y) = 11y And (21y) × (-10y) = -210y^{2} Replacing the middle term -11xy by -10xy + 21 xy, we have: 14x^{2 }+ 11xy – 15y^{2 }= 14x^{2 }– 10xy + 21xy – 15y^{2} = (14x^{2 }– 10xy) + (21xy – 15y^{2}) = 2x(7x – 5y) + 3y(7x – 5y) = (7x – 5y)(2x + 3y) Question: 15 Resolve of the following quadratic equation trinomials into factors: 6a^{2 }+ 17ab – 3b^{2} Solution: The given expression is 6a^{2 }+ 17ab – 3b^{2}. (Co-efficient of a^{2 }= 6, co-efficient of a = 17b and the constant term = – 3b^{2}) We will split the co-efficient of x into two parts such that their sum is 17b and their product equals to the product of the co-efficient of a^{2} and the constant term, i.e., 6 × (-3b^{2}) = -18b^{2} Now, (18b) + (-b) = 17b And (18b) × (-b) = -18b^{2} Replacing the middle term 17ab by -ab + 18ab, we have: 6a^{2 }+ 17ab – 3b^{2 }= 6a^{2 }-ab + 18ab - 3b^{2} = (6a^{2 }– ab) + (18ab – 3b^{2}) = a(6a – b) + 3b(6a – b) = (a + 3b)(6a – b) Question: 16 Resolve of the following quadratic equation trinomials into factors: 36a^{2 }+ 12abc – 15b^{2}c^{2} Solution: The given expression is 36a^{2 }+ 12abc – 15b^{2}c^{2}. (Co-efficient of a^{2 }= 36, co-efficient of a = 12bc and the constant term = -15b^{2 }c^{2}) We will split the co-efficient of x into two parts such that their sum is 17b and their product equals to the product of the co-efficient of a^{2} and the constant term, i.e., 36 × ( -15b^{2} c^{2}) = -540b^{2}c^{2} Now, (-18bc) + 30bc = 12bc And (-18bc) × (30bc) = - 540b^{2}c^{2} Replacing the middle term 12abc by -18abc + 30abc, we have: 36a^{2 }+ 12abc – 15b^{2}c^{2 }= 36a^{2 }-18abc + 30abc – 15b^{2}c^{2} = (36a^{2 }-18abc) + (30abc – 15b^{2}c^{2}) = 18a(2a – bc) + 15bc( 2a – bc) = 3(6a + 5bc)(2a – bc) Question: 17 Resolve of the following quadratic equation trinomials into factors: 15x^{2 }– 16xyz – 15y^{2}z^{2} Solution: The given expression is 15x^{2 }– 16xyz – 15y^{2}z^{2}. (Co-efficient of x^{2 }= 15, co-efficient of x = -16yz and the constant term = -15y^{2 }z^{2}) We will split the co-efficient of x into two parts such that their sum is -16yz and their product equals to the product of the co-efficient of x^{2} and the constant term, i.e., 15 × ( -15y^{2} z^{2}) = -225y^{2}z^{2} Now, (-25yz) + 9yz = -16yx And (-25yz) × (9yz) = -225y^{2} z^{2} Replacing the middle term -16xyz by -25xyz + 9xyz, we have: 15x^{2 }- 16xyz – 15y^{2}z^{2 }= 15x^{2 }– 25xyz + 9xyz – 15y^{2}z^{2} = (15x^{2 }- 25xyz) + (9xyz – 15y^{2}z^{2}) = 5x(3x – 5yz) + 3yz(3x – 5yz) = (3x – 5yz)(5x + 3yz) Question: 18 Resolve of the following quadratic equation trinomials into factors: (x – 2y)^{2} – 5(x – 2y) + 6 Solution: The given expression is a^{2} – 5a + 6. Assuming a = x – 2y, we have: (x – 2y)^{2} – 5( x – 2y) + 6 = a^{2} – 5a + 6 (Co-efficient of a^{2} = 1, co-efficient of a = -5 and the constant term = 6) Now, we will split the co-efficient of a into two parts such that their sum is -5 and their product equals to the product of the co-efficient of a^{2} and the constant term, i.e., 1 × 6 = 6. Clearly, (-2) + (-3) = – 5 And, (-2) × (-3) = 6 Replacing the middle term -5a by -2a – 3a, we have: a^{2} – 5a + 6 = a^{2} – 2a – 3a + 6 = (a^{2} – 2a) – (3a – 6) = a( a – 2) – 3 (a – 2) = (a – 2)(a – 3) Replacing a by (x – 2y), we get: (a – 3)(a – 2) = (x – 2y – 3)(x -2y – 2) Question: 19 Resolve of the following quadratic equation trinomials into factors: (2a – b )^{2} + 2(2a – b) – 8 Solution: Assuming x = 2a – b, we have: (2a – b )^{2} + 2(2a – b) – 8 = x^{2} + 2x – 8 The given expression becomes x^{2} + 2x – 8 (Co-efficient of x^{2} = 1 and that of x = 2; constant term = -8) Now, we will split the co-efficient of x into two parts such that their sum is 2 and their product equals the product of the co-efficient of x^{2} and the constant term, i.e., 1 × (- 8) = – 8 Clearly, (-2) + 4 = 2 And, (-2) × 4 = -8 Replacing the middle term 2x by -2x + 4x, we get: x^{2} + 2x – 8 = x^{2} – 2x + 4x – 8 = (x^{2} – 2x) + (4x – 8) = x(x – 2) + 4 (x – 2) = (x – 2)(x + 4) Replacing x by 2a – b, we get: (x + 4)(x – 2) = (2a – b + 4)(2a – b – 2)
Resolve of the following quadratic equation trinomials into factors:
2x^{2 }+ 5x + 3
The given expression is 2x^{2 }+ 5x + 3.
(Co-efficient of x^{2 }= 2, co-efficient of x = 5 and the constant term = 3)
We will split the co-efficient of x into two parts such that their sum is 5 and their product equals to the product of the co-efficient of x^{2} and the constant term, i.e., 2 × 3 = 6.
Now,
2 + 3 = 5 And 2 × 3 = 6
Replacing the middle term 5x by 2x + 3x, we have:
2x^{2 }+ 5x + 3 = 2x^{2 }+ 2x + 3x + 3
= (2x^{2 }+ 2x) + (3x + 3)
= 2x (x + 1) + 3 (x + 1)
= (2x + 3) (x + 1)
2x^{2 }- 3x - 2
The given expression is 2x^{2 }– 3x – 2.
(Co-efficient of x^{2 }= 2, co-efficient of x = -3 and the constant term = -2)
We will split the co-efficient of x into two parts such that their sum is -3 and their product equals to the product of the co-efficient of x^{2} and the constant term, i.e., 2 × (-2) = - 4
(- 4) + 1 = - 3 And (- 4) × 1 = – 4
Replacing the middle term 3x by – 4x + x, we have:
2x^{2 }- 3x - 2 = 2x^{2 }- 4x + x -2
= (2x^{2 }– 4x) + (x - 2)
= 2x(x - 2) + 1(x – 2)
= (x – 2)(2x + 1)
3x^{2 }+ 10x + 3
The given expression is 3x^{2 }+ 10x + 3.
(Co-efficient of x^{2 }= 3, co-efficient of x = 10 and the constant term = 3)
We will split the co-efficient of x into two parts such that their sum is 10 and their product equals to the product of the co-efficient of x^{2} and the constant term, i.e., 3 × 3 = 9
9 + 1 = 10 And 9 × 1 = 9
Replacing the middle term 10x by 9x + x, we have:
3x^{2 }+ 10x + 3 = 3x^{2 }+ 9x + x + 3
= (3x^{2 }+ 9x) + (x + 3)
= 3x(x + 3) + 1(x + 3)
= (x + 3)(3x + 1)
7x – 6 – 2x^{2}
The given expression is 7x – 6 – 2x^{2}.
(Co-efficient of x^{2 }= -2, co-efficient of x = 7 and the constant term = – 6)
We will split the co-efficient of x into two parts such that their sum is 7 and their product equals to the product of the co-efficient of x^{2} and the constant term, i.e., (-2) × (- 6) = 12
4 + 3 = 7 And 4 × 3 = 12
Replacing the middle term 7x by 4x + 3x, we have:
7x - 6 - 2x^{2 }= -2x^{2 }+ 4x + 3x - 6
= (-2x^{2 }+ 4x) + (3x – 6)
= 2x (2 - x) – 3(2 - x)
= (2x – 3) (2 – x)
7x^{2 }– 19x – 6
The given expression is 7x^{2 }– 19x – 6.
(Co-efficient of x^{2 }= 7, co-efficient of x = -19 and the constant term = – 6)
We will split the co-efficient of x into two parts such that their sum is -19 and their product equals to the product of the co-efficient of x^{2} and the constant term, i.e., 7 × (– 6) = 9
(-21) + 2 = -19 And (-21) × 2 = - 42
Replacing the middle term -19x by -21x + 2x, we have:
7x^{2 }– 19x – 6 = 7x^{2 }– 21x + 2x -6
= (7x^{2 }– 21x) + (2x – 6)
= 7x(x – 3) + 2(x – 3)
= (x – 3) (7x + 2)
28 – 31x – 5x^{2}
The given expression is 28 – 31x – 5x^{2}.
(Co-efficient of x^{2 }= -5, co-efficient of x = -31 and the constant term = 28)
We will split the co-efficient of x into two parts such that their sum is -31 and their product equals to the product of the co-efficient of x^{2} and the constant term, i.e., ( -5 ) × ( 28 ) = -140
(-35) + 4 = -31 And (-35) × 4 = -140
Replacing the middle term -31x by -35x + 4x, we have:
28 – 31x – 5x^{2 }= -5x^{2 }– 35x + 4x + 28
= (-5x^{2 }– 35x) + (4x + 28)
= -5x(x + 7) + 4(x + 7)
= (4 - 5x)(x + 7)
3 + 23y – 8y^{2}
The given expression is 3 + 23y – 8y^{2}.
(Co-efficient of y^{2 }= -8, co-efficient of y = 23 and the constant term = 3)
We will split the co-efficient of x into two parts such that their sum is 23 and their product equals to the product of the co-efficient of x^{2} and the constant term, i.e., ( -8 ) × 3 = -24
(-1) + 24 = 23 And (-1) × 24 = -24
Replacing the middle term 23y by -y + 24y, we have:
3 + 23y – 8y^{2 }= -8y^{2 }– y + 24y + 3
= (-8y^{2 }– y) + (24y + 3)
= – y(8y + 1) + 3(8y + 1)
= (8y + 1)( y + z)
11x^{2 }– 54x + 63
The given expression is 11x^{2 }– 54x + 63.
(Co-efficient of x^{2 }= 11, co-efficient of x = – 54 and the constant term = 63)
We will split the co-efficient of x into two parts such that their sum is -19 and their product equals to the product of the co-efficient of x^{2} and the constant term, i.e., 11 × 63 = 693
(-33) + (-21) =- 54 And (-33) × (-21) = 693
Replacing the middle term -54x by -33x – 21x, we have:
11x^{2 }– 54x + 63 = 11x^{2 }– 33x – 21x + 63
= (11x^{2 }– 33x) + (-21x + 63)
= 11x(x – 3) - 21(x – 3)
= (x - 3)( 11x - 21)
7x – 6x^{2 }+ 20
The given expression is 7x – 6x^{2 }+ 20.
(Co-efficient of x^{2 }= -6, co-efficient of x = 7 and the constant term = 20)
We will split the co-efficient of x into two parts such that their sum is -19 and their product equals to the product of the co-efficient of x^{2} and the constant term, i.e., (– 6) × 20 = – 120
(15) + (- 8) = 7 And (15) × (-8) = -120
Replacing the middle term 7x by 15x - 8x, we have:
7x – 6x^{2 }+ 20 = – 6x^{2 }+ 15x - 8x + 20
= (-6x^{2 }+ 15x) + (- 8x + 20)
= 3x (-2x + 5) + 4(- 2x + 5)
= (-2x + 5)(3x + 4)
3x^{2 }+ 22x + 35
The given expression is 3x^{2 }+ 22x + 35.
(Co-efficient of x^{2 }= 3, co-efficient of x = 22 and the constant term = 35)
We will split the co-efficient of x into two parts such that their sum is -19 and their product equals to the product of the co-efficient of x^{2} and the constant term, i.e., 3 × 35 = 105
(15) + (7) = 22 And (15) × (7) = 105
Replacing the middle term 22x by 15x + 7x, we have:
3x^{2 }+ 22x + 35 = 3x^{2 }+ 15x + 7x + 35
= (3x^{2 }+ 15x) + (7x + 35)
= 3x(x + 5) + 7(x + 5)
= (x + 5)( 3x + 7)
12x^{2 }– 17xy + 6y^{2}
The given expression is 12x^{2 }– 17xy + 6y^{2}.
(Co-efficient of x^{2 }= 12, co-efficient of x = -17y and the constant term = 6y^{2})
We will split the co-efficient of x into two parts such that their sum is -17y and their product equals to the product of the co-efficient of x^{2} and the constant term, i.e., 12 × 6y^{2} = 72y^{2}
(-9y) + (-8y) = -17y And (-9y) × (-8y) = 72y^{2}
Replacing the middle term -17xy by -9xy – 8xy, we have:
12x^{2 }– 17xy + 6y^{2 }= 12x^{2 }– 9xy – 8xy + 6y^{2}
= (12x^{2 }– 9xy) – (8xy + 6y^{2})
= 3x(4x – 3y) -2y(4x – 3y)
= (4x – 3y)( 3x – 2y)
6x^{2 }– 5xy – 6y^{2}
The given expression is 6x^{2 }– 5xy – 6y^{2}.
(Co-efficient of x^{2 }= 6, co-efficient of x = -5y and the constant term = – 6y^{2})
We will split the co-efficient of x into two parts such that their sum is -17y and their product equals to the product of the co-efficient of x^{2} and the constant term, i.e., 6 × (– 6y^{2}) = –36y^{2}
(-9y) + (4y) = - 5y And (-9y) × (4y) = -36y^{2}
Replacing the middle term -5xy by -9xy + 4xy, we have:
6x^{2 }– 5xy – 6y^{2 }= 6x^{2 }– 9xy + 4xy – 6y^{2}
= (6x^{2 }– 9xy) + (4xy – 6y^{2})
= 3x(2x – 3y) + 2y(2x – 3y)
= (2x – 3y)(3x + 2y)
6x^{2 }– 13xy + 2y^{2}
The given expression is 6x^{2 }– 13xy + 2y^{2}.
(Co-efficient of x^{2 }= 6, co-efficient of x = -13y and the constant term = 2y^{2})
We will split the co-efficient of x into two parts such that their sum is -13y and their product equals to the product of the co-efficient of x^{2} and the constant term, i.e., 6 × ( 2y^{2} ) = 12y^{2}
(-12y) + (-y) = -13y And (-12y) × (-y) = 12y^{2}
Replacing the middle term -13xy by -12xy – xy, we have:
6x^{2 }– 13xy + 2y^{2 }= 6x^{2 }– 12xy – xy + 2y^{2}
= (6x^{2 }– 12xy) – (xy – 2y^{2})
= 6x(x – 2y) – y(x – 2y)
= (x – 2y)(6x – y)
14x^{2 }+ 11xy – 15y^{2}
The given expression is 14x^{2 }+ 11xy – 15y^{2}.
(Co-efficient of x^{2 }= 14, co-efficient of x = 11y and the constant term = -15y^{2})
We will split the co-efficient of x into two parts such that their sum is 11y and their product equals to the product of the co-efficient of x^{2} and the constant term, i.e., 14 × (-15y^{2}) = -210y^{2}
(21y) + (-10y) = 11y And (21y) × (-10y) = -210y^{2}
Replacing the middle term -11xy by -10xy + 21 xy, we have:
14x^{2 }+ 11xy – 15y^{2 }= 14x^{2 }– 10xy + 21xy – 15y^{2}
= (14x^{2 }– 10xy) + (21xy – 15y^{2})
= 2x(7x – 5y) + 3y(7x – 5y)
= (7x – 5y)(2x + 3y)
6a^{2 }+ 17ab – 3b^{2}
The given expression is 6a^{2 }+ 17ab – 3b^{2}.
(Co-efficient of a^{2 }= 6, co-efficient of a = 17b and the constant term = – 3b^{2})
We will split the co-efficient of x into two parts such that their sum is 17b and their product equals to the product of the co-efficient of a^{2} and the constant term, i.e., 6 × (-3b^{2}) = -18b^{2}
(18b) + (-b) = 17b And (18b) × (-b) = -18b^{2}
Replacing the middle term 17ab by -ab + 18ab, we have:
6a^{2 }+ 17ab – 3b^{2 }= 6a^{2 }-ab + 18ab - 3b^{2}
= (6a^{2 }– ab) + (18ab – 3b^{2})
= a(6a – b) + 3b(6a – b)
= (a + 3b)(6a – b)
36a^{2 }+ 12abc – 15b^{2}c^{2}
The given expression is 36a^{2 }+ 12abc – 15b^{2}c^{2}.
(Co-efficient of a^{2 }= 36, co-efficient of a = 12bc and the constant term = -15b^{2 }c^{2})
We will split the co-efficient of x into two parts such that their sum is 17b and their product equals to the product of the co-efficient of a^{2} and the constant term, i.e., 36 × ( -15b^{2} c^{2}) = -540b^{2}c^{2}
(-18bc) + 30bc = 12bc And (-18bc) × (30bc) = - 540b^{2}c^{2}
Replacing the middle term 12abc by -18abc + 30abc, we have:
36a^{2 }+ 12abc – 15b^{2}c^{2 }= 36a^{2 }-18abc + 30abc – 15b^{2}c^{2}
= (36a^{2 }-18abc) + (30abc – 15b^{2}c^{2})
= 18a(2a – bc) + 15bc( 2a – bc)
= 3(6a + 5bc)(2a – bc)
15x^{2 }– 16xyz – 15y^{2}z^{2}
The given expression is 15x^{2 }– 16xyz – 15y^{2}z^{2}.
(Co-efficient of x^{2 }= 15, co-efficient of x = -16yz and the constant term = -15y^{2 }z^{2})
We will split the co-efficient of x into two parts such that their sum is -16yz and their product equals to the product of the co-efficient of x^{2} and the constant term, i.e., 15 × ( -15y^{2} z^{2}) = -225y^{2}z^{2}
(-25yz) + 9yz = -16yx And (-25yz) × (9yz) = -225y^{2} z^{2}
Replacing the middle term -16xyz by -25xyz + 9xyz, we have:
15x^{2 }- 16xyz – 15y^{2}z^{2 }= 15x^{2 }– 25xyz + 9xyz – 15y^{2}z^{2}
= (15x^{2 }- 25xyz) + (9xyz – 15y^{2}z^{2})
= 5x(3x – 5yz) + 3yz(3x – 5yz)
= (3x – 5yz)(5x + 3yz)
(x – 2y)^{2} – 5(x – 2y) + 6
The given expression is a^{2} – 5a + 6.
Assuming a = x – 2y, we have:
(x – 2y)^{2} – 5( x – 2y) + 6 = a^{2} – 5a + 6
(Co-efficient of a^{2} = 1, co-efficient of a = -5 and the constant term = 6)
Now, we will split the co-efficient of a into two parts such that their sum is -5 and their product equals to the product of the co-efficient of a^{2} and the constant term, i.e., 1 × 6 = 6.
Clearly,
(-2) + (-3) = – 5 And, (-2) × (-3) = 6
Replacing the middle term -5a by -2a – 3a, we have:
a^{2} – 5a + 6 = a^{2} – 2a – 3a + 6
= (a^{2} – 2a) – (3a – 6)
= a( a – 2) – 3 (a – 2)
= (a – 2)(a – 3)
Replacing a by (x – 2y), we get:
(a – 3)(a – 2) = (x – 2y – 3)(x -2y – 2)
(2a – b )^{2} + 2(2a – b) – 8
Assuming x = 2a – b, we have:
(2a – b )^{2} + 2(2a – b) – 8 = x^{2} + 2x – 8
The given expression becomes x^{2} + 2x – 8
(Co-efficient of x^{2} = 1 and that of x = 2; constant term = -8)
Now, we will split the co-efficient of x into two parts such that their sum is 2 and their product equals the product of the co-efficient of x^{2} and the constant term, i.e., 1 × (- 8) = – 8
(-2) + 4 = 2 And, (-2) × 4 = -8
Replacing the middle term 2x by -2x + 4x, we get:
x^{2} + 2x – 8 = x^{2} – 2x + 4x – 8
= (x^{2} – 2x) + (4x – 8)
= x(x – 2) + 4 (x – 2)
= (x – 2)(x + 4)
Replacing x by 2a – b, we get:
(x + 4)(x – 2) = (2a – b + 4)(2a – b – 2)
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Chapter 7: Factorization Exercise – 7.5...