Resolve of the following quadratic equation trinomials into factors:
2x2 + 5x + 3
The given expression is 2x2 + 5x + 3.
(Co-efficient of x2 = 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 x2 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:
2x2 + 5x + 3 = 2x2 + 2x + 3x + 3
= (2x2 + 2x) + (3x + 3)
= 2x (x + 1) + 3 (x + 1)
= (2x + 3) (x + 1)
Resolve of the following quadratic equation trinomials into factors:
2x2 - 3x - 2
The given expression is 2x2 – 3x – 2.
(Co-efficient of x2 = 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 x2 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:
2x2 - 3x - 2 = 2x2 - 4x + x -2
= (2x2 – 4x) + (x - 2)
= 2x(x - 2) + 1(x – 2)
= (x – 2)(2x + 1)
Resolve of the following quadratic equation trinomials into factors:
3x2 + 10x + 3
The given expression is 3x2 + 10x + 3.
(Co-efficient of x2 = 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 x2 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:
3x2 + 10x + 3 = 3x2 + 9x + x + 3
= (3x2 + 9x) + (x + 3)
= 3x(x + 3) + 1(x + 3)
= (x + 3)(3x + 1)
Resolve of the following quadratic equation trinomials into factors:
7x – 6 – 2x2
The given expression is 7x – 6 – 2x2.
(Co-efficient of x2 = -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 x2 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 - 2x2 = -2x2 + 4x + 3x - 6
= (-2x2 + 4x) + (3x – 6)
= 2x (2 - x) – 3(2 - x)
= (2x – 3) (2 – x)
Resolve of the following quadratic equation trinomials into factors:
7x2 – 19x – 6
The given expression is 7x2 – 19x – 6.
(Co-efficient of x2 = 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 x2 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:
7x2 – 19x – 6 = 7x2 – 21x + 2x -6
= (7x2 – 21x) + (2x – 6)
= 7x(x – 3) + 2(x – 3)
= (x – 3) (7x + 2)
Resolve of the following quadratic equation trinomials into factors:
28 – 31x – 5x2
The given expression is 28 – 31x – 5x2.
(Co-efficient of x2 = -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 x2 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 – 5x2 = -5x2 – 35x + 4x + 28
= (-5x2 – 35x) + (4x + 28)
= -5x(x + 7) + 4(x + 7)
= (4 - 5x)(x + 7)
Resolve of the following quadratic equation trinomials into factors:
3 + 23y – 8y2
The given expression is 3 + 23y – 8y2.
(Co-efficient of y2 = -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 x2 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 – 8y2 = -8y2 – y + 24y + 3
= (-8y2 – y) + (24y + 3)
= – y(8y + 1) + 3(8y + 1)
= (8y + 1)( y + z)
Resolve of the following quadratic equation trinomials into factors:
11x2 – 54x + 63
The given expression is 11x2 – 54x + 63.
(Co-efficient of x2 = 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 x2 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:
11x2 – 54x + 63 = 11x2 – 33x – 21x + 63
= (11x2 – 33x) + (-21x + 63)
= 11x(x – 3) - 21(x – 3)
= (x - 3)( 11x - 21)
Resolve of the following quadratic equation trinomials into factors:
7x – 6x2 + 20
The given expression is 7x – 6x2 + 20.
(Co-efficient of x2 = -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 x2 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 – 6x2 + 20 = – 6x2 + 15x - 8x + 20
= (-6x2 + 15x) + (- 8x + 20)
= 3x (-2x + 5) + 4(- 2x + 5)
= (-2x + 5)(3x + 4)
Resolve of the following quadratic equation trinomials into factors:
3x2 + 22x + 35
The given expression is 3x2 + 22x + 35.
(Co-efficient of x2 = 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 x2 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:
3x2 + 22x + 35 = 3x2 + 15x + 7x + 35
= (3x2 + 15x) + (7x + 35)
= 3x(x + 5) + 7(x + 5)
= (x + 5)( 3x + 7)
Resolve of the following quadratic equation trinomials into factors:
12x2 – 17xy + 6y2
The given expression is 12x2 – 17xy + 6y2.
(Co-efficient of x2 = 12, co-efficient of x = -17y and the constant term = 6y2)
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 x2 and the constant term, i.e., 12 × 6y2 = 72y2
Now,
(-9y) + (-8y) = -17y And (-9y) × (-8y) = 72y2
Replacing the middle term -17xy by -9xy – 8xy, we have:
12x2 – 17xy + 6y2 = 12x2 – 9xy – 8xy + 6y2
= (12x2 – 9xy) – (8xy + 6y2)
= 3x(4x – 3y) -2y(4x – 3y)
= (4x – 3y)( 3x – 2y)
Resolve of the following quadratic equation trinomials into factors:
6x2 – 5xy – 6y2
The given expression is 6x2 – 5xy – 6y2.
(Co-efficient of x2 = 6, co-efficient of x = -5y and the constant term = – 6y2)
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 x2 and the constant term, i.e., 6 × (– 6y2) = –36y2
Now,
(-9y) + (4y) = - 5y And (-9y) × (4y) = -36y2
Replacing the middle term -5xy by -9xy + 4xy, we have:
6x2 – 5xy – 6y2 = 6x2 – 9xy + 4xy – 6y2
= (6x2 – 9xy) + (4xy – 6y2)
= 3x(2x – 3y) + 2y(2x – 3y)
= (2x – 3y)(3x + 2y)
Resolve of the following quadratic equation trinomials into factors:
6x2 – 13xy + 2y2
The given expression is 6x2 – 13xy + 2y2.
(Co-efficient of x2 = 6, co-efficient of x = -13y and the constant term = 2y2)
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 x2 and the constant term, i.e., 6 × ( 2y2 ) = 12y2
Now,
(-12y) + (-y) = -13y And (-12y) × (-y) = 12y2
Replacing the middle term -13xy by -12xy – xy, we have:
6x2 – 13xy + 2y2 = 6x2 – 12xy – xy + 2y2
= (6x2 – 12xy) – (xy – 2y2)
= 6x(x – 2y) – y(x – 2y)
= (x – 2y)(6x – y)
Resolve of the following quadratic equation trinomials into factors:
14x2 + 11xy – 15y2
The given expression is 14x2 + 11xy – 15y2.
(Co-efficient of x2 = 14, co-efficient of x = 11y and the constant term = -15y2)
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 x2 and the constant term, i.e., 14 × (-15y2) = -210y2
Now,
(21y) + (-10y) = 11y And (21y) × (-10y) = -210y2
Replacing the middle term -11xy by -10xy + 21 xy, we have:
14x2 + 11xy – 15y2 = 14x2 – 10xy + 21xy – 15y2
= (14x2 – 10xy) + (21xy – 15y2)
= 2x(7x – 5y) + 3y(7x – 5y)
= (7x – 5y)(2x + 3y)
Resolve of the following quadratic equation trinomials into factors:
6a2 + 17ab – 3b2
The given expression is 6a2 + 17ab – 3b2.
(Co-efficient of a2 = 6, co-efficient of a = 17b and the constant term = – 3b2)
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 a2 and the constant term, i.e., 6 × (-3b2) = -18b2
Now,
(18b) + (-b) = 17b And (18b) × (-b) = -18b2
Replacing the middle term 17ab by -ab + 18ab, we have:
6a2 + 17ab – 3b2 = 6a2 -ab + 18ab - 3b2
= (6a2 – ab) + (18ab – 3b2)
= a(6a – b) + 3b(6a – b)
= (a + 3b)(6a – b)
Resolve of the following quadratic equation trinomials into factors:
36a2 + 12abc – 15b2c2
The given expression is 36a2 + 12abc – 15b2c2.
(Co-efficient of a2 = 36, co-efficient of a = 12bc and the constant term = -15b2 c2)
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 a2 and the constant term, i.e., 36 × ( -15b2 c2) = -540b2c2
Now,
(-18bc) + 30bc = 12bc And (-18bc) × (30bc) = - 540b2c2
Replacing the middle term 12abc by -18abc + 30abc, we have:
36a2 + 12abc – 15b2c2 = 36a2 -18abc + 30abc – 15b2c2
= (36a2 -18abc) + (30abc – 15b2c2)
= 18a(2a – bc) + 15bc( 2a – bc)
= 3(6a + 5bc)(2a – bc)
Resolve of the following quadratic equation trinomials into factors:
15x2 – 16xyz – 15y2z2
The given expression is 15x2 – 16xyz – 15y2z2.
(Co-efficient of x2 = 15, co-efficient of x = -16yz and the constant term = -15y2 z2)
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 x2 and the constant term, i.e., 15 × ( -15y2 z2) = -225y2z2
Now,
(-25yz) + 9yz = -16yx And (-25yz) × (9yz) = -225y2 z2
Replacing the middle term -16xyz by -25xyz + 9xyz, we have:
15x2 - 16xyz – 15y2z2 = 15x2 – 25xyz + 9xyz – 15y2z2
= (15x2 - 25xyz) + (9xyz – 15y2z2)
= 5x(3x – 5yz) + 3yz(3x – 5yz)
= (3x – 5yz)(5x + 3yz)
Resolve of the following quadratic equation trinomials into factors:
(x – 2y)2 – 5(x – 2y) + 6
The given expression is a2 – 5a + 6.
Assuming a = x – 2y, we have:
(x – 2y)2 – 5( x – 2y) + 6 = a2 – 5a + 6
(Co-efficient of a2 = 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 a2 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:
a2 – 5a + 6 = a2 – 2a – 3a + 6
= (a2 – 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)
Resolve of the following quadratic equation trinomials into factors:
(2a – b )2 + 2(2a – b) – 8
Assuming x = 2a – b, we have:
(2a – b )2 + 2(2a – b) – 8 = x2 + 2x – 8
The given expression becomes x2 + 2x – 8
(Co-efficient of x2 = 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 x2 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:
x2 + 2x – 8 = x2 – 2x + 4x – 8
= (x2 – 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)