Middle Term Splitting Questions: Middle Term Splitting Questions are one of the most common but also important types of questions in Maths. Middle Term Splitting Questions are asked at various stages of a person’s educational journey.
Middle Term Splitting Questions are required from class 7th to CAT Exams. Thus, in this article, we provide you with 25 Middle Term Splitting Questions with solutions. The level of difficulty of the questions will increase gradually. These Middle Term Splitting questions are unique and prepared by our team.
To factorize this expression, we need to find two numbers α and β such that α + β = –12 and αβ = 36
4x2 – 6x – 6x + 9
2x(x – 3) – 3(x – 3)
(2x – 3)(x – 3)
To factorize this expression, we need to find two numbers α and β such that α + β = –28 and αβ = 140
10y2 – 14y – 10y + 14
2y(5y – 7) – 2(5y – 7)
(2y – 2)(5y – 7)
2(y – 1)(5y – 7)
To factorize this expression, we need to find two numbers α and β such that α + β = –10 and αβ = 25
x2 – 5x – 5x + 25
x(x – 5) – 5(x – 5)
(x – 5)(x – 5)
(x – 5)2
View Solution To factorise the above equation, we need to find two numbers α and β such that α + β = 8 and αβ = 16 y2 + 4y + 2y + 16 y(y + 4) + 2(y + 4) (y + 2)(y + 4)
View Solution To factorize this expression, we need to find two numbers α and β such that α + β = -4 and αβ = -12 z2 – 6z + 2z – 12 z(z – 6) + 2(z – 6) (z + 2)(z – 6)
View Solution To factorize this expression, we need to find two numbers α and β such that α + β = 34 and αβ = 225 25x2 + 9x + 25x + 9 x(25x + 9) + 1(25x + 9) (25x + 9)(x + 1)
View Solution To factorize this expression, we need to find two numbers α and β such that α + β = 2 and αβ = 1 x4 + x2y2 + x2y2 + y4 x2(x2 + y2) + y2(x2 + y2) (x2 + y2)(x2 + y2) (x2 + y2)2
View Solution To factorize this expression, we need to find two numbers α and β such that α + β = 84 and αβ = 1764 49a2 + 42ab + 42ab + 36b2 7a(7a + 6b) + 6b(7a + 6b) (7a + 6b)(7a + 6b) (7a + 6b)2
View Solution a2 + b2 + 2ab – 4ab a2 + b2 – 2ab a2 – ab + b2 – ab a(a –b) – b(a – b) (a – b)(a – b) (a – b)2
View Solution To factorize the expression 121x2 – 88xy + 16y2, we can look for two numbers α and β such that α + β = -88 and αβ = 16 * 121. The numbers that satisfy these conditions are -44 and -44 (-44 + -44 = -88 and -44 * -44 = 16 * 121 = 1936). 121x2 – 44xy – 44xy + 16y2 (121x2 – 44xy) – (44xy – 16y2) 11x(11x – 4y) – 4y(11x – 4y) (11x – 4y)(11x – 4y) (11x – 4y)2
View Solution To factorize this expression, we need to find two numbers α and β such that α + β = -6 and αβ = 9 1 – 3x – 3x + 9x2 (1 – 3x) – 3x(1 – 3x) (1 – 3x)(1 – 3x) (1 – 3x)2
View Solution To factorize this expression, we need to find two numbers α and β such that α + β = 0.9 and αβ = 0.2 0.4x2 + 0.4x + 0.5x + 0.5 0.4 x(x+1) + 0.5 (x+1) (0.4x + 0.5)(x+1)
View Solution We need to find two numbers α and β such that α + β = -80 and αβ = 16 * 100. To factorize the expression, we split the middle term -80xy into two terms: -64xy and -16xy. 100x2 – 64xy – 16xy + 16y2 (100x2 – 64xy) + (-16xy + 16y2) 4x(25x – 16y) – 16y(25x – 16y) (25x – 16y)(4x – 16y)
View Solution To factorize this expression, we need to find two numbers α and β such that α + β = 5 and αβ = -14. We can factorize the expression as follows: x2y2 – 2xyz + 7xyz – 14z2 (x2y2 – 2xyz) + (7xyz – 14z2) xy(xy – 2z) + 7z(xy – 2z) (xy – 2z)(xy + 7z)
View Solution 3(xy)2 – (xy)(yz) – 24(yz)2 To factorize this expression, we need to find two numbers α and β such that α + β = 1 and αβ = 72 3(xy)2 – 9(xy)(yz) + 8(xy)(yz) – 24(yz)2 3xy2(x – 3z) + 8y2z(x – 3z) y2[3x(x – 3z) + 8z(x – 3z)]
y2(x – 3z)(3x + 8z) y2(x – 3z)(3x + 8z)
View Solution To factorize this expression, we need to find two numbers α and β such that α + β = -6 and αβ = 9 11(a + b)2 + 22(a+b)(c + d) – (a+b)(c + d) – 2(c + d)2 11(a + b)[(a+b) + 2(c + d)] – (c + d)[(a + b) + 2(c + d)]
(a + b + 2c + 2d)(11a + 11b – c – d)
View Solution Writing the above equation as: x2 + (2/x)2 + 2 × x × (2/x) Using the formula (a + b)2 = a2 + 2ab + b2, we get (x + 2/x)2
View Solution √3a2 + 3a + 2a + 2√3 √3a(a + √3) + 2(a + √3) (√3a + 2)(a + √3)
View Solution Rewriting the above equation as (2x)2 + 1/(4x)2 + 2 × 2x × 1/4x Using the formula (a + b)2 = a2 + 2ab + b2, we get (2x + 1/4x)2
View Solution Rewriting the above equation as (s2 + r2 + 2sr) + 1/4(s + r)2 + 2 × (s + r)2 × 1/2(s + r)2 (s + r)2 + 1/{2(s + r)2} + 2 × (s + r)2 × 1/2(s + r)2 Using the formula (a + b)2 = a2 + 2ab + b2, we get {(s + r) + 1/2(s + r)}2
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