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.

## 20 Middle Term Splitting Questions

###### Ques 1) Factorise 4x^{2} – 12x + 9

View SolutionTo factorize this expression, we need to find two numbers α and β such that α + β = –12 and αβ = 36

4x^{2} – 6x – 6x + 9

2x(x – 3) – 3(x – 3)

(2x – 3)(x – 3)

###### Ques 2) Factorise 10y^{2} – 28y + 14

View SolutionTo factorize this expression, we need to find two numbers α and β such that α + β = –28 and αβ = 140

10y^{2} – 14y – 10y + 14

2y(5y – 7) – 2(5y – 7)

(2y – 2)(5y – 7)

2(y – 1)(5y – 7)

###### Ques 3) Factorise x^{2} – 10x + 25

View SolutionTo factorize this expression, we need to find two numbers α and β such that α + β = –10 and αβ = 25

x^{2} – 5x – 5x + 25

x(x – 5) – 5(x – 5)

(x – 5)(x – 5)

(x – 5)^{2}

###### Ques 4) Factorise y^{2} + 8y + 16

View Solution

To factorise the above equation, we need to find two numbers α and β such that α + β = 8 and αβ = 16

y^{2} + 4y + 2y + 16

y(y + 4) + 2(y + 4)

(y + 2)(y + 4)

###### Ques 5) Factorise z^{2} – 4z –12

View Solution

To factorize this expression, we need to find two numbers α and β such that α + β = -4 and αβ = -12

z^{2} – 6z + 2z – 12

z(z – 6) + 2(z – 6)

(z + 2)(z – 6)

###### Ques 6) Factorise 25x^{2} + 34x + 9

View Solution

To factorize this expression, we need to find two numbers α and β such that α + β = 34 and αβ = 225

25x^{2} + 9x + 25x + 9

x(25x + 9) + 1(25x + 9)

(25x + 9)(x + 1)

###### Ques 7) Factorise x^{4} + 2x^{2}y^{2} + y^{4}

View Solution

To factorize this expression, we need to find two numbers α and β such that α + β = 2 and αβ = 1

x^{4} + x^{2}y^{2} + x^{2}y^{2} + y^{4}

x^{2}(x^{2} + y^{2}) + y^{2}(x^{2} + y^{2})

(x^{2} + y^{2})(x^{2} + y^{2})

(x^{2} + y^{2})^{2}

###### Ques 8) Factorise 49a^{2} + 84ab + 36b^{2}

View Solution

To factorize this expression, we need to find two numbers α and β such that α + β = 84 and αβ = 1764

49a^{2} + 42ab + 42ab + 36b^{2}

7a(7a + 6b) + 6b(7a + 6b)

(7a + 6b)(7a + 6b)

(7a + 6b)^{2}

###### Ques 9) Factorise (a + b)^{2} – 4ab

View Solution

a^{2} + b^{2} + 2ab – 4ab

a^{2} + b^{2} – 2ab

a^{2} – ab + b^{2} – ab

a(a –b) – b(a – b)

(a – b)(a – b)

(a – b)^{2}

###### Ques 10) Factorise 121x^{2} – 88xy + 16y^{2}

View Solution

To factorize the expression 121x^{2} – 88xy + 16y^{2}, 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).

121x^{2} – 44xy – 44xy + 16y^{2}

(121x^{2} – 44xy) – (44xy – 16y^{2})

11x(11x – 4y) – 4y(11x – 4y)

(11x – 4y)(11x – 4y)

(11x – 4y)^{2}

###### Ques 11) Factorise 1 – 6x + 9x^{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 + 9x^{2}

(1 – 3x) – 3x(1 – 3x)

(1 – 3x)(1 – 3x)

(1 – 3x)^{2}

###### Ques 12) Factorise 0.4 x^{2} + 0.9x + 0.5

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

0.4x^{2} + 0.4x + 0.5x + 0.5

0.4 x(x+1) + 0.5 (x+1)

(0.4x + 0.5)(x+1)

###### Ques 13) Factorise 100x^{2} – 80xy + 16y^{2}

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.

100x^{2} – 64xy – 16xy + 16y^{2}

(100x^{2} – 64xy) + (-16xy + 16y^{2})

4x(25x – 16y) – 16y(25x – 16y)

(25x – 16y)(4x – 16y)

###### Ques 14) Factorise x^{2}y^{2} + 5xyz – 14z^{2}

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:

x^{2}y^{2} – 2xyz + 7xyz – 14z^{2}

(x^{2}y^{2} – 2xyz) + (7xyz – 14z^{2})

xy(xy – 2z) + 7z(xy – 2z)

(xy – 2z)(xy + 7z)

###### Ques 15) Factorise 3x^{2}y^{2} – xy^{2}z – 24y^{2}z^{2}

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}

3xy^{2}(x – 3z) + 8y^{2}z(x – 3z)

y^{2}[3x(x – 3z) + 8z(x – 3z)]

y^{2}(x – 3z)(3x + 8z)

y^{2}(x – 3z)(3x + 8z)

###### Ques 16) Factorise 11(a + b)^{2} + 21(a+b)(c + d) – 2(c + d)^{2}

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) + 2(c + d)][11(a + b) – (c + d)]

(a + b + 2c + 2d)(11a + 11b – c – d)

###### Ques 17) Factorise x^{2} + 4/x^{2} + 4

View Solution

Writing the above equation as:

x^{2} + (2/x)^{2} + 2 × x × (2/x)

Using the formula (a + b)^{2} = a^{2} + 2ab + b^{2}, we get

(x + 2/x)^{2}

###### Ques 18) Factorise √3a2 + 5a + 2√3

View Solution

√3a^{2} + 3a + 2a + 2√3

√3a(a + √3) + 2(a + √3)

(√3a + 2)(a + √3)

###### Ques 19) Factorise 4x^{2} + 1/16x^{2} + 1

View Solution

Rewriting the above equation as

(2x)^{2} + 1/(4x)^{2} + 2 × 2x × 1/4x

Using the formula (a + b)^{2} = a^{2} + 2ab + b^{2}, we get

(2x + 1/4x)^{2}

###### Ques 20) Factorise s^{2} + r^{2} + 2sr + 1/4(s + r)^{2} +1

View Solution

Rewriting the above equation as

(s^{2} + r^{2} + 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} = a^{2 }+ 2ab + b^{2}, we get

{(s + r) + 1/2(s + r)}^{2}

We hope that you might have clearly understood the 20 Middle Term Splitting Questions

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