factorising-a-quadratic-by-splitting-the-middle-term MCQs for UPSC Prelims

25 practice questions on factorising-a-quadratic-by-splitting-the-middle-term from the Exploring Algebraic Identities section of the UPSC Prelims syllabus. 25 come with a written explanation. Try the sample set below - the answer stays hidden until you ask for it.

8 Easy 12 Medium 5 Hard

Sample questions

Q1
hard
A rectangle has area $x^2-11x+28$ square units, and one of its sides has length $x-4$ units. What is the length of the other side?
  1. A $x-24$
  2. B $x+7$
  3. C $x-7$
  4. D $x-4$
Show answer and explanation

Correct answer: C - $x-7$

Factorising, the constants must multiply to $28$ and add to $-11$, so $x^2-11x+28=(x-4)(x-7)$ and the other side is $x-7$. Checking at $x=10$: the area is $100-110+28=18$, one side is $6$, and $18\div 6=3=10-7$. A side of $x+7$ would give $102$ at $x=10$, and $x-4$ or $x-24$ do not pair with $x-4$ to give the right constant term.
Q2
hard
For which positive value of $k$ is $16x^2+kxy+9y^2$ a perfect-square trinomial?
  1. A $24$
  2. B $48$
  3. C $144$
  4. D $12$
Show answer and explanation

Correct answer: A - $24$

The outer terms are the squares of $4x$ and $3y$, so the middle term must be $2\times 4x\times 3y=24xy$, giving $k=24$ and $(4x+3y)^2$. At $x=y=1$ that is $16+24+9=49=7^2$. Using $12$ forgets the doubling and gives $37$, $48$ doubles twice and gives $73$, and $144$ squares the product $12$ instead of doubling it.
Q3
medium
Factorise $9m^2-30mn+25n^2$.
  1. A $(3m+5n)^2$
  2. B $(9m-25n)^2$
  3. C $(3m-5n)(3m+5n)$
  4. D $(3m-5n)^2$
Show answer and explanation

Correct answer: D - $(3m-5n)^2$

The outer terms are the squares of $3m$ and $5n$, and $2\times 3m\times 5n=30mn$ with a minus sign, so the expression is $(3m-5n)^2$. At $m=n=1$ it equals $9-30+25=4$, and $(-2)^2=4$. The plus version gives $64$, the pair of different brackets gives $-16$, and keeping $9$ and $25$ inside the bracket gives $256$.
Q4
easy
Factorise $r^2-r-42$.
  1. A $(r+6)(r-7)$
  2. B $(r+6)(r+7)$
  3. C $(r-6)(r+7)$
  4. D $(r-6)(r-7)$
Show answer and explanation

Correct answer: A - $(r+6)(r-7)$

The two constants must multiply to $-42$ and add to $-1$, and $6$ with $-7$ does both. At $r=1$ the expression is $1-1-42=-42$, and $7\times(-6)=-42$. Swapping the signs gives a middle term of $+r$ and the value $-40$, while making both signs the same gives $56$ or $30$.
Q5
easy
Which of these expressions is a perfect-square trinomial?
  1. A $x^2+10x-25$
  2. B $x^2+10x+25$
  3. C $x^2+7x+12$
  4. D $x^2+10x+16$
Show answer and explanation

Correct answer: B - $x^2+10x+25$

A perfect-square trinomial has the form $a^2+2ab+b^2$, and $x^2+10x+25$ fits with $a=x$ and $b=5$, since $2\times x\times 5=10x$; it factorises as $(x+5)^2$. The expression ending in $16$ would need a middle term of $8x$, the one ending in $12$ factorises as $(x+3)(x+4)$ with two different brackets, and a negative constant can never be the square $b^2$.
Q6
hard
Factorise $2x^2-2x-84$ as far as possible.
  1. A $(x+6)(x-7)$
  2. B $2(x-6)(x+7)$
  3. C $2(x+6)(x+7)$
  4. D $2(x+6)(x-7)$
Show answer and explanation

Correct answer: D - $2(x+6)(x-7)$

Every term has a factor $2$, so the expression is $2(x^2-x-42)$, and $x^2-x-42=(x+6)(x-7)$ because $6$ and $-7$ multiply to $-42$ and add to $-1$. At $x=1$ the expression equals $-84$, and $2\times 7\times(-6)=-84$. Swapping the signs gives $-80$, dropping the factor $2$ gives $-42$, and making both signs positive gives $112$.
Q7
medium
Factorise $10a^2-9a+2$.
  1. A $(5a-1)(2a-2)$
  2. B $(10a-1)(a-2)$
  3. C $(5a+2)(2a+1)$
  4. D $(5a-2)(2a-1)$
Show answer and explanation

Correct answer: D - $(5a-2)(2a-1)$

The product needed is $10\times 2=20$ and the sum is $-9$, giving $-4$ and $-5$, so $10a^2-4a-5a+2=2a(5a-2)-1(5a-2)=(5a-2)(2a-1)$. At $a=1$ the expression is $10-9+2=3$, and $3\times 1=3$. The other brackets give $0$, $-9$ and $21$ at $a=1$, matching middle terms of $-12a$, $-21a$ and $+9a$.
Q8
easy
Factorise $49g^2+14gh+h^2$.
  1. A $(7g+h)(7g-h)$
  2. B $(49g+h)^2$
  3. C $(7g+h)^2$
  4. D $(7g-h)^2$
Show answer and explanation

Correct answer: C - $(7g+h)^2$

The outer terms are the squares of $7g$ and $h$, and the middle term $2\times 7g\times h=14gh$ matches, so the expression is $(7g+h)^2$. At $g=h=1$ it equals $49+14+1=64$, and $8^2=64$. The version with a minus sign gives $36$, keeping $49$ inside the bracket gives $2500$, and the pair of different brackets gives $48$.

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