Solution by factorisation — Class 10 CBSE Mathematics MCQs with Solutions
Free Class 10 CBSE Mathematics Solution by factorisation MCQs with step-by-step solutions (8 questions). Part of Quadratic Equations. Practise online on Prepizo — no login needed.
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Questions with solutions
Q1 — Solution by factorisation · easy · theory
The roots of the equation $x^2 - 3x - 10 = 0$ are:
A. $-5$ and $2$
B. $5$ and $-2$ ✓ Correct
C. $5$ and $2$
D. $-5$ and $-2$
Solution: Splitting the middle term: $x^2 - 5x + 2x - 10 = 0$ gives $x(x-5) + 2(x-5) = 0$, i.e. $(x-5)(x+2) = 0$. So $x = 5$ or $x = -2$. Check: $25 - 15 - 10 = 0$ and $4 + 6 - 10 = 0$.
Q2 — Solution by factorisation · easy · theory
The roots of $x^2 - 7x = 0$ are:
A. $7$ only
B. $7$ and $-7$
C. $0$ and $-7$
D. $0$ and $7$ ✓ Correct
Solution: Taking $x$ common: $x(x - 7) = 0$, so $x = 0$ or $x = 7$. A common mistake is to divide both sides by $x$, which loses the root $x = 0$ — a quadratic equation always has two roots (counting repetition).
Q3 — Solution by factorisation · medium · theory
By splitting the middle term, the roots of $6x^2 - x - 2 = 0$ are:
A. $\frac{2}{3}$ and $-\frac{1}{2}$ ✓ Correct
B. $-\frac{2}{3}$ and $\frac{1}{2}$
C. $\frac{3}{2}$ and $-2$
D. $\frac{2}{3}$ and $\frac{1}{2}$
Solution: We need two numbers whose product is $6 \times (-2) = -12$ and sum is $-1$: they are $-4$ and $3$. So $6x^2 - 4x + 3x - 2 = 0$ gives $2x(3x - 2) + 1(3x - 2) = 0$, i.e. $(3x - 2)(2x + 1) = 0$. Hence $x = \frac{2}{3}$ or $x = -\frac{1}{2}$.
Q4 — Solution by factorisation · easy · theory
The solutions of the equation $x^2 - 49 = 0$ are:
A. $x = 7$ only
B. $x = -7$ only
C. $x = 7$ and $x = -7$ ✓ Correct
D. $x = 49$ and $x = -49$
Solution: Using the difference of squares, $x^2 - 49 = (x - 7)(x + 7) = 0$, so $x = 7$ or $x = -7$. Remember that taking the square root of both sides gives both the positive and the negative root — writing only $x = 7$ misses a solution.
Q5 — Solution by factorisation · medium · theory
The roots of $\sqrt{2}x^2 + 7x + 5\sqrt{2} = 0$ are:
A. $\sqrt{2}$ and $\frac{5}{\sqrt{2}}$
B. $-\sqrt{2}$ and $-\frac{5}{\sqrt{2}}$ ✓ Correct
C. $-\sqrt{2}$ and $\frac{5}{\sqrt{2}}$
D. $\sqrt{2}$ and $-\frac{5}{\sqrt{2}}$
Solution: Split the middle term: $\sqrt{2}x^2 + 2x + 5x + 5\sqrt{2} = 0$ (since $2 \times 5 = 10 = \sqrt{2} \times 5\sqrt{2}$). Grouping: $\sqrt{2}x(x + \sqrt{2}) + 5(x + \sqrt{2}) = 0$, so $(x + \sqrt{2})(\sqrt{2}x + 5) = 0$. Hence $x = -\sqrt{2}$ or $x = -\frac{5}{\sqrt{2}}$; both roots are negative because all coefficients are positive.
Q6 — Solution by factorisation · hard · theory
The difference of the squares of two positive numbers is 180 and the square of the smaller number is 8 times the larger number. The numbers are:
A. $16$ and $10$
B. $18$ and $10$
C. $20$ and $12$
D. $18$ and $12$ ✓ Correct
Solution: Let the larger number be $x$. The square of the smaller number is $8x$, so $x^2 - 8x = 180$, i.e. $x^2 - 8x - 180 = 0$. Factorising: $(x - 18)(x + 10) = 0$, so $x = 18$ (rejecting $-10$ as the numbers are positive). Then the smaller number is $\sqrt{8 \times 18} = \sqrt{144} = 12$; check: $18^2 - 12^2 = 324 - 144 = 180$.
Q7 — Solution by factorisation · hard · theory
A cottage industry produces a certain number of pottery articles in a day. The cost of production of each article (in ₹) was 3 more than twice the number of articles produced. If the total cost of production on that day was ₹90, the number of articles produced was:
A. $6$ ✓ Correct
B. $15$
C. $5$
D. $9$
Solution: Let the number of articles be $x$; then the cost of each article is $₹(2x + 3)$. Total cost: $x(2x + 3) = 90$, i.e. $2x^2 + 3x - 90 = 0$. Factorising: $2x^2 + 15x - 12x - 90 = (2x + 15)(x - 6) = 0$, so $x = 6$ (rejecting the negative root). Check: each article costs ₹15 and $6 \times 15 = 90$.
Q8 — Solution by factorisation · hard · theory
The values of $x$ satisfying $\frac{1}{x+4} - \frac{1}{x-7} = \frac{11}{30}$, where $x \neq -4, 7$, are:
A. $-1$ and $-2$
B. $1$ and $-2$
C. $1$ and $2$ ✓ Correct
D. $3$ and $-1$
Solution: Combining the fractions: $\frac{(x-7) - (x+4)}{(x+4)(x-7)} = \frac{-11}{(x+4)(x-7)} = \frac{11}{30}$, so $(x+4)(x-7) = -30$. Expanding: $x^2 - 3x - 28 = -30$, i.e. $x^2 - 3x + 2 = 0$, which factorises as $(x-1)(x-2) = 0$. So $x = 1$ or $x = 2$; check for $x = 1$: $\frac{1}{5} + \frac{1}{6} = \frac{11}{30}$.