Assertion (A) : The polynomial $p(y) = y^2 + 4y + 3$ has two zeroes.

Reason (R) : A quadratic polynomial can have at most two zeroes.

(a) Both, Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A).

(b) Both, Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A).

(c) Assertion (A) is true, but Reason (R) is false.

(d) Assertion (A) is false, but Reason (R) is true.

Q: What are the zeroes of $y^2 + 4y + 3$?

Factorising gives $(y+1)(y+3) = 0$, so the zeroes are $y = -1$ and $y = -3$.

Polynomials

Zeroes of a Quadratic Polynomial

Options

(A)Both A and R true, R is correct explanation of A

(B)Both A and R true, R is not correct explanation of A

(C)A true, R false

(D)A false, R true

Official Answer

(a) Both Assertion (A) and Reason (R) are true, and Reason (R) correctly explains Assertion (A) — $p(y)$ has exactly 2 zeroes $(-1, -3)$ , the maximum possible for a quadratic.

zeroes of polynomialquadratic polynomialfactorisationdegree of polynomialassertion reasonmaximum zeroes

Marking Scheme

11 mark: correctly selecting option (a); no partial credit in MCQ format.

Hint

Factorise $y^2 + 4y + 3 = (y+1)(y+3)$ to find the two zeroes, then recall that a degree-n polynomial has at most n zeroes.

Quick Oral Answer

$y^2 + 4y + 3$ factorises as $(y+1)(y+3)$ , giving zeroes -1 and -3, so it has two zeroes — the Assertion is true. Since a quadratic can have at most two zeroes, the Reason is also true and correctly explains the Assertion, so the answer is option (a).

Analysis & Explanation

Tests the zero-count rule for quadratic polynomials.

Concept

A degree-n polynomial has at most n real zeroes (Fundamental Theorem of Algebra, real case).
$p(y) = y^2 + 4y + 3$ factorises as $(y+1)(y+3)$ .

Key points

Zeroes of p(y): $y = -1$ and $y = -3$ — exactly two distinct real zeroes ⟹ A is TRUE.
A quadratic (degree 2) can have at most 2 zeroes ⟹ R is TRUE, and it correctly explains why A holds (this quadratic attains the maximum of 2 zeroes).
Hence option (a): both true, R is the correct explanation of A.

Common mistakes

Wrongly assuming R must be unrelated just because it's a general statement — here it directly explains A.
Confusing this with Q19's pattern, where R was true-sounding but factually wrong; here both statements are independently correct and logically linked.

Common Mistakes

1Forgetting to actually factorise/solve $p(y)$ and instead assuming the number of zeroes without verification.
2Confusing 'at most two zeroes' (Reason, a general true rule) with 'exactly two zeroes' and doubting whether R explains A.
3Choosing option (b) out of hesitation, not recognising that R is indeed the textbook justification for why a quadratic can have 2 zeroes.

Interesting Facts

The rule 'a polynomial of degree n has at most n zeroes' is a direct consequence of the Fundamental Theorem of Algebra and is one of the most frequently tested one-line facts in CBSE Class 10 Algebra.

$y^2 + 4y + 3$ can also be solved instantly using the quadratic formula: $y = \frac{-4 \pm \sqrt{16-12}}{2} = \frac{-4 \pm 2}{2}$ , giving -1 and -3, matching the factorisation method.

Graphically, the two zeroes -1 and -3 are exactly the two points where the upward parabola $y = y^2 + 4y + 3$ crosses the x-axis.

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Frequently Asked Questions

How many zeroes can a polynomial of degree n have?

A polynomial of degree n can have at most n real zeroes; it may have fewer if some roots are repeated or complex/imaginary.

What are the zeroes of $y^2 + 4y + 3$ ?

Factorising gives $(y+1)(y+3) = 0$ , so the zeroes are $y = -1$ and $y = -3$ .