Chapter 2: Polynomials
10 Important Multiple Choice Questions (MCQs)
Quiz Results Summary
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If one zero of the quadratic polynomial $x^2 + 3x + k$ is $2$, then the value of $k$ is:
Correct Answer: (B) $-10$
Explanation: Since $x = 2$ is a zero of $p(x) = x^2 + 3x + k$, we have $p(2) = 0$.
$\implies (2)^2 + 3(2) + k = 0 \implies 4 + 6 + k = 0 \implies 10 + k = 0 \implies k = -10$.
A quadratic polynomial, whose zeroes are $-3$ and $4$, is:
Correct Answer: (C) $\frac{x^2}{2} - \frac{x}{2} - 6$
Explanation: Sum of zeroes $S = -3 + 4 = 1$. Product of zeroes $P = (-3) \times 4 = -12$.
General polynomial form: $p(x) = k(x^2 - Sx + P) = k(x^2 - x - 12)$.
For $k = \frac{1}{2}$, we get $p(x) = \frac{x^2}{2} - \frac{x}{2} - 6$.
If the zeroes of the quadratic polynomial $ax^2 + bx + c$ (where $c \neq 0$) are equal, then:
Correct Answer: (C) $c$ and $a$ have the same sign
Explanation: For equal zeroes, discriminant $D = b^2 - 4ac = 0 \implies b^2 = 4ac$.
Since $b^2 > 0$ for non-zero real numbers, $4ac$ must be positive ($\implies ac > 0$).
Product $ac > 0$ implies $a$ and $c$ must both be positive or both negative (i.e., same sign).
The total number of quadratic polynomials having zeroes as $-2$ and $5$ is:
Correct Answer: (D) More than 3
Explanation: Polynomials are of the form $p(x) = k(x^2 - 3x - 10)$ where $k \in \mathbb{R}$ and $k \neq 0$. As $k$ can take infinite real values, infinitely many polynomials exist.
If one of the zeroes of a cubic polynomial $ax^3 + bx^2 + cx + d$ is $0$, the product of the other two zeroes is:
Correct Answer: (B) $\frac{c}{a}$
Explanation: Let zeroes be $\alpha, \beta, \gamma$. Given $\alpha = 0$.
For cubic polynomial: $\alpha\beta + \beta\gamma + \gamma\alpha = \frac{c}{a}$.
Substitute $\alpha = 0 \implies 0 \cdot \beta + \beta\gamma + \gamma \cdot 0 = \frac{c}{a} \implies \beta\gamma = \frac{c}{a}$.
If one zero of $3x^2 + 8x + k$ is the reciprocal of the other, then the value of $k$ is:
Correct Answer: (A) $3$
Explanation: Let zeroes be $\alpha$ and $\frac{1}{\alpha}$.
Product of zeroes = $\alpha \times \frac{1}{\alpha} = 1$.
From formula, Product of zeroes = $\frac{\text{Constant}}{\text{Coeff. of } x^2} = \frac{k}{3}$.
Equating both: $\frac{k}{3} = 1 \implies k = 3$.
The zeroes of the quadratic polynomial $x^2 + 99x + 127$ are:
Correct Answer: (B) both negative
Explanation: Here $a = 1, b = 99, c = 127$.
$\text{Sum of zeroes } (\alpha + \beta) = -\frac{b}{a} = -99 < 0$.
$\text{Product of zeroes } (\alpha\beta) = \frac{c}{a} = 127 > 0$.
Since product is positive, both zeroes have the same sign. Since sum is negative, both zeroes must be negative.
If $\alpha$ and $\beta$ are zeroes of $p(x) = 2x^2 + 5x + k$ such that $\alpha^2 + \beta^2 + \alpha\beta = \frac{21}{4}$, then $k$ equals:
Correct Answer: (A) $2$
Explanation: $\alpha + \beta = -\frac{5}{2}$ and $\alpha\beta = \frac{k}{2}$.
$\alpha^2 + \beta^2 + \alpha\beta = (\alpha + \beta)^2 - \alpha\beta = \left(-\frac{5}{2}\right)^2 - \frac{k}{2} = \frac{25}{4} - \frac{k}{2}$.
Given: $\frac{25}{4} - \frac{k}{2} = \frac{21}{4} \implies \frac{k}{2} = \frac{25}{4} - \frac{21}{4} = \frac{4}{4} = 1 \implies k = 2$.
The graph of a polynomial $y = p(x)$ intersects the $x$-axis at 3 points and touches it at 2 points. The total number of zeroes of $p(x)$ is:
Correct Answer: (C) $5$
Explanation: Total zeroes equal the total number of points where the graph intersects OR touches the $x$-axis.
Total Zeroes = $3 \text{ (intersections)} + 2 \text{ (touch points)} = 5$.
If $\alpha$ and $\beta$ are zeroes of $x^2 - p(x + 1) - c$, then $(\alpha + 1)(\beta + 1)$ equals:
Correct Answer: (C) $1 - c$
Explanation: Expand $p(x) = x^2 - px - p - c = x^2 - px - (p + c)$.
Here $a = 1, b = -p, d = -(p + c)$.
$\alpha + \beta = p$, and $\alpha\beta = -(p + c)$.
Now $(\alpha + 1)(\beta + 1) = \alpha\beta + \alpha + \beta + 1 = -(p + c) + p + 1 = -p - c + p + 1 = 1 - c$.
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