Polynomials — Class 10 CBSE Mathematics MCQs with Solutions

Solution: By definition, a zero of a polynomial $p(x)$ is a value $a$ for which $p(a) = 0$. This is the fundamental concept: when you substitute the zero into the polynomial, the output is exactly zero. The other options confuse zeroes with other polynomial properties (domain, constant term, degree).

Solution: When a polynomial's graph crosses the x-axis at a point, the x-coordinate of that point is a real zero. Two distinct crossing points mean two distinct real zeroes. Touching the axis (not crossing) would indicate a repeated zero; complex zeroes don't appear as x-intercepts on a real graph.

Solution: The relationship between zeroes and coefficients is derived from the factored form: $p(x) = a(x - \alpha)(x - \beta)$. Expanding gives $p(x) = ax^2 - a(\alpha + \beta)x + a\alpha\beta$. Comparing with $ax^2 + bx + c$, we get $b = -a(\alpha + \beta)$, so $\alpha + \beta = -\frac{b}{a}$. This is a KEY relationship to memorize and understand.

Solution: A cubic polynomial always has degree 3 and always has at least one real zero (because odd-degree polynomials have end behaviours that force them to cross the x-axis). By the Fundamental Theorem of Algebra, it has exactly 3 zeroes counting multiplicity (in complex numbers). However, these three zeroes need not be DISTINCT — one could be repeated, or some could be non-real complex conjugates. So a cubic need not have three distinct real zeroes.

Solution: A fundamental theorem in algebra states that a polynomial of degree $n$ has at most $n$ zeroes (counting multiplicity). This is because if $p(x)$ has $n$ distinct zeroes $r_1, \ldots, r_n$, then $p(x) = a(x - r_1) \cdots (x - r_n)$, which has degree exactly $n$. More zeroes would require a higher-degree polynomial.

Solution: The Remainder Theorem states that when a polynomial $p(x)$ is divided by the linear divisor $(x - a)$, the remainder is the constant $p(a)$. This is derived from the division algorithm: $p(x) = (x - a) \cdot q(x) + r$. Since $r$ must have degree less than 1, $r$ is a constant. Setting $x = a$ gives $p(a) = 0 + r$, so $r = p(a)$.

Solution: A binomial $(x - a)$ is a factor of a polynomial if and only if $a$ is a zero of that polynomial. So $(x - 3)$ is a factor when $3$ is a zero, meaning substituting $x = 3$ makes the polynomial equal zero. This is the conceptual link between factors and zeroes.

Solution: By the Factor Theorem, if $(x - 2)$ divides $p(x)$, then $p(2) = 0$, so $2$ is a zero. Similarly, if $(x + 3) = (x - (-3))$ divides $p(x)$, then $p(-3) = 0$, so $-3$ is a zero. Note the sign carefully: divisor $(x - a)$ corresponds to zero $a$. We can conclude at least these two are zeroes, but $p(x)$ might have others if it has degree greater than 2.

Solution: We have $\frac{\alpha}{\beta} + \frac{\beta}{\alpha} = \frac{\alpha^2 + \beta^2}{\alpha\beta}$. Using Vieta's formulas: $\alpha + \beta = -p$ and $\alpha\beta = q$. We know $\alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta = p^2 - 2q$. So $\frac{\alpha}{\beta} + \frac{\beta}{\alpha} = \frac{p^2 - 2q}{q^2}$. This shows how symmetric expressions in the zeroes can always be reduced to expressions in the coefficients — a powerful technique in polynomial theory.

Solution: By the Remainder Theorem: when $p(x)$ is divided by $(x - a)$, the remainder is $p(a) = r_1$. When divided by $(x - b)$, the remainder is $p(b) = r_2$. Therefore, $p(a) - p(b) = r_1 - r_2$. This directly applies the Remainder Theorem to relate polynomial values to division remainders, showing the deep connection between these concepts.

Solution: If a cubic $p(x)$ is divisible by $(x - 1)$, $(x - 2)$, and $(x - 3)$ — three distinct linear factors — then $p(x)$ has exactly these three zeroes. A cubic with three specified zeroes must have the form $p(x) = a(x - 1)(x - 2)(x - 3)$ where $a$ is any non-zero constant (the leading coefficient). Different values of $a$ give different cubics, all satisfying the divisibility conditions. Divisibility determines the factors but not their scalar multiple.

Solution: When dividing by $(x^2 - 1)$, the remainder must have degree less than 2, so it's at most linear: $r(x) = 2x + 3$. The division algorithm gives: $p(x) = (x^2 - 1)q(x) + (2x + 3)$. Setting $x = 1$: $p(1) = (1 - 1)q(1) + 2(1) + 3 = 0 + 5 = 5$. Setting $x = -1$: $p(-1) = (1 - 1)q(-1) + 2(-1) + 3 = 0 + 1 = 1$. The remainder formula directly determines polynomial values at the divisor's zeroes, regardless of the coefficients in the quotient.

Solution: If $p(2) = 0$, then $4 + 2b + c = 0$ (by substitution). If the remainder when dividing by $(x + 1)$ is $-2$, then by the Remainder Theorem, $p(-1) = -2$. Substituting: $1 - b + c = -2$, so $-b + c = -3$. Together these form a system that uniquely determines $b$ and $c$. This combines the Factor Theorem (if $p(2) = 0$, then $(x - 2)$ is a factor) with the Remainder Theorem, showing how multiple conditions constrain polynomial coefficients.

Solution: If $(x - a)$ is a factor, we can write $p(x) = (x - a)h(x)$ for some polynomial $h(x)$. If $(x - b)$ is also a factor, then $(x - b)$ divides $p(x) = (x - a)h(x)$. Since $(x - b)$ and $(x - a)$ are distinct linear factors (as $a \neq b$), they share no common factor. Therefore, $(x - b)$ must divide $h(x)$, giving $h(x) = (x - b)g(x)$. Thus, $p(x) = (x - a)(x - b)g(x)$, proving that $(x - a)(x - b)$ divides $p(x)$. This is a subtle but powerful divisibility argument.

Solution: By Vieta's formulas, the product of the zeroes equals the constant term divided by the leading coefficient: $\alpha \beta = \frac{q}{1} = q$. If the product is zero, then $q = 0$. This means the polynomial is $x^2 + px = x(x + p)$, which has zeroes $0$ and $-p$. Notice: when the constant term is zero, one zero is always $0$; when the coefficient of $x$ is zero, the non-zero zeroes are opposites.

Solution: By the division algorithm, $p(x) = g(x) \cdot q(x) + r(x)$ where $r(x)$ is either zero or has degree less than $m$. The presence of a non-zero remainder does NOT determine a specific relationship between $n$ and $m$. For example, dividing $x^3 + 1$ (degree 3) by $x^2 + 1$ (degree 2) gives a linear remainder; dividing $x + 1$ (degree 1) by $x^2 + 1$ (degree 2) also gives a linear remainder. The only requirement is $\deg(r) < m$.

Solution: From $x^2 - px + q$: $\alpha + \beta = p$ and $\alpha\beta = q$. New sum: $\frac{\alpha}{\beta} + \frac{\beta}{\alpha} = \frac{\alpha^2 + \beta^2}{\alpha\beta} = \frac{p^2 - 2q}{q}$. New product: $\frac{\alpha}{\beta} \cdot \frac{\beta}{\alpha} = 1$. Polynomial: $qx^2 - (p^2 - 2q)x + q$.

Solution: From $2x^2 + 5x + k$: $\alpha + \beta = -\frac{5}{2}$ and $\alpha\beta = \frac{k}{2}$. Given: $\alpha^2 + \beta^2 + \alpha\beta = (\alpha + \beta)^2 - \alpha\beta = \frac{25}{4} - \frac{k}{2} = \frac{21}{4}$. Solve: $\frac{k}{2} = 1$, so $k = 2$.

Solution: From $\alpha + \beta = 6$ and $3\alpha + 2\beta = 20$: Substitute $\beta = 6 - \alpha$ into the second equation. $3\alpha + 2(6 - \alpha) = 20$. Solve: $\alpha = 8$, $\beta = -2$. Product: $k = 8 \times (-2) = -16$.

Solution: Since $\alpha$ is a zero: $a\alpha^2 + b\alpha + c = 0$. Therefore $a\alpha + b = -\frac{c}{\alpha}$. Similarly, $a\beta + b = -\frac{c}{\beta}$. Compute: $\frac{1}{-c/\alpha} + \frac{1}{-c/\beta} = -\frac{\alpha + \beta}{c} = -\frac{-b/a}{c} = \frac{b}{ac}$.

Solution: From $\alpha + \beta = 5$ and $\alpha - \beta = 1$: Add to get $2\alpha = 6$, so $\alpha = 3$. Subtract to get $2\beta = 4$, so $\beta = 2$. Product: $k = 3 \times 2 = 6$.

Solution: Expand: $p(x) = x^2 - kx - (k + c)$. Sum: $\alpha + \beta = k$. Product: $\alpha\beta = -(k + c)$. Calculate: $(\alpha+1)(\beta+1) = \alpha\beta + (\alpha + \beta) + 1 = -(k + c) + k + 1 = 1 - c$.

Solution: Let zeros be $2m$ and $3m$. Sum: $5m = a$, so $m = \frac{a}{5}$. Product: $6m^2 = b$. Substitute: $6(\frac{a}{5})^2 = b$. Therefore $6a^2 = 25b$.

Solution: From $2x^2 - 4x + 5$: $\alpha + \beta = 2$ and $\alpha\beta = \frac{5}{2}$. Calculate: $\frac{1}{\alpha^2} + \frac{1}{\beta^2} = \frac{\alpha^2 + \beta^2}{(\alpha\beta)^2} = \frac{4 - 5}{25/4} = \frac{-1}{25/4} = -\frac{4}{25}$.

Solution: Product of zeros = 1. Using $\frac{c}{a} = 1$: $\frac{4k}{k^2 + 4} = 1$. Solve: $k^2 + 4 = 4k$. Rearrange: $k^2 - 4k + 4 = 0$. Therefore $(k - 2)^2 = 0$, so $k = 2$.

Solution: Simplify: $\sqrt{\frac{\alpha}{\beta}} + \sqrt{\frac{\beta}{\alpha}} = \frac{\alpha + \beta}{\sqrt{\alpha\beta}} = \frac{-b/a}{\sqrt{c/a}} = -\frac{b}{\sqrt{ac}}$.

Solution: Factorize: $x^2 - 3x + 2 = (x - 1)(x - 2)$, so $\alpha = 1, \beta = 2$ (or vice versa). New zeros: $\frac{1+1}{2} = 1$ and $\frac{2+1}{1} = 3$. Sum = 4, Product = 3. Polynomial: $x^2 - 4x + 3$.

Solution: Let zeros be $k$ and $k+1$. Difference: $(k+1) - k = 1$. Using $(\alpha - \beta)^2 = (\alpha + \beta)^2 - 4\alpha\beta$: $1 = b^2 - 4c$.

Solution: Using $\alpha^2 + \beta^2 = p^2 - 2q$: $\alpha^4 + \beta^4 = (\alpha^2 + \beta^2)^2 - 2(\alpha\beta)^2 = (p^2 - 2q)^2 - 2q^2 = p^4 - 4p^2q + 2q^2$. Both A and C are equivalent.

Solution: From $x^2 - 8x + k$: $\alpha + \beta = 8$ and $\alpha\beta = k$. Using $\alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta$: $40 = 64 - 2k$. Solve: $k = 12$.