Division algorithm — Class 10 CBSE Mathematics MCQs with Solutions

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: By the Factor Theorem, if $(x - 1)$ divides $p(x)$, then $p(1) = 0$. Similarly, if $(x + 2)$ divides $p(x)$, then $p(-2) = 0$. Divisibility directly implies the corresponding points are zeroes. This relationship holds regardless of the values of $p$ and $q$ — divisibility *determines* that these values must be zeroes.

Solution: The Remainder Theorem states that the remainder when $p(x)$ is divided by $(x - a)$ equals $p(a)$. Here, dividing by $(x - 1)$ gives remainder 10, so $p(1) = 10$. This means we can set up the equation $1 + 2 + k + 6 = 10$, which gives $k = 1$. The theorem directly converts a divisibility statement into a condition on the polynomial's value.

Solution: If $x = 2$ is a zero, then by the Factor Theorem, $(x - 2)$ is a factor, making $p(2) = 0$. If the remainder when dividing by $(x + 1)$ is 5, then by the Remainder Theorem, $p(-1) = 5$. These two statements follow directly from applying the relevant theorems. Note: a cubic might have one real zero and two complex conjugate zeroes, so we can't say it has three real zeroes.

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 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: Since $x = 1$ is a zero, $p(1) = 0$. Substitute: $a(1)^2 - 3(a - 1)(1) - 1 = 0$. Simplify: $a - 3a + 3 - 1 = 0$. Therefore $-2a + 2 = 0$, so $a = 1$.

Solution: Since $x = 2$ is a zero, $p(2) = 0$. Substitute: $(2)^2 + 3(2) + k = 0$. Simplify: $4 + 6 + k = 0$, so $k = -10$.

Solution: By the Factor Theorem, if $(x-2)$ is a factor, then $p(2) = 0$. Substitute: $2^2 - 5(2) + p = 0$. Simplify: $4 - 10 + p = 0$, so $p = 6$.

Solution: Set $ax + b = 0$. Solve: $ax = -b$, so $x = -\frac{b}{a}$.

Solution: A polynomial cannot contain negative or fractional powers of the variable. Option B: $x + 2x^{-1}$ contains $x^{-1}$ (negative exponent), so it is not a polynomial.

Solution: If $(x+a)$ is a factor, then $p(-a) = 0$. Substitute: $2(-a)^2 + 2a(-a) + 5(-a) + 10 = 0$. Simplify: $2a^2 - 2a^2 - 5a + 10 = 0$, so $-5a + 10 = 0$ and $a = 2$.

Solution: Since $x^2 - 1 = (x - 1)(x + 1)$, both $x = 1$ and $x = -1$ are roots. At $x = 1$: $a + b + c + d + e = 0$. At $x = -1$: $a - b + c - d + e = 0$, which simplifies to $(a + c + e) - (b + d) = 0$. Combining: $a + c + e = 0$ and $b + d = 0$.