Polynomials — Class 10 CBSE Mathematics MCQs with Solutions

Solution: A zero of $p(x)$ is a value of $x$ where $p(x) = 0$, which is exactly a point where the graph of $y = p(x)$ meets the x-axis. Since the graph cuts the x-axis at 3 distinct points, $p(x)$ has exactly 3 zeroes.

Solution: For $ax^2 + bx + c$, the sum of zeroes is $-\frac{b}{a}$. Here $a = 1$ and $b = -7$, so the sum $= -\frac{(-7)}{1} = 7$. Indeed the zeroes are $2$ and $5$, and $2 + 5 = 7$.

Solution: For $ax^2 + bx + c$, the product of zeroes is $\frac{c}{a}$. Here $\frac{c}{a} = \frac{6}{2} = 3$. Note that $4$ is the sum of the zeroes, not the product.

Solution: Sum of zeroes $= 3 + (-4) = -1$ and product $= 3 \times (-4) = -12$. A required polynomial is $x^2 - (\text{sum})x + \text{product} = x^2 - (-1)x + (-12) = x^2 + x - 12$. Check: $(x - 3)(x + 4) = x^2 + x - 12$.

Solution: Setting $x^2 - 3 = 0$ gives $x^2 = 3$, so $x = \pm\sqrt{3}$. We can also write $x^2 - 3 = (x - \sqrt{3})(x + \sqrt{3})$, showing both zeroes clearly.

Solution: Factorising, $x^2 + 3x + 2 = (x + 1)(x + 2)$. Dividing by $x + 1$ therefore gives quotient $x + 2$ with remainder $0$.

Solution: The division algorithm states that for any polynomials $p(x)$ and $g(x) \ne 0$, we can write $p(x) = g(x)q(x) + r(x)$ where either $r(x) = 0$ or $\deg r(x) < \deg g(x)$. If the remainder had degree equal to or more than $g(x)$, the division could be carried further.

Solution: By the division algorithm, $p(x) = g(x) \cdot q(x) + r(x) = (x - 2)(x + 3) + 5$. Expanding, $(x - 2)(x + 3) = x^2 + x - 6$, so $p(x) = x^2 + x - 6 + 5 = x^2 + x - 1$.

Solution: If the remainder is zero, the division algorithm gives $p(x) = g(x) \cdot q(x)$. This means $g(x)$ divides $p(x)$ exactly, i.e. $g(x)$ is a factor of $p(x)$.

Solution: When the remainder is zero, $p(x) = g(x) \cdot q(x)$, and degrees add on multiplication: $\deg p = \deg g + \deg q$. So $\deg q = 4 - 2 = 2$.

Solution: Here $\alpha + \beta = 1$ and $\alpha\beta = -2$. Sum of new zeroes $= 2(\alpha + \beta) + 2 = 4$; product $= (2\alpha + 1)(2\beta + 1) = 4\alpha\beta + 2(\alpha + \beta) + 1 = -8 + 2 + 1 = -5$. The required polynomial is $x^2 - 4x - 5$.

Solution: For $ax^3 + bx^2 + cx + d$: $\alpha\beta + \beta\gamma + \gamma\alpha = \frac{c}{a} = \frac{-14}{2} = -7$ and $\alpha\beta\gamma = -\frac{d}{a} = -\frac{8}{2} = -4$. So $\frac{1}{\alpha} + \frac{1}{\beta} + \frac{1}{\gamma} = \frac{\alpha\beta + \beta\gamma + \gamma\alpha}{\alpha\beta\gamma} = \frac{-7}{-4} = \frac{7}{4}$.

Solution: Sum of zeroes: $(a - d) + a + (a + d) = 3a = -\frac{(-3)}{1} = 3$, so $a = 1$. Product of zeroes: $(a - d) \cdot a \cdot (a + d) = a(a^2 - d^2) = -\frac{1}{1} = -1$, giving $1 - d^2 = -1$, so $d^2 = 2$ and $d = \pm\sqrt{2}$.

Solution: If the zeroes are $\alpha$ and $\frac{1}{\alpha}$, their product is $1$. So $\frac{4k}{k^2 + 4} = 1$, which gives $k^2 - 4k + 4 = 0$, i.e. $(k - 2)^2 = 0$. Hence $k = 2$ only; $k = -2$ gives product $\frac{-8}{8} = -1$, which does not work.

Solution: From the polynomial, $\alpha + \beta = 6$. Writing $3\alpha + 2\beta = 2(\alpha + \beta) + \alpha = 12 + \alpha = 20$ gives $\alpha = 8$, so $\beta = -2$. Then $k = \alpha\beta = 8 \times (-2) = -16$.

Solution: Since $\pm\sqrt{\frac{5}{3}}$ are zeroes, $\left(x^2 - \frac{5}{3}\right)$, and hence $3x^2 - 5$, is a factor. Dividing $3x^4 + 6x^3 - 2x^2 - 10x - 5$ by $3x^2 - 5$ gives $x^2 + 2x + 1 = (x + 1)^2$. So the other two zeroes are $-1$ and $-1$.

Solution: The factor from the given zeroes is $\left(x - (2 + \sqrt{3})\right)\left(x - (2 - \sqrt{3})\right) = x^2 - 4x + 1$. Dividing the quartic by $x^2 - 4x + 1$ gives quotient $x^2 - 2x - 35 = (x - 7)(x + 5)$. Hence the remaining zeroes are $7$ and $-5$.

Solution: Dividing $x^4 + x^3 + 8x^2 + ax + b$ by $x^2 + 1$ gives quotient $x^2 + x + 7$ and remainder $(a - 1)x + (b - 7)$. For exact divisibility the remainder must be identically zero, so $a - 1 = 0$ and $b - 7 = 0$, i.e. $a = 1,\ b = 7$. Check: $(x^2 + 1)(x^2 + x + 7) = x^4 + x^3 + 8x^2 + x + 7$.

Solution: If $q(x) = 0$, the division algorithm gives $p(x) = g(x) \cdot 0 + r(x)$, so the remainder equals $p(x)$ itself. Since the remainder must satisfy $\deg r(x) < \deg g(x)$, we conclude $\deg p(x) < \deg g(x)$. For example, dividing $x + 1$ by $x^2 + 3$ gives quotient $0$ and remainder $x + 1$.

Solution: The remainder on dividing by a linear polynomial is the value at its zero. So $R_1 = (-1)^3 + 2(-1)^2 - 5a(-1) - 7 = 5a - 6$ and $R_2 = 2^3 + a(2)^2 - 12(2) + 6 = 4a - 10$. Then $2R_1 + R_2 = 10a - 12 + 4a - 10 = 14a - 22 = 6$, giving $a = 2$.

Solution: Here $\alpha + \beta = 5$ and $\alpha\beta = 6$. Using the identity $\alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta = 25 - 12 = 13$. Choosing $25$ forgets to subtract $2\alpha\beta$, while $37$ comes from wrongly adding $2\alpha\beta$.

Solution: Since $x = 2$ is a zero, substituting gives $k(2)^2 + 3(2) + k = 0$, i.e. $4k + 6 + k = 0$. So $5k = -6$, which gives $k = -\frac{6}{5}$.

Solution: From the coefficients, $\alpha + \beta = k + 6$ and $\alpha\beta = 2(2k - 1) = 4k - 2$. The condition $\alpha + \beta = \frac{1}{2}\alpha\beta$ gives $k + 6 = \frac{4k - 2}{2} = 2k - 1$. Solving, $k = 7$.

Solution: A required polynomial is $x^2 - (\text{sum})x + \text{product} = x^2 - \sqrt{2}x + \frac{1}{3}$. Multiplying throughout by $3$ to clear the fraction gives $3x^2 - 3\sqrt{2}x + 1$. Check: sum $= \frac{3\sqrt{2}}{3} = \sqrt{2}$ and product $= \frac{1}{3}$.

Solution: We have $\alpha + \beta = -\frac{5}{2}$ and $\alpha\beta = \frac{1}{2}$. So $\frac{1}{\alpha} + \frac{1}{\beta} = \frac{\alpha + \beta}{\alpha\beta} = \frac{-5/2}{1/2} = -5$.

Solution: By long division, the first term of the quotient is $3x$; subtracting $3x(x^2 + 2x + 1) = 3x^3 + 6x^2 + 3x$ leaves $-5x^2 - x + 5$. The next quotient term is $-5$; subtracting $-5(x^2 + 2x + 1) = -5x^2 - 10x - 5$ leaves $9x + 10$. Since $\deg(9x + 10) < \deg(x^2 + 2x + 1)$, the remainder is $9x + 10$.

Solution: Long division by $x^2 + 5$ gives quotient $x^2 + 2x + 3$ and remainder $2x + 3$. Check: $(x^2 + 5)(x^2 + 2x + 3) + (2x + 3) = x^4 + 2x^3 + 8x^2 + 10x + 15 + 2x + 3 = x^4 + 2x^3 + 8x^2 + 12x + 18$. Hence $p = 2$ and $q = 3$.

Solution: Dividing $x^3 + 5x^2 + 5x + 8$ by $x^2 + 3x - 2$ gives quotient $x + 2$ and remainder $x + 12$. Subtracting the remainder makes the division exact: $x^3 + 5x^2 + 5x + 8 - (x + 12) = (x^2 + 3x - 2)(x + 2)$, which you can verify by expanding.

Solution: On dividing $x^4 + 2x^3 - 2x^2 + x - 1$ by $x^2 + 2x - 3$, the quotient is $x^2 + 1$ and the remainder is $2 - x$. To make the remainder zero we must add its negative, i.e. add $x - 2$. Check: $x^4 + 2x^3 - 2x^2 + x - 1 + (x - 2) = (x^2 + 2x - 3)(x^2 + 1)$.

Solution: From $p(x) = g(x)q(x) + r(x)$ we get $g(x) = \frac{p(x) - r(x)}{q(x)}$. Now $p(x) - r(x) = x^3 - 3x^2 + x + 2 - (-2x + 4) = x^3 - 3x^2 + 3x - 2$. Dividing this by $x - 2$ gives $g(x) = x^2 - x + 1$.