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Chapter 2: Polynomials

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This page provides comprehensive Chapter 2: Polynomials - HOTS Worksheet - SJMaths. High Order Thinking Skills (HOTS) worksheet for Class 10 Polynomials.

HOTS (High Order Thinking Skills)

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  1. Question 1: If $\alpha$ and $\beta$ are the zeroes of the quadratic polynomial $f(x) = x^2 - px + q$, prove that $\frac{\alpha^2}{\beta^2} + \frac{\beta^2}{\alpha^2} = \frac{p^4}{q^2} - \frac{4p^2}{q} + 2$.
    Solution:
    Step 1: From the polynomial $x^2 - px + q$, we have sum of zeroes $\alpha + \beta = p$ and product $\alpha\beta = q$.
    Step 2: Simplify the LHS: $\frac{\alpha^2}{\beta^2} + \frac{\beta^2}{\alpha^2} = \frac{\alpha^4 + \beta^4}{\alpha^2\beta^2}$.
    Step 3: Express $\alpha^4 + \beta^4$ in terms of sum and product:
    $\alpha^2 + \beta^2 = (\alpha+\beta)^2 - 2\alpha\beta = p^2 - 2q$.
    $\alpha^4 + \beta^4 = (\alpha^2+\beta^2)^2 - 2\alpha^2\beta^2 = (p^2 - 2q)^2 - 2q^2$.
    Step 4: Expand the numerator:
    $(p^2 - 2q)^2 - 2q^2 = p^4 - 4p^2q + 4q^2 - 2q^2 = p^4 - 4p^2q + 2q^2$.
    Step 5: Divide by denominator $(\alpha\beta)^2 = q^2$:
    $\frac{p^4 - 4p^2q + 2q^2}{q^2} = \frac{p^4}{q^2} - \frac{4p^2q}{q^2} + \frac{2q^2}{q^2} = \frac{p^4}{q^2} - \frac{4p^2}{q} + 2$.
    Hence Proved.
  2. Question 2: If the zeroes of the cubic polynomial $x^3 - 3x^2 + x + 1$ are $a - b$, $a$, $a + b$, find $a$ and $b$.
    Solution:
    Step 1: Let the zeroes be $\alpha = a-b, \beta = a, \gamma = a+b$.
    Step 2: Sum of zeroes $\alpha+\beta+\gamma = -\frac{\text{coeff of } x^2}{\text{coeff of } x^3} = -\frac{-3}{1} = 3$.
    $(a-b) + a + (a+b) = 3 \Rightarrow 3a = 3 \Rightarrow a = 1$.
    Step 3: Product of zeroes $\alpha\beta\gamma = -\frac{\text{constant term}}{\text{coeff of } x^3} = -\frac{1}{1} = -1$.
    $(a-b)(a)(a+b) = -1 \Rightarrow a(a^2 - b^2) = -1$.
    Step 4: Substitute $a=1$:
    $1(1^2 - b^2) = -1 \Rightarrow 1 - b^2 = -1 \Rightarrow b^2 = 2 \Rightarrow b = \pm\sqrt{2}$.
    Thus, $a=1, b=\pm\sqrt{2}$.
  3. Question 3: If the polynomial $x^4 - 6x^3 + 16x^2 - 25x + 10$ is divided by another polynomial $x^2 - 2x + k$, the remainder comes out to be $x + a$, find $k$ and $a$.
    Solution:
    Step 1: Perform long division of $x^4 - 6x^3 + 16x^2 - 25x + 10$ by $x^2 - 2x + k$.
    Step 2: After dividing, the remainder is found to be $(2k - 9)x + (k^2 - 8k + 10)$.
    Step 3: Compare the remainder with the given form $x + a$.
    Coefficient of $x$: $2k - 9 = 1 \Rightarrow 2k = 10 \Rightarrow k = 5$.
    Constant term: $a = k^2 - 8k + 10$.
    Step 4: Substitute $k=5$ to find $a$:
    $a = (5)^2 - 8(5) + 10 = 25 - 40 + 10 = -5$.
    Therefore, $k=5$ and $a=-5$.
  4. Question 4: Find the zeroes of the polynomial $f(x) = x^3 - 5x^2 - 16x + 80$, if its two zeroes are equal in magnitude but opposite in sign.
    Solution:
    Step 1: Let the zeroes be $\alpha, -\alpha$ and $\gamma$.
    Step 2: Sum of zeroes $= \alpha + (-\alpha) + \gamma = -\frac{-5}{1} = 5$.
    $\Rightarrow \gamma = 5$.
    Step 3: Product of zeroes $= \alpha(-\alpha)\gamma = -\frac{80}{1} = -80$.
    $-\alpha^2(5) = -80 \Rightarrow \alpha^2 = 16 \Rightarrow \alpha = \pm 4$.
    Step 4: The zeroes are $4, -4, 5$.
  5. Question 5: If $\alpha$ and $\beta$ are the zeroes of the quadratic polynomial $f(x) = 2x^2 - 5x + 7$, find a polynomial whose zeroes are $2\alpha + 3\beta$ and $3\alpha + 2\beta$.
    Solution:
    Step 1: From $2x^2 - 5x + 7$, $\alpha+\beta = \frac{5}{2}$ and $\alpha\beta = \frac{7}{2}$.
    Step 2: Sum of new zeroes $S = (2\alpha + 3\beta) + (3\alpha + 2\beta) = 5(\alpha+\beta) = 5(\frac{5}{2}) = \frac{25}{2}$.
    Step 3: Product of new zeroes $P = (2\alpha + 3\beta)(3\alpha + 2\beta) = 6\alpha^2 + 4\alpha\beta + 9\alpha\beta + 6\beta^2 = 6(\alpha^2+\beta^2) + 13\alpha\beta$.
    Using $\alpha^2+\beta^2 = (\alpha+\beta)^2 - 2\alpha\beta = (\frac{5}{2})^2 - 2(\frac{7}{2}) = \frac{25}{4} - 7 = -\frac{3}{4}$.
    $P = 6(-\frac{3}{4}) + 13(\frac{7}{2}) = -\frac{9}{2} + \frac{91}{2} = \frac{82}{2} = 41$.
    Step 4: Required polynomial is $k[x^2 - Sx + P] = k[x^2 - \frac{25}{2}x + 41]$.
    Taking $k=2$, we get $2x^2 - 25x + 82$.
  6. Question 6: If $\alpha$ and $\beta$ are the zeroes of the quadratic polynomial $p(s) = 3s^2 - 6s + 4$, find the value of $\frac{\alpha}{\beta} + \frac{\beta}{\alpha} + 2(\frac{1}{\alpha} + \frac{1}{\beta}) + 3\alpha\beta$.
    Solution:
    Step 1: $\alpha+\beta = -\frac{-6}{3} = 2$ and $\alpha\beta = \frac{4}{3}$.
    Step 2: Evaluate term 1: $\frac{\alpha}{\beta} + \frac{\beta}{\alpha} = \frac{\alpha^2+\beta^2}{\alpha\beta} = \frac{(\alpha+\beta)^2-2\alpha\beta}{\alpha\beta} = \frac{2^2 - 2(4/3)}{4/3} = \frac{4 - 8/3}{4/3} = \frac{4/3}{4/3} = 1$.
    Step 3: Evaluate term 2: $2(\frac{1}{\alpha} + \frac{1}{\beta}) = 2(\frac{\alpha+\beta}{\alpha\beta}) = 2(\frac{2}{4/3}) = 2(\frac{6}{4}) = 3$.
    Step 4: Evaluate term 3: $3\alpha\beta = 3(\frac{4}{3}) = 4$.
    Step 5: Total value $= 1 + 3 + 4 = 8$.
  7. Question 7: If the square of the difference of the zeroes of the quadratic polynomial $f(x) = x^2 + px + 45$ is equal to 144, find the value of $p$.
    Solution:
    Step 1: Given $(\alpha - \beta)^2 = 144$.
    Step 2: Use the identity $(\alpha - \beta)^2 = (\alpha + \beta)^2 - 4\alpha\beta$.
    Step 3: From polynomial, $\alpha + \beta = -p$ and $\alpha\beta = 45$.
    Step 4: Substitute values: $(-p)^2 - 4(45) = 144$.
    $p^2 - 180 = 144 \Rightarrow p^2 = 324$.
    Step 5: $p = \pm\sqrt{324} = \pm 18$.
  8. Question 8: Find the value of $k$ for which the polynomial $x^4 + 10x^3 + 25x^2 + 15x + k$ is exactly divisible by $x + 7$.
    Solution:
    Step 1: Let $P(x) = x^4 + 10x^3 + 25x^2 + 15x + k$. Since it is divisible by $x+7$, by Factor Theorem, $P(-7) = 0$.
    Step 2: Substitute $x = -7$:
    $(-7)^4 + 10(-7)^3 + 25(-7)^2 + 15(-7) + k = 0$.
    $2401 + 10(-343) + 25(49) - 105 + k = 0$.
    $2401 - 3430 + 1225 - 105 + k = 0$.
    Step 3: Simplify: $(2401 + 1225) - (3430 + 105) + k = 0$.
    $3626 - 3535 + k = 0 \Rightarrow 91 + k = 0$.
    Step 4: $k = -91$.
  9. Question 9: If $\alpha$ and $\beta$ are the zeroes of the quadratic polynomial $f(x) = x^2 - x - 2$, find a quadratic polynomial whose zeroes are $\frac{2\alpha}{\beta}$ and $\frac{2\beta}{\alpha}$.
    Solution:
    Step 1: $\alpha+\beta = 1, \alpha\beta = -2$.
    Step 2: Sum of new zeroes $S = \frac{2\alpha}{\beta} + \frac{2\beta}{\alpha} = 2(\frac{\alpha^2+\beta^2}{\alpha\beta})$.
    $\alpha^2+\beta^2 = (\alpha+\beta)^2 - 2\alpha\beta = (1)^2 - 2(-2) = 1 + 4 = 5$.
    $S = 2(\frac{5}{-2}) = -5$.
    Step 3: Product of new zeroes $P = (\frac{2\alpha}{\beta})(\frac{2\beta}{\alpha}) = 4$.
    Step 4: Required polynomial is $k[x^2 - Sx + P] = k[x^2 - (-5)x + 4]$.
    For $k=1$, polynomial is $x^2 + 5x + 4$.
  10. Question 10: Obtain all other zeroes of $3x^4 + 6x^3 - 2x^2 - 10x - 5$, if two of its zeroes are $\sqrt{\frac{5}{3}}$ and $-\sqrt{\frac{5}{3}}$.
    Solution:
    Step 1: Since $x = \sqrt{5/3}$ and $x = -\sqrt{5/3}$ are zeroes, $(x - \sqrt{5/3})(x + \sqrt{5/3}) = x^2 - 5/3$ is a factor.
    To simplify, $3x^2 - 5$ is a factor.
    Step 2: Perform long division of $3x^4 + 6x^3 - 2x^2 - 10x - 5$ by $3x^2 - 5$.
    The quotient obtained is $x^2 + 2x + 1$.
    Step 3: Factorize the quotient to find remaining zeroes.
    $x^2 + 2x + 1 = (x+1)^2$.
    Step 4: The other zeroes are $-1, -1$.
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