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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 9 Polynomials.

HOTS (High Order Thinking Skills) Worksheet

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  1. Question 1: If $a+b+c=5$ and $ab+bc+ca=10$, prove that $a^3+b^3+c^3-3abc = -25$.
    Solution:
    $a^2+b^2+c^2 = (a+b+c)^2 - 2(ab+bc+ca) = 5^2 - 2(10) = 25 - 20 = 5$.
    $a^3+b^3+c^3-3abc = (a+b+c)(a^2+b^2+c^2 - (ab+bc+ca))$
    $= 5(5 - 10) = 5(-5) = -25$.
  2. Question 2: Factorise $(a^2 - b^2)^3 + (b^2 - c^2)^3 + (c^2 - a^2)^3$.
    Solution: Let $X=a^2-b^2, Y=b^2-c^2, Z=c^2-a^2$.
    $X+Y+Z=0$. So $X^3+Y^3+Z^3 = 3XYZ$.
    Answer: $3(a^2-b^2)(b^2-c^2)(c^2-a^2) = 3(a-b)(a+b)(b-c)(b+c)(c-a)(c+a)$.
  3. Question 3: If $x = 2y + 6$, find the value of $x^3 - 8y^3 - 36xy - 216$.
    Solution: $x - 2y - 6 = 0$. Let $a=x, b=-2y, c=-6$.
    $a+b+c=0 \Rightarrow a^3+b^3+c^3 = 3abc$.
    $x^3 + (-2y)^3 + (-6)^3 = 3(x)(-2y)(-6) \Rightarrow x^3 - 8y^3 - 216 = 36xy$.
    So, $x^3 - 8y^3 - 36xy - 216 = 0$.
  4. Question 4: Find the value of $k$ if $x-1$ is a factor of $4x^3 + 3x^2 - 4x + k$.
    Solution: $p(1) = 0 \Rightarrow 4(1)^3 + 3(1)^2 - 4(1) + k = 0$.
    $4 + 3 - 4 + k = 0 \Rightarrow 3 + k = 0 \Rightarrow k = -3$.
  5. Question 5: If polynomials $ax^3 + 3x^2 - 13$ and $2x^3 - 5x + a$ leave the same remainder when divided by $x-2$, find $a$.
    Solution: $p(2) = q(2)$.
    $a(8) + 3(4) - 13 = 2(8) - 5(2) + a$
    $8a + 12 - 13 = 16 - 10 + a \Rightarrow 8a - 1 = 6 + a \Rightarrow 7a = 7 \Rightarrow a = 1$.
  6. Question 6: Factorise $x^3 + 13x^2 + 32x + 20$.
    Solution: $p(-1) = -1 + 13 - 32 + 20 = 0$, so $(x+1)$ is a factor.
    Dividing by $(x+1)$, we get $x^2 + 12x + 20$.
    Factorising quadratic: $(x+10)(x+2)$.
    Answer: $(x+1)(x+2)(x+10)$.
  7. Question 7: If $a+b+c=9$ and $a^2+b^2+c^2=35$, find the value of $a^3+b^3+c^3-3abc$.
    Solution: $(a+b+c)^2 = a^2+b^2+c^2 + 2(ab+bc+ca) \Rightarrow 81 = 35 + 2(ab+bc+ca) \Rightarrow ab+bc+ca = 23$.
    Value $= (a+b+c)(a^2+b^2+c^2 - (ab+bc+ca)) = 9(35 - 23) = 9(12) = 108$.
  8. Question 8: Simplify $\frac{(a^2-b^2)^3 + (b^2-c^2)^3 + (c^2-a^2)^3}{(a-b)^3 + (b-c)^3 + (c-a)^3}$.
    Solution: Numerator $= 3(a^2-b^2)(b^2-c^2)(c^2-a^2)$. Denominator $= 3(a-b)(b-c)(c-a)$.
    Ratio $= \frac{3(a-b)(a+b)(b-c)(b+c)(c-a)(c+a)}{3(a-b)(b-c)(c-a)} = (a+b)(b+c)(c+a)$.
  9. Question 9: If $x+y=12$ and $xy=27$, find $x^3+y^3$.
    Solution: $x^3+y^3 = (x+y)(x^2-xy+y^2) = (x+y)((x+y)^2 - 3xy)$.
    $= 12(12^2 - 3(27)) = 12(144 - 81) = 12(63) = 756$.
  10. Question 10: Prove that $(x+y)^3 - (x-y)^3 - 6y(x^2-y^2) = 8y^3$.
    Solution: LHS $= (x^3+y^3+3xy(x+y)) - (x^3-y^3-3xy(x-y)) - 6x^2y + 6y^3$
    $= x^3+y^3+3x^2y+3xy^2 - x^3+y^3+3x^2y-3xy^2 - 6x^2y + 6y^3$
    $= 2y^3 + 6x^2y - 6x^2y + 6y^3 = 8y^3$.
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