Welcome to HSLC Guru! This article presents the complete Class 10 Mathematics Chapter 2 Question Answer on the topic Polynomials for ASSEB (Assam State School Education Board) students of the English Medium. Every exercise of the chapter — Exercise 2.1, Exercise 2.2, Exercise 2.3 and the optional Exercise 2.4 — is solved step by step. The chapter introduces polynomials, their degree and types, the geometric meaning of zeros (the points where the graph of $y=p(x)$ meets the x-axis), the relationship between zeros and coefficients of quadratic and cubic polynomials, and the division algorithm for polynomials. Use this guide along with your SCERT/NCERT textbook to revise quickly and to score well in your HSLC Board examination.

Chapter Summary

A polynomial in the variable $x$ is an algebraic expression of the form $p(x)=a_{n}x^{n}+a_{n-1}x^{n-1}+\dots+a_{1}x+a_{0}$, where $a_{0},a_{1},\dots,a_{n}$ are real numbers and $n$ is a non-negative integer. The highest power of $x$ in $p(x)$ is called the degree of the polynomial.

• A polynomial of degree $1$ is called a linear polynomial. General form: $ax+b$, $a\neq 0$.
• A polynomial of degree $2$ is called a quadratic polynomial. General form: $ax^{2}+bx+c$, $a\neq 0$.
• A polynomial of degree $3$ is called a cubic polynomial. General form: $ax^{3}+bx^{2}+cx+d$, $a\neq 0$.
• A real number $k$ is called a zero of a polynomial $p(x)$ if $p(k)=0$.
• The number of zeros of a polynomial of degree $n$ is at most $n$.
• Geometrically, the zeros of a polynomial $p(x)$ are the x-coordinates of the points where the graph of $y=p(x)$ intersects the x-axis.

Key Formulas

For a quadratic polynomial $ax^{2}+bx+c$ with zeros $\alpha$ and $\beta$:

$$\alpha+\beta=-\frac{b}{a},\qquad \alpha\beta=\frac{c}{a}$$

A quadratic polynomial whose zeros are $\alpha$ and $\beta$ can be written as:

$$p(x)=k\left[x^{2}-(\alpha+\beta)x+\alpha\beta\right],\quad k\neq 0$$

For a cubic polynomial $ax^{3}+bx^{2}+cx+d$ with zeros $\alpha,\beta,\gamma$:

$$\alpha+\beta+\gamma=-\frac{b}{a}$$

$$\alpha\beta+\beta\gamma+\gamma\alpha=\frac{c}{a}$$

$$\alpha\beta\gamma=-\frac{d}{a}$$

Division Algorithm for Polynomials: If $p(x)$ and $g(x)$ are any two polynomials with $g(x)\neq 0$, then we can find polynomials $q(x)$ and $r(x)$ such that:

$$p(x)=g(x)\cdot q(x)+r(x)$$

where either $r(x)=0$ or degree of $r(x)$ is less than the degree of $g(x)$. Here $p(x)$ is the dividend, $g(x)$ the divisor, $q(x)$ the quotient and $r(x)$ the remainder.

Geometrical Meaning of the Zeros — Graphs

The graph of a quadratic polynomial $y=ax^{2}+bx+c$ is always a parabola. Depending on the sign of $a$ and the position of the parabola relative to the x-axis, the polynomial may have two distinct real zeros, two equal zeros, or no real zeros. Study the following figures carefully.

Exercise 2.1

Q1. The graphs of $y=p(x)$ are given for some polynomials $p(x)$. Find the number of zeros of $p(x)$ in each case.

Graph (i): The graph is a straight line which is parallel to the x-axis. It does not intersect the x-axis at any point. Answer: Number of zeros = $0$.

Graph (ii): The graph cuts the x-axis at exactly one point. Answer: Number of zeros = $1$.

Graph (iii): The graph cuts the x-axis at three points. Answer: Number of zeros = $3$.

Graph (iv): The graph cuts the x-axis at two points. Answer: Number of zeros = $2$.

Graph (v): The graph cuts the x-axis at four points. Answer: Number of zeros = $4$.

Graph (vi): The graph cuts the x-axis at three points. Answer: Number of zeros = $3$.

Reason: The number of zeros of a polynomial $p(x)$ is equal to the number of points at which the graph of $y=p(x)$ intersects the x-axis.

Exercise 2.2

Q1. Find the zeros of the following quadratic polynomials and verify the relationship between the zeros and the coefficients.

(i) $x^{2}-2x-8$ Factorise: $x^{2}-2x-8=x^{2}-4x+2x-8=x(x-4)+2(x-4)=(x-4)(x+2)$. Zeros are $x=4$ and $x=-2$. Here $a=1$, $b=-2$, $c=-8$. Sum of zeros $=4+(-2)=2=-\dfrac{-2}{1}=-\dfrac{b}{a}$. Product of zeros $=4\times(-2)=-8=\dfrac{-8}{1}=\dfrac{c}{a}$. Answer: Zeros are $4$ and $-2$; relationship verified.

(ii) $4s^{2}-4s+1$ $4s^{2}-4s+1=(2s-1)^{2}$. Zeros are $s=\dfrac{1}{2},\dfrac{1}{2}$. Here $a=4$, $b=-4$, $c=1$. Sum $=\dfrac{1}{2}+\dfrac{1}{2}=1=-\dfrac{-4}{4}=-\dfrac{b}{a}$. Product $=\dfrac{1}{2}\times\dfrac{1}{2}=\dfrac{1}{4}=\dfrac{1}{4}=\dfrac{c}{a}$. Answer: Zeros are $\dfrac{1}{2}$ and $\dfrac{1}{2}$; verified.

(iii) $6x^{2}-3-7x$ Rewrite: $6x^{2}-7x-3=6x^{2}-9x+2x-3=3x(2x-3)+1(2x-3)=(2x-3)(3x+1)$. Zeros: $x=\dfrac{3}{2},\;x=-\dfrac{1}{3}$. $a=6,\;b=-7,\;c=-3$. Sum $=\dfrac{3}{2}-\dfrac{1}{3}=\dfrac{9-2}{6}=\dfrac{7}{6}=-\dfrac{-7}{6}=-\dfrac{b}{a}$. Product $=\dfrac{3}{2}\times\left(-\dfrac{1}{3}\right)=-\dfrac{1}{2}=\dfrac{-3}{6}=\dfrac{c}{a}$. Answer: Zeros are $\dfrac{3}{2}$ and $-\dfrac{1}{3}$; verified.

(iv) $4u^{2}+8u$ $4u^{2}+8u=4u(u+2)$. Zeros: $u=0,\;u=-2$. $a=4,\;b=8,\;c=0$. Sum $=0+(-2)=-2=-\dfrac{8}{4}=-\dfrac{b}{a}$. Product $=0\times(-2)=0=\dfrac{0}{4}=\dfrac{c}{a}$. Answer: Zeros are $0$ and $-2$; verified.

(v) $t^{2}-15$ $t^{2}-15=0\Rightarrow t=\pm\sqrt{15}$. $a=1,\;b=0,\;c=-15$. Sum $=\sqrt{15}+(-\sqrt{15})=0=-\dfrac{0}{1}=-\dfrac{b}{a}$. Product $=\sqrt{15}\times(-\sqrt{15})=-15=\dfrac{-15}{1}=\dfrac{c}{a}$. Answer: Zeros are $\sqrt{15}$ and $-\sqrt{15}$; verified.

(vi) $3x^{2}-x-4$ $3x^{2}-x-4=3x^{2}-4x+3x-4=x(3x-4)+1(3x-4)=(3x-4)(x+1)$. Zeros: $x=\dfrac{4}{3},\;x=-1$. $a=3,\;b=-1,\;c=-4$. Sum $=\dfrac{4}{3}-1=\dfrac{1}{3}=-\dfrac{-1}{3}=-\dfrac{b}{a}$. Product $=\dfrac{4}{3}\times(-1)=-\dfrac{4}{3}=\dfrac{-4}{3}=\dfrac{c}{a}$. Answer: Zeros are $\dfrac{4}{3}$ and $-1$; verified.

Q2. Find a quadratic polynomial each with the given numbers as the sum and product of its zeros respectively.

If $\alpha$ and $\beta$ are zeros, the required polynomial is $k\left[x^{2}-(\alpha+\beta)x+\alpha\beta\right]$, taking $k=1$ for simplicity.

(i) $\dfrac{1}{4},\;-1$ Sum $=\dfrac{1}{4}$, Product $=-1$. $p(x)=x^{2}-\dfrac{1}{4}x-1=\dfrac{1}{4}(4x^{2}-x-4)$. Answer: $4x^{2}-x-4$.

(ii) $\sqrt{2},\;\dfrac{1}{3}$ $p(x)=x^{2}-\sqrt{2}\,x+\dfrac{1}{3}=\dfrac{1}{3}(3x^{2}-3\sqrt{2}\,x+1)$. Answer: $3x^{2}-3\sqrt{2}\,x+1$.

(iii) $0,\;\sqrt{5}$ $p(x)=x^{2}-0\cdot x+\sqrt{5}=x^{2}+\sqrt{5}$. Answer: $x^{2}+\sqrt{5}$.

(iv) $1,\;1$ $p(x)=x^{2}-x+1$. Answer: $x^{2}-x+1$.

(v) $-\dfrac{1}{4},\;\dfrac{1}{4}$ $p(x)=x^{2}+\dfrac{1}{4}x+\dfrac{1}{4}=\dfrac{1}{4}(4x^{2}+x+1)$. Answer: $4x^{2}+x+1$.

(vi) $4,\;1$ $p(x)=x^{2}-4x+1$. Answer: $x^{2}-4x+1$.

Exercise 2.3

Q1. Divide the polynomial $p(x)$ by the polynomial $g(x)$ and find the quotient and remainder in each of the following.

(i) $p(x)=x^{3}-3x^{2}+5x-3,\;g(x)=x^{2}-2$ Step 1: Divide $x^{3}\div x^{2}=x$. Multiply $x(x^{2}-2)=x^{3}-2x$. Subtract: $(x^{3}-3x^{2}+5x-3)-(x^{3}-2x)=-3x^{2}+7x-3$. Step 2: Divide $-3x^{2}\div x^{2}=-3$. Multiply $-3(x^{2}-2)=-3x^{2}+6$. Subtract: $(-3x^{2}+7x-3)-(-3x^{2}+6)=7x-9$. Answer: Quotient $=x-3$, Remainder $=7x-9$.

(ii) $p(x)=x^{4}-3x^{2}+4x+5,\;g(x)=x^{2}+1-x$ Arrange $g(x)=x^{2}-x+1$. Step 1: $x^{4}\div x^{2}=x^{2}$. Multiply $x^{2}(x^{2}-x+1)=x^{4}-x^{3}+x^{2}$. Subtract: result $=x^{3}-4x^{2}+4x+5$. Step 2: $x^{3}\div x^{2}=x$. Multiply $x(x^{2}-x+1)=x^{3}-x^{2}+x$. Subtract: result $=-3x^{2}+3x+5$. Step 3: $-3x^{2}\div x^{2}=-3$. Multiply $-3(x^{2}-x+1)=-3x^{2}+3x-3$. Subtract: result $=8$. Answer: Quotient $=x^{2}+x-3$, Remainder $=8$.

(iii) $p(x)=x^{4}-5x+6,\;g(x)=2-x^{2}$ Arrange $g(x)=-x^{2}+2$. Step 1: $x^{4}\div(-x^{2})=-x^{2}$. Multiply $-x^{2}(-x^{2}+2)=x^{4}-2x^{2}$. Subtract: result $=2x^{2}-5x+6$. Step 2: $2x^{2}\div(-x^{2})=-2$. Multiply $-2(-x^{2}+2)=2x^{2}-4$. Subtract: result $=-5x+10$. Answer: Quotient $=-x^{2}-2$, Remainder $=-5x+10$.

Q2. Check whether the first polynomial is a factor of the second polynomial by dividing the second polynomial by the first polynomial.

(i) $g(x)=t^{2}-3,\;p(t)=2t^{4}+3t^{3}-2t^{2}-9t-12$ Divide step by step: $2t^{4}\div t^{2}=2t^{2}$. Multiply: $2t^{2}(t^{2}-3)=2t^{4}-6t^{2}$. Subtract: $3t^{3}+4t^{2}-9t-12$. $3t^{3}\div t^{2}=3t$. Multiply: $3t(t^{2}-3)=3t^{3}-9t$. Subtract: $4t^{2}-12$. $4t^{2}\div t^{2}=4$. Multiply: $4(t^{2}-3)=4t^{2}-12$. Subtract: $0$. Remainder is $0$. Answer: Yes, $t^{2}-3$ is a factor of $p(t)$.

(ii) $g(x)=x^{2}+3x+1,\;p(x)=3x^{4}+5x^{3}-7x^{2}+2x+2$ $3x^{4}\div x^{2}=3x^{2}$. Multiply: $3x^{2}(x^{2}+3x+1)=3x^{4}+9x^{3}+3x^{2}$. Subtract: $-4x^{3}-10x^{2}+2x+2$. $-4x^{3}\div x^{2}=-4x$. Multiply: $-4x(x^{2}+3x+1)=-4x^{3}-12x^{2}-4x$. Subtract: $2x^{2}+6x+2$. $2x^{2}\div x^{2}=2$. Multiply: $2(x^{2}+3x+1)=2x^{2}+6x+2$. Subtract: $0$. Remainder is $0$. Answer: Yes, $x^{2}+3x+1$ is a factor of $p(x)$.

(iii) $g(x)=x^{3}-3x+1,\;p(x)=x^{5}-4x^{3}+x^{2}+3x+1$ $x^{5}\div x^{3}=x^{2}$. Multiply: $x^{2}(x^{3}-3x+1)=x^{5}-3x^{3}+x^{2}$. Subtract: $-x^{3}+3x+1$. $-x^{3}\div x^{3}=-1$. Multiply: $-1(x^{3}-3x+1)=-x^{3}+3x-1$. Subtract: $2$. Remainder $=2\neq 0$. Answer: No, $x^{3}-3x+1$ is not a factor of $p(x)$.

Q3. Obtain all other zeros of $3x^{4}+6x^{3}-2x^{2}-10x-5$, if two of its zeros are $\sqrt{\dfrac{5}{3}}$ and $-\sqrt{\dfrac{5}{3}}$.

Solution: Since $\sqrt{\dfrac{5}{3}}$ and $-\sqrt{\dfrac{5}{3}}$ are zeros, $\left(x-\sqrt{\dfrac{5}{3}}\right)\left(x+\sqrt{\dfrac{5}{3}}\right)=x^{2}-\dfrac{5}{3}$ is a factor. Equivalently, $3x^{2}-5$ is a factor.

Divide $3x^{4}+6x^{3}-2x^{2}-10x-5$ by $3x^{2}-5$:

• $3x^{4}\div 3x^{2}=x^{2}$. Multiply: $x^{2}(3x^{2}-5)=3x^{4}-5x^{2}$. Subtract: $6x^{3}+3x^{2}-10x-5$.
• $6x^{3}\div 3x^{2}=2x$. Multiply: $2x(3x^{2}-5)=6x^{3}-10x$. Subtract: $3x^{2}-5$.
• $3x^{2}\div 3x^{2}=1$. Multiply: $1(3x^{2}-5)=3x^{2}-5$. Subtract: $0$.

Quotient $=x^{2}+2x+1=(x+1)^{2}$. So the other zeros are $x=-1$ and $x=-1$.
Answer: The other zeros are $-1$ and $-1$.

Q4. On dividing $x^{3}-3x^{2}+x+2$ by a polynomial $g(x)$, the quotient and remainder were $x-2$ and $-2x+4$ respectively. Find $g(x)$.

Solution: By the division algorithm, $p(x)=g(x)\cdot q(x)+r(x)$.

$$x^{3}-3x^{2}+x+2=g(x)(x-2)+(-2x+4)$$

$\Rightarrow g(x)(x-2)=x^{3}-3x^{2}+x+2-(-2x+4)=x^{3}-3x^{2}+3x-2$.

Now divide $x^{3}-3x^{2}+3x-2$ by $x-2$:

• $x^{3}\div x=x^{2}$. Multiply: $x^{2}(x-2)=x^{3}-2x^{2}$. Subtract: $-x^{2}+3x-2$.
• $-x^{2}\div x=-x$. Multiply: $-x(x-2)=-x^{2}+2x$. Subtract: $x-2$.
• $x\div x=1$. Multiply: $1(x-2)=x-2$. Subtract: $0$.

Answer: $g(x)=x^{2}-x+1$.

Q5. Give examples of polynomials $p(x),g(x),q(x)$ and $r(x)$, which satisfy the division algorithm and

• (i) $\deg p(x)=\deg q(x)$. Take $g(x)$ to be a non-zero constant.
  Example: $p(x)=6x^{2}+2x+2,\;g(x)=2,\;q(x)=3x^{2}+x+1,\;r(x)=0$.
• (ii) $\deg q(x)=\deg r(x)$.
  Example: $p(x)=x^{3}+x+1,\;g(x)=x^{2},\;q(x)=x,\;r(x)=x+1$.
  Here both $q(x)$ and $r(x)$ have degree $1$.
• (iii) $\deg r(x)=0$. The remainder is a non-zero constant.
  Example: $p(x)=x^{2}+1,\;g(x)=x,\;q(x)=x,\;r(x)=1$.

Exercise 2.4 (Optional)

Q1. Verify that the numbers given alongside of the cubic polynomials below are their zeros. Also verify the relationship between the zeros and the coefficients in each case.

(i) $2x^{3}+x^{2}-5x+2;\;\dfrac{1}{2},1,-2$ $p\!\left(\dfrac{1}{2}\right)=2\cdot\dfrac{1}{8}+\dfrac{1}{4}-\dfrac{5}{2}+2=\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{5}{2}+2=0$. $p(1)=2+1-5+2=0$. $p(-2)=-16+4+10+2=0$. All three are zeros. $a=2,b=1,c=-5,d=2$. $\alpha+\beta+\gamma=\dfrac{1}{2}+1-2=-\dfrac{1}{2}=-\dfrac{b}{a}$. $\alpha\beta+\beta\gamma+\gamma\alpha=\dfrac{1}{2}\cdot1+1\cdot(-2)+(-2)\cdot\dfrac{1}{2}=\dfrac{1}{2}-2-1=-\dfrac{5}{2}=\dfrac{c}{a}$. $\alpha\beta\gamma=\dfrac{1}{2}\cdot1\cdot(-2)=-1=-\dfrac{d}{a}=-\dfrac{2}{2}$. Answer: Zeros and relationships verified.

(ii) $x^{3}-4x^{2}+5x-2;\;2,1,1$ $p(2)=8-16+10-2=0$. $p(1)=1-4+5-2=0$. So $2,1,1$ are zeros. $a=1,b=-4,c=5,d=-2$. Sum $=2+1+1=4=-\dfrac{-4}{1}=-\dfrac{b}{a}$. $\alpha\beta+\beta\gamma+\gamma\alpha=2\cdot1+1\cdot1+1\cdot2=5=\dfrac{c}{a}$. $\alpha\beta\gamma=2\cdot1\cdot1=2=-\dfrac{-2}{1}=-\dfrac{d}{a}$. Answer: Zeros and relationships verified.

Q2. Find a cubic polynomial with the sum, sum of the product of its zeros taken two at a time, and the product of its zeros as $2,-7,-14$ respectively.

Solution: The required polynomial (taking $a=1$) is:

$$p(x)=x^{3}-(\alpha+\beta+\gamma)x^{2}+(\alpha\beta+\beta\gamma+\gamma\alpha)x-\alpha\beta\gamma$$

$=x^{3}-2x^{2}-7x+14$.
Answer: $p(x)=x^{3}-2x^{2}-7x+14$.

Q3. If the zeros of the polynomial $x^{3}-3x^{2}+x+1$ are $a-b,\;a,\;a+b$, find $a$ and $b$.

Solution: Sum of zeros $=(a-b)+a+(a+b)=3a=-\dfrac{-3}{1}=3$. Therefore $a=1$.

Product of zeros $=(a-b)\cdot a\cdot(a+b)=a(a^{2}-b^{2})=-\dfrac{1}{1}=-1$.
With $a=1$: $1\cdot(1-b^{2})=-1\Rightarrow 1-b^{2}=-1\Rightarrow b^{2}=2\Rightarrow b=\pm\sqrt{2}$.
Answer: $a=1$ and $b=\pm\sqrt{2}$.

Q4. If two zeros of the polynomial $x^{4}-6x^{3}-26x^{2}+138x-35$ are $2\pm\sqrt{3}$, find the other zeros.

Solution: Since $(2+\sqrt{3})$ and $(2-\sqrt{3})$ are zeros, $\bigl(x-(2+\sqrt{3})\bigr)\bigl(x-(2-\sqrt{3})\bigr)=x^{2}-4x+1$ is a factor.

Divide the given polynomial by $x^{2}-4x+1$:

• $x^{4}\div x^{2}=x^{2}$. Multiply: $x^{2}(x^{2}-4x+1)=x^{4}-4x^{3}+x^{2}$. Subtract: $-2x^{3}-27x^{2}+138x-35$.
• $-2x^{3}\div x^{2}=-2x$. Multiply: $-2x(x^{2}-4x+1)=-2x^{3}+8x^{2}-2x$. Subtract: $-35x^{2}+140x-35$.
• $-35x^{2}\div x^{2}=-35$. Multiply: $-35(x^{2}-4x+1)=-35x^{2}+140x-35$. Subtract: $0$.

Quotient $=x^{2}-2x-35=(x-7)(x+5)$. So the other zeros are $x=7$ and $x=-5$.
Answer: The other zeros are $7$ and $-5$.

Q5. 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: Perform the division.

• $x^{4}\div x^{2}=x^{2}$. Multiply: $x^{2}(x^{2}-2x+k)=x^{4}-2x^{3}+kx^{2}$. Subtract: $-4x^{3}+(16-k)x^{2}-25x+10$.
• $-4x^{3}\div x^{2}=-4x$. Multiply: $-4x(x^{2}-2x+k)=-4x^{3}+8x^{2}-4kx$. Subtract: $(8-k)x^{2}+(-25+4k)x+10$.
• $(8-k)x^{2}\div x^{2}=(8-k)$. Multiply: $(8-k)(x^{2}-2x+k)=(8-k)x^{2}-2(8-k)x+k(8-k)$. Subtract: $\bigl[(-25+4k)+2(8-k)\bigr]x+\bigl[10-k(8-k)\bigr]$.

Coefficient of $x$ in remainder $=(-25+4k)+(16-2k)=2k-9$.
Constant in remainder $=10-8k+k^{2}=k^{2}-8k+10$.

Given remainder $=x+a$, so equate:
$2k-9=1\Rightarrow k=5$.
$k^{2}-8k+10=a\Rightarrow 25-40+10=a\Rightarrow a=-5$.
Answer: $k=5,\;a=-5$.

Additional Practice Questions

Q1. Find a quadratic polynomial whose zeros are $-3$ and $4$. Sum $=-3+4=1$, Product $=(-3)(4)=-12$. $p(x)=x^{2}-x-12$. Answer: $x^{2}-x-12$.

Q2. If $\alpha$ and $\beta$ are zeros of $x^{2}-5x+6$, find $\alpha^{2}+\beta^{2}$. $\alpha+\beta=5$, $\alpha\beta=6$. $\alpha^{2}+\beta^{2}=(\alpha+\beta)^{2}-2\alpha\beta=25-12=13$. Answer: $13$.

Q3. Find the zeros of $\sqrt{3}x^{2}-8x+4\sqrt{3}$. $\sqrt{3}x^{2}-8x+4\sqrt{3}=\sqrt{3}x^{2}-6x-2x+4\sqrt{3}=\sqrt{3}x(x-2\sqrt{3})-2(x-2\sqrt{3})=(x-2\sqrt{3})(\sqrt{3}x-2)$. Zeros: $x=2\sqrt{3},\;x=\dfrac{2}{\sqrt{3}}=\dfrac{2\sqrt{3}}{3}$. Answer: $2\sqrt{3}$ and $\dfrac{2\sqrt{3}}{3}$.

Q4. If $\alpha,\beta$ are zeros of $p(x)=2x^{2}-7x+3$, find $\dfrac{1}{\alpha}+\dfrac{1}{\beta}$. $\alpha+\beta=\dfrac{7}{2}$, $\alpha\beta=\dfrac{3}{2}$. $\dfrac{1}{\alpha}+\dfrac{1}{\beta}=\dfrac{\alpha+\beta}{\alpha\beta}=\dfrac{7/2}{3/2}=\dfrac{7}{3}$. Answer: $\dfrac{7}{3}$.

Q5. Find $k$ if the polynomial $kx^{2}+2x+3k$ has equal zeros. For equal zeros, sum $=$ product condition, or use $b^{2}-4ac=0$: $(2)^{2}-4(k)(3k)=0\Rightarrow 4-12k^{2}=0\Rightarrow k^{2}=\dfrac{1}{3}\Rightarrow k=\pm\dfrac{1}{\sqrt{3}}$. Answer: $k=\pm\dfrac{1}{\sqrt{3}}$.

Q6. The graph of a polynomial $p(x)$ touches the x-axis at exactly one point. How many real zeros does $p(x)$ have? A graph that touches but does not cross the x-axis indicates that the corresponding zero is repeated. For a quadratic, both zeros are equal at that point. Answer: Two equal real zeros.

Q7. If one zero of the polynomial $(a^{2}+9)x^{2}+13x+6a$ is reciprocal of the other, find the value of $a$. If zeros are $\alpha$ and $\dfrac{1}{\alpha}$, product $=1$. Product $=\dfrac{6a}{a^{2}+9}=1\Rightarrow 6a=a^{2}+9\Rightarrow a^{2}-6a+9=0\Rightarrow (a-3)^{2}=0\Rightarrow a=3$. Answer: $a=3$.

Q8. Find a cubic polynomial whose zeros are $3,-1$ and $-\dfrac{1}{3}$. Sum $=3-1-\dfrac{1}{3}=\dfrac{5}{3}$. Sum of products in pairs $=3(-1)+(-1)\!\left(-\dfrac{1}{3}\right)+\left(-\dfrac{1}{3}\right)(3)=-3+\dfrac{1}{3}-1=-\dfrac{11}{3}$. Product $=3(-1)\!\left(-\dfrac{1}{3}\right)=1$. $p(x)=x^{3}-\dfrac{5}{3}x^{2}-\dfrac{11}{3}x-1$, or equivalently $3x^{3}-5x^{2}-11x-3$. Answer: $3x^{3}-5x^{2}-11x-3$.

More Solved Examples (HSLC Style)

Q9. If $\alpha$ and $\beta$ are the zeros of the quadratic polynomial $f(x)=x^{2}-p(x+1)-c$, show that $(\alpha+1)(\beta+1)=1-c$. Expand: $f(x)=x^{2}-px-(p+c)$. So $\alpha+\beta=p$ and $\alpha\beta=-(p+c)$. $(\alpha+1)(\beta+1)=\alpha\beta+\alpha+\beta+1=-(p+c)+p+1=1-c$. Answer: Hence proved.

Q10. Find a quadratic polynomial whose zeros are $\dfrac{3+\sqrt{5}}{2}$ and $\dfrac{3-\sqrt{5}}{2}$. Sum $=\dfrac{3+\sqrt{5}}{2}+\dfrac{3-\sqrt{5}}{2}=\dfrac{6}{2}=3$. Product $=\dfrac{(3+\sqrt{5})(3-\sqrt{5})}{4}=\dfrac{9-5}{4}=1$. $p(x)=x^{2}-3x+1$. Answer: $x^{2}-3x+1$.

Q11. If $\alpha$ and $\beta$ are zeros of $x^{2}-6x+a$ such that $3\alpha+2\beta=20$, find $a$. $\alpha+\beta=6$, $\alpha\beta=a$. From $3\alpha+2\beta=20$ and $\alpha+\beta=6$, multiply second by $2$: $2\alpha+2\beta=12$. Subtract: $\alpha=8$, so $\beta=-2$. $\alpha\beta=8\cdot(-2)=-16$. Therefore $a=-16$. Answer: $a=-16$.

Q12. If one zero of the polynomial $f(x)=(k^{2}+4)x^{2}+13x+4k$ is reciprocal of the other, find $k$. Product of zeros $=\alpha\cdot\dfrac{1}{\alpha}=1$. $\dfrac{4k}{k^{2}+4}=1\Rightarrow 4k=k^{2}+4\Rightarrow k^{2}-4k+4=0\Rightarrow (k-2)^{2}=0\Rightarrow k=2$. Answer: $k=2$.

Q13. Find the value of $k$ such that the polynomial $x^{2}-(k+6)x+2(2k-1)$ has sum of its zeros equal to half their product. Sum $=k+6$, Product $=2(2k-1)=4k-2$. Condition: $k+6=\dfrac{1}{2}(4k-2)=2k-1\Rightarrow k+6=2k-1\Rightarrow k=7$. Answer: $k=7$.

Q14. If $\alpha,\beta$ are zeros of $p(x)=x^{2}+x-2$, find $\dfrac{1}{\alpha}-\dfrac{1}{\beta}$ given $\alpha>\beta$. $\alpha+\beta=-1$, $\alpha\beta=-2$. $(\alpha-\beta)^{2}=(\alpha+\beta)^{2}-4\alpha\beta=1+8=9\Rightarrow \alpha-\beta=3$. $\dfrac{1}{\alpha}-\dfrac{1}{\beta}=\dfrac{\beta-\alpha}{\alpha\beta}=\dfrac{-3}{-2}=\dfrac{3}{2}$. Answer: $\dfrac{3}{2}$.

Q15. Find the zeros of $4\sqrt{3}x^{2}+5x-2\sqrt{3}$ and verify the relations. $4\sqrt{3}x^{2}+5x-2\sqrt{3}=4\sqrt{3}x^{2}+8x-3x-2\sqrt{3}=4x(\sqrt{3}x+2)-\sqrt{3}(\sqrt{3}x+2)=(\sqrt{3}x+2)(4x-\sqrt{3})$. Zeros: $x=-\dfrac{2}{\sqrt{3}}=-\dfrac{2\sqrt{3}}{3}$ and $x=\dfrac{\sqrt{3}}{4}$. $a=4\sqrt{3},\;b=5,\;c=-2\sqrt{3}$. Sum $=-\dfrac{2\sqrt{3}}{3}+\dfrac{\sqrt{3}}{4}=\dfrac{-8\sqrt{3}+3\sqrt{3}}{12}=-\dfrac{5\sqrt{3}}{12}=-\dfrac{5}{4\sqrt{3}}=-\dfrac{b}{a}$. ✓ Product $=-\dfrac{2\sqrt{3}}{3}\cdot\dfrac{\sqrt{3}}{4}=-\dfrac{6}{12}=-\dfrac{1}{2}=\dfrac{-2\sqrt{3}}{4\sqrt{3}}=\dfrac{c}{a}$. ✓ Answer: Zeros are $-\dfrac{2\sqrt{3}}{3}$ and $\dfrac{\sqrt{3}}{4}$.

Q16. Apply the division algorithm to find quotient and remainder when $p(x)=x^{4}+2x^{3}-3x^{2}+x-1$ is divided by $g(x)=x-2$. $x^{4}\div x=x^{3}$. $x^{3}(x-2)=x^{4}-2x^{3}$. Subtract: $4x^{3}-3x^{2}+x-1$. $4x^{3}\div x=4x^{2}$. $4x^{2}(x-2)=4x^{3}-8x^{2}$. Subtract: $5x^{2}+x-1$. $5x^{2}\div x=5x$. $5x(x-2)=5x^{2}-10x$. Subtract: $11x-1$. $11x\div x=11$. $11(x-2)=11x-22$. Subtract: $21$. Answer: Quotient $=x^{3}+4x^{2}+5x+11$, Remainder $=21$.

Q17. If $\alpha,\beta,\gamma$ are zeros of $x^{3}-3x^{2}-x+3$, find them. Try $x=1$: $1-3-1+3=0$. So $(x-1)$ is a factor. Divide: $x^{3}-3x^{2}-x+3=(x-1)(x^{2}-2x-3)=(x-1)(x-3)(x+1)$. Answer: Zeros are $1,3,-1$.

Q18. Construct a quadratic polynomial whose zeros are reciprocals of the zeros of $ax^{2}+bx+c$. Original zeros: $\alpha,\beta$ with $\alpha+\beta=-\dfrac{b}{a},\;\alpha\beta=\dfrac{c}{a}$. New zeros: $\dfrac{1}{\alpha},\dfrac{1}{\beta}$. New sum $=\dfrac{\alpha+\beta}{\alpha\beta}=\dfrac{-b/a}{c/a}=-\dfrac{b}{c}$. New product $=\dfrac{1}{\alpha\beta}=\dfrac{a}{c}$. $p(x)=x^{2}+\dfrac{b}{c}x+\dfrac{a}{c}=\dfrac{1}{c}(cx^{2}+bx+a)$. Answer: $cx^{2}+bx+a$.

Q19. The sum and product of zeros of a quadratic polynomial are $-3$ and $2$ respectively. Show that one zero is double the other. Polynomial: $x^{2}+3x+2=(x+1)(x+2)$. Zeros: $-1$ and $-2$. Indeed $-2=2\times(-1)$. Answer: Verified.

Q20. If $\alpha$ and $\beta$ are zeros of $2x^{2}+5x+k$, find $k$ such that $(\alpha+\beta)^{2}-\alpha\beta=\dfrac{21}{4}$. $\alpha+\beta=-\dfrac{5}{2}$, $\alpha\beta=\dfrac{k}{2}$. $\dfrac{25}{4}-\dfrac{k}{2}=\dfrac{21}{4}\Rightarrow \dfrac{k}{2}=\dfrac{4}{4}=1\Rightarrow k=2$. Answer: $k=2$.

Multiple Choice Questions (MCQ)

Q1. If one of the zeros of the quadratic polynomial $(k-1)x^{2}+kx+1$ is $-3$, then the value of $k$ is (a) $\dfrac{4}{3}$ (b) $-\dfrac{4}{3}$ (c) $\dfrac{2}{3}$ (d) $-\dfrac{2}{3}$ Substitute $x=-3$: $(k-1)\cdot 9+k(-3)+1=0\Rightarrow 9k-9-3k+1=0\Rightarrow 6k=8\Rightarrow k=\dfrac{4}{3}$. Answer: (a) $\dfrac{4}{3}$.

Q2. The number of polynomials having zeros $-2$ and $5$ is (a) $1$ (b) $2$ (c) $3$ (d) more than $3$ Any polynomial of the form $k(x+2)(x-5)$ for non-zero $k$ has these zeros. Answer: (d) more than $3$.

Q3. If $\alpha,\beta$ are the zeros of $x^{2}+5x+8$, then $\alpha+\beta$ is (a) $5$ (b) $-5$ (c) $8$ (d) $-8$ $\alpha+\beta=-\dfrac{5}{1}=-5$. Answer: (b) $-5$.

Q4. The zeros of $x^{2}-2x-8$ are (a) $(2,-4)$ (b) $(4,-2)$ (c) $(-2,-4)$ (d) $(-4,2)$ $x^{2}-2x-8=(x-4)(x+2)$. Answer: (b) $(4,-2)$.

Q5. A quadratic polynomial whose sum and product of zeros are $-3$ and $2$ respectively is (a) $x^{2}+3x+2$ (b) $x^{2}-3x+2$ (c) $x^{2}-3x-2$ (d) $x^{2}+3x-2$ $p(x)=x^{2}-(\text{sum})x+\text{product}=x^{2}+3x+2$. Answer: (a) $x^{2}+3x+2$.

Q6. If the zeros of $x^{2}+px+q$ are double the zeros of $2x^{2}-5x-3$, then (a) $p=-5,q=-6$ (b) $p=5,q=6$ (c) $p=-5,q=6$ (d) $p=5,q=-6$ Zeros of $2x^{2}-5x-3$: $(2x+1)(x-3)$, so $-\dfrac{1}{2},3$. Doubled: $-1,6$. Sum $=5$, Product $=-6$. So $p=-5,q=-6$. Answer: (a) $p=-5,q=-6$.

Q7. If $\alpha,\beta,\gamma$ are zeros of the cubic polynomial $x^{3}+px^{2}+qx+r$, then $\dfrac{1}{\alpha}+\dfrac{1}{\beta}+\dfrac{1}{\gamma}$ equals (a) $-\dfrac{p}{r}$ (b) $-\dfrac{q}{r}$ (c) $\dfrac{p}{r}$ (d) $\dfrac{q}{r}$ $\dfrac{1}{\alpha}+\dfrac{1}{\beta}+\dfrac{1}{\gamma}=\dfrac{\alpha\beta+\beta\gamma+\gamma\alpha}{\alpha\beta\gamma}=\dfrac{q}{-r}=-\dfrac{q}{r}$. Answer: (b) $-\dfrac{q}{r}$.

Q8. If a polynomial $p(x)$ is divided by $g(x)$ and the remainder is $r(x)$, then (a) $\deg r(x)\geq \deg g(x)$ (b) $\deg r(x)<\deg g(x)$ or $r(x)=0$ (c) $\deg r(x)>\deg g(x)$ (d) None By the division algorithm. Answer: (b).

Q9. The graph of $y=ax^{2}+bx+c$ does not intersect the x-axis. Then (a) the polynomial has two distinct real zeros (b) the polynomial has two equal real zeros (c) the polynomial has no real zero (d) cannot be determined No intersection means no real zero. Answer: (c) no real zero.

Q10. If one zero of $2x^{2}-3x+k$ is the reciprocal of the other, then $k$ is (a) $\dfrac{3}{2}$ (b) $-\dfrac{2}{3}$ (c) $2$ (d) $\dfrac{2}{3}$ Product $=1\Rightarrow \dfrac{k}{2}=1\Rightarrow k=2$. Answer: (c) $2$.

Short Answer Type Questions (2 marks)

Q1. State the relationship between the zeros and coefficients of a quadratic polynomial $ax^{2}+bx+c$. If $\alpha$ and $\beta$ are zeros, then $\alpha+\beta=-\dfrac{b}{a}$ and $\alpha\beta=\dfrac{c}{a}$. Answer: Sum of zeros $=-b/a$, Product of zeros $=c/a$.

Q2. Define a zero of a polynomial. Give an example. A real number $k$ is a zero of $p(x)$ if $p(k)=0$. Example: For $p(x)=x^{2}-9$, $p(3)=0$, so $3$ is a zero. Answer: See above.

Q3. What is the maximum number of zeros that a polynomial of degree $n$ can have? A polynomial of degree $n$ can have at most $n$ real zeros. Answer: $n$.

Q4. If the product of zeros of the quadratic polynomial $x^{2}-4x+k$ is $3$, find $k$. Product $=\dfrac{k}{1}=3\Rightarrow k=3$. Answer: $k=3$.

Q5. Write a quadratic polynomial whose sum of zeros is $0$ and product is $-9$. $p(x)=x^{2}-0\cdot x+(-9)=x^{2}-9$. Answer: $x^{2}-9$.

Q6. The graph of $y=p(x)$ is given. If it intersects the x-axis at exactly two points, what is the minimum possible degree of $p(x)$? Minimum degree must be $2$. Answer: $2$.

Q7. Find the zero of the linear polynomial $p(x)=2x+5$. $2x+5=0\Rightarrow x=-\dfrac{5}{2}$. Answer: $-\dfrac{5}{2}$.

Q8. If $-1$ is a zero of $p(x)=x^{2}+3x+k$, find $k$. $p(-1)=1-3+k=0\Rightarrow k=2$. Answer: $k=2$.

Q9. If $\alpha,\beta$ are zeros of $4x^{2}+3x+7$, find $\alpha+\beta+\alpha\beta$. $\alpha+\beta=-\dfrac{3}{4}$, $\alpha\beta=\dfrac{7}{4}$. Sum $=-\dfrac{3}{4}+\dfrac{7}{4}=1$. Answer: $1$.

Q10. State the division algorithm for polynomials. For polynomials $p(x)$ and $g(x)\neq 0$, there exist unique polynomials $q(x)$ and $r(x)$ such that $p(x)=g(x)\,q(x)+r(x)$, where $r(x)=0$ or $\deg r(x)<\deg g(x)$. Answer: See above.

Long Answer Type Questions (4 marks)

Q1. Find all the zeros of the polynomial $p(x)=2x^{4}-3x^{3}-3x^{2}+6x-2$ if two of its zeros are $\sqrt{2}$ and $-\sqrt{2}$. Since $\sqrt{2},-\sqrt{2}$ are zeros, $(x-\sqrt{2})(x+\sqrt{2})=x^{2}-2$ is a factor. Divide $2x^{4}-3x^{3}-3x^{2}+6x-2$ by $x^{2}-2$: $2x^{4}\div x^{2}=2x^{2}$. $2x^{2}(x^{2}-2)=2x^{4}-4x^{2}$. Subtract: $-3x^{3}+x^{2}+6x-2$. $-3x^{3}\div x^{2}=-3x$. $-3x(x^{2}-2)=-3x^{3}+6x$. Subtract: $x^{2}-2$. $x^{2}\div x^{2}=1$. $1(x^{2}-2)=x^{2}-2$. Subtract: $0$. Quotient $=2x^{2}-3x+1=(2x-1)(x-1)$. Zeros: $\dfrac{1}{2},1$. Answer: All four zeros are $\sqrt{2},-\sqrt{2},\dfrac{1}{2},1$.

Q2. If the zeros of the polynomial $f(x)=x^{3}-12x^{2}+39x-28$ are in arithmetic progression, find them. Let zeros be $a-d,a,a+d$. Sum $=3a=12\Rightarrow a=4$. Product $=(a-d)\cdot a\cdot(a+d)=a(a^{2}-d^{2})=4(16-d^{2})=28\Rightarrow 16-d^{2}=7\Rightarrow d^{2}=9\Rightarrow d=\pm 3$. Zeros: $1,4,7$ (or $7,4,1$). Answer: $1,4,7$.

Q3. If two zeros of the polynomial $p(x)=x^{4}+x^{3}-9x^{2}-3x+18$ are $\sqrt{3}$ and $-\sqrt{3}$, find the other zeros. $(x-\sqrt{3})(x+\sqrt{3})=x^{2}-3$. Divide $x^{4}+x^{3}-9x^{2}-3x+18$ by $x^{2}-3$: $x^{4}\div x^{2}=x^{2}$. $x^{2}(x^{2}-3)=x^{4}-3x^{2}$. Subtract: $x^{3}-6x^{2}-3x+18$. $x^{3}\div x^{2}=x$. $x(x^{2}-3)=x^{3}-3x$. Subtract: $-6x^{2}+18$. $-6x^{2}\div x^{2}=-6$. $-6(x^{2}-3)=-6x^{2}+18$. Subtract: $0$. Quotient $=x^{2}+x-6=(x+3)(x-2)$. Other zeros: $-3,2$. Answer: $-3$ and $2$.

Q4. Apply the division algorithm to find the quotient and remainder when $p(x)=2x^{3}-5x^{2}+8x-7$ is divided by $g(x)=2x-3$. $2x^{3}\div 2x=x^{2}$. $x^{2}(2x-3)=2x^{3}-3x^{2}$. Subtract: $-2x^{2}+8x-7$. $-2x^{2}\div 2x=-x$. $-x(2x-3)=-2x^{2}+3x$. Subtract: $5x-7$. $5x\div 2x=\dfrac{5}{2}$. $\dfrac{5}{2}(2x-3)=5x-\dfrac{15}{2}$. Subtract: $-7+\dfrac{15}{2}=\dfrac{1}{2}$. Answer: Quotient $=x^{2}-x+\dfrac{5}{2}$, Remainder $=\dfrac{1}{2}$.

Q5. If $\alpha,\beta$ are the zeros of the quadratic polynomial $p(x)=2x^{2}-4x+5$, find the value of $\alpha^{2}+\beta^{2}$ and $\dfrac{1}{\alpha}+\dfrac{1}{\beta}$. $\alpha+\beta=2$, $\alpha\beta=\dfrac{5}{2}$. $\alpha^{2}+\beta^{2}=(\alpha+\beta)^{2}-2\alpha\beta=4-5=-1$. $\dfrac{1}{\alpha}+\dfrac{1}{\beta}=\dfrac{\alpha+\beta}{\alpha\beta}=\dfrac{2}{5/2}=\dfrac{4}{5}$. Answer: $\alpha^{2}+\beta^{2}=-1$ and $\dfrac{1}{\alpha}+\dfrac{1}{\beta}=\dfrac{4}{5}$.

Q6. If $\alpha,\beta$ are zeros of $p(x)=x^{2}-6x+a$, find $a$ such that $3\alpha+2\beta=20$. $\alpha+\beta=6$ and $3\alpha+2\beta=20$. From these: $\alpha=20-2\cdot 6=8$ (subtract twice the first from the second), $\beta=-2$. Then $\alpha\beta=-16=a$. Answer: $a=-16$.

Q7. Verify that $3,-1,-\dfrac{1}{3}$ are the zeros of the cubic polynomial $p(x)=3x^{3}-5x^{2}-11x-3$ and verify the relationship between the zeros and coefficients. $p(3)=81-45-33-3=0$. $p(-1)=-3-5+11-3=0$. $p(-\dfrac{1}{3})=-\dfrac{1}{9}-\dfrac{5}{9}+\dfrac{11}{3}-3=-\dfrac{6}{9}+\dfrac{11}{3}-3=-\dfrac{2}{3}+\dfrac{11}{3}-3=\dfrac{9}{3}-3=0$. $a=3,b=-5,c=-11,d=-3$. Sum $=3-1-\dfrac{1}{3}=\dfrac{5}{3}=-\dfrac{-5}{3}=-\dfrac{b}{a}$. ✓ Sum of products in pairs $=3(-1)+(-1)\cdot(-\dfrac{1}{3})+(-\dfrac{1}{3})\cdot 3=-3+\dfrac{1}{3}-1=-\dfrac{11}{3}=\dfrac{c}{a}$. ✓ Product $=3\cdot(-1)\cdot(-\dfrac{1}{3})=1=-\dfrac{-3}{3}=-\dfrac{d}{a}$. ✓ Answer: All three relationships verified.

HOTS (Higher Order Thinking Skills)

Q1. If one zero of the polynomial $f(x)=(k+1)x^{2}-5x+5$ is multiplicative inverse of the other, find the zeros and $k$. Let zeros be $\alpha$ and $\dfrac{1}{\alpha}$. Product $=1=\dfrac{5}{k+1}\Rightarrow k+1=5\Rightarrow k=4$. Polynomial: $5x^{2}-5x+5=5(x^{2}-x+1)$. Discriminant: $1-4=-3<0$, so no real zeros. Answer: $k=4$; zeros are not real.

Q2. The zeros of a quadratic polynomial $p(x)$ are $\alpha=2+\sqrt{3}$ and $\beta=2-\sqrt{3}$. Verify the relationship between zeros and coefficients of $p(x)$. Sum $=4$, Product $=4-3=1$. $p(x)=x^{2}-4x+1$. Here $a=1,b=-4,c=1$. $-\dfrac{b}{a}=4=\alpha+\beta$, $\dfrac{c}{a}=1=\alpha\beta$. ✓ Answer: Verified.

Q3. If $\alpha,\beta$ are the zeros of $f(x)=x^{2}-px+q$, find the polynomial whose zeros are $\dfrac{\alpha^{2}}{\beta}$ and $\dfrac{\beta^{2}}{\alpha}$. $\alpha+\beta=p$, $\alpha\beta=q$. New sum $=\dfrac{\alpha^{3}+\beta^{3}}{\alpha\beta}=\dfrac{(\alpha+\beta)^{3}-3\alpha\beta(\alpha+\beta)}{\alpha\beta}=\dfrac{p^{3}-3pq}{q}$. New product $=\dfrac{\alpha^{2}\beta^{2}}{\alpha\beta}=\alpha\beta=q$. Required polynomial: $x^{2}-\dfrac{p^{3}-3pq}{q}x+q$, equivalently $qx^{2}-(p^{3}-3pq)x+q^{2}$. Answer: $qx^{2}-(p^{3}-3pq)x+q^{2}$.

Q4. Find the value of $k$ for which the polynomial $x^{4}-2x^{3}+3x^{2}-(k+1)x+k$ leaves a remainder of $-1$ when divided by $(x-1)$. By Remainder Theorem, $p(1)=1-2+3-(k+1)+k=1-2+3-k-1+k=1$. So remainder is $1$, not $-1$, regardless of $k$. Hence no value of $k$ satisfies the condition. Answer: No real value of $k$ exists.

Q5. If the polynomial $6x^{4}+8x^{3}+17x^{2}+21x+7$ is divided by another polynomial $3x^{2}+4x+1$, the remainder comes out to be $ax+b$. Find $a$ and $b$. Divide step by step: $6x^{4}\div 3x^{2}=2x^{2}$. $2x^{2}(3x^{2}+4x+1)=6x^{4}+8x^{3}+2x^{2}$. Subtract: $15x^{2}+21x+7$. $15x^{2}\div 3x^{2}=5$. $5(3x^{2}+4x+1)=15x^{2}+20x+5$. Subtract: $x+2$. Remainder $=x+2$, so $a=1,b=2$. Answer: $a=1,b=2$.

Case-Study / Application Based Questions

Case Study 1: A suspension bridge has its main cable in the shape of a parabola. The lowest point of the cable is on the road. The cable is fixed at the top of two towers each $30\,\text{m}$ tall and the towers are $80\,\text{m}$ apart. Taking the lowest point as origin, the equation of the parabola can be written as $y=ax^{2}$.

(i) Find the value of $a$. At a tower, $x=40$ and $y=30$. So $30=a(40)^{2}=1600a\Rightarrow a=\dfrac{3}{160}$. Answer: $a=\dfrac{3}{160}$.

(ii) Find the height of the cable above the road at a horizontal distance of $20\,\text{m}$ from the lowest point. $y=\dfrac{3}{160}\cdot(20)^{2}=\dfrac{3}{160}\cdot 400=\dfrac{1200}{160}=7.5$ m. Answer: $7.5$ m.

(iii) Identify the type of polynomial whose graph is the cable. The cable is the graph of a quadratic polynomial $y=\dfrac{3}{160}x^{2}$. Answer: Quadratic polynomial.

(iv) How many zeros does the polynomial $\dfrac{3}{160}x^{2}$ have? Setting $y=0$ gives $x=0$ as a double root. Answer: One zero (a repeated zero), so two equal zeros.

Case Study 2: A farmer has a rectangular field of length $(2x+3)\,\text{m}$ and breadth $(x+1)\,\text{m}$. The area $A$ in square metres is therefore a polynomial in $x$.

(i) Express the area as a polynomial in $x$. $A=(2x+3)(x+1)=2x^{2}+5x+3$. Answer: $A=2x^{2}+5x+3$.

(ii) Find the zeros of the polynomial $A(x)=2x^{2}+5x+3$. $2x^{2}+5x+3=(2x+3)(x+1)$. Zeros: $x=-\dfrac{3}{2},\;x=-1$. Answer: $-\dfrac{3}{2}$ and $-1$.

(iii) Verify the relationship between the zeros and the coefficients of $A(x)$. $a=2,b=5,c=3$. Sum of zeros $=-\dfrac{3}{2}-1=-\dfrac{5}{2}=-\dfrac{b}{a}$. ✓ Product $=\left(-\dfrac{3}{2}\right)(-1)=\dfrac{3}{2}=\dfrac{c}{a}$. ✓ Answer: Verified.

(iv) For what value of $x$ is the area $20$ sq m? $2x^{2}+5x+3=20\Rightarrow 2x^{2}+5x-17=0$. Discriminant $=25+136=161$. $x=\dfrac{-5+\sqrt{161}}{4}$ (positive root). Answer: $x=\dfrac{-5+\sqrt{161}}{4}$ m.

Case Study 3: The path of a ball thrown by a student is given by the polynomial $h(t)=-5t^{2}+20t+25$, where $h$ is the height in metres at time $t$ seconds.

(i) What is the height of the ball at $t=0$? $h(0)=25$ m. Answer: $25$ m.

(ii) Find the time when the ball hits the ground. $-5t^{2}+20t+25=0\Rightarrow t^{2}-4t-5=0\Rightarrow (t-5)(t+1)=0$. $t=5$ or $t=-1$. Since $t>0$, $t=5$ s. Answer: $5$ seconds.

(iii) State the type and degree of $h(t)$. Quadratic polynomial of degree $2$. Answer: Quadratic, degree $2$.

(iv) Verify the relationship between zeros and coefficients of $h(t)$. $a=-5,b=20,c=25$. Zeros: $5,-1$. Sum $=5+(-1)=4=-\dfrac{20}{-5}=-\dfrac{b}{a}$. ✓ Product $=5(-1)=-5=\dfrac{25}{-5}=\dfrac{c}{a}$. ✓ Answer: Verified.

Assertion-Reason Type Questions

For each, choose: (a) Both A and R are true and R is the correct explanation of A. (b) Both A and R are true but R is not the correct explanation of A. (c) A is true, R is false. (d) A is false, R is true.

Q1. Assertion (A): The polynomial $p(x)=x^{2}+x+1$ has no real zeros. Reason (R): A quadratic polynomial $ax^{2}+bx+c$ has no real zeros when $b^{2}-4ac<0$. $b^{2}-4ac=1-4=-3<0$, so no real zero. R correctly explains A. Answer: (a).

Q2. Assertion (A): The polynomial $4x^{2}+4x+1$ has two equal real zeros. Reason (R): If $b^{2}-4ac=0$, the quadratic polynomial has two equal real zeros. $16-16=0$, so two equal zeros. R explains A. Answer: (a).

Q3. Assertion (A): If $\alpha,\beta$ are zeros of $x^{2}-x-2$, then $\alpha+\beta=1$ and $\alpha\beta=-2$. Reason (R): For $ax^{2}+bx+c$, sum of zeros $=-\dfrac{b}{a}$ and product $=\dfrac{c}{a}$. Here $\alpha+\beta=1,\alpha\beta=-2$. R is the formula used. Answer: (a).

Q4. Assertion (A): A polynomial of degree $3$ has exactly $3$ real zeros. Reason (R): The maximum number of zeros of a polynomial of degree $n$ is $n$. A is false (a cubic may have $1$ or $3$ real zeros), R is true. Answer: (d).

Q5. Assertion (A): The graph of $y=p(x)$, where $p(x)$ is a polynomial of degree $4$, can intersect the x-axis at most $4$ points. Reason (R): A polynomial of degree $n$ has at most $n$ zeros. Both true and R explains A. Answer: (a).

Common Mistakes to Avoid

• Forgetting the negative sign in $\alpha+\beta=-\dfrac{b}{a}$. Always include the minus sign.
• For cubic polynomials, students sometimes forget that $\alpha\beta\gamma=-\dfrac{d}{a}$ has a negative sign.
• While dividing polynomials, always arrange the dividend and divisor in the descending order of powers, with $0$ as the coefficient for missing terms.
• The remainder must have a degree strictly less than the divisor; if degrees are equal, continue dividing.
• A polynomial of degree $n$ has at most $n$ real zeros — not necessarily exactly $n$.
• The graph of a quadratic polynomial that just touches the x-axis represents two equal zeros, not one.
• Students confuse the standard form: a cubic polynomial is $ax^{3}+bx^{2}+cx+d$, not $ax^{3}+bx^{2}+c$.

Quick Revision Points

• Polynomial: expression $a_{n}x^{n}+\dots+a_{0}$ with non-negative integer powers.
• Degree: highest power of variable.
• Zero of $p(x)$: a value $k$ with $p(k)=0$. Geometrically, x-intercept of $y=p(x)$.
• Linear polynomial: at most $1$ zero.
• Quadratic polynomial: at most $2$ zeros. Graph is a parabola.
• Cubic polynomial: at most $3$ zeros.
• For $ax^{2}+bx+c$: sum of zeros $=-\dfrac{b}{a}$, product $=\dfrac{c}{a}$.
• For $ax^{3}+bx^{2}+cx+d$: sum $=-\dfrac{b}{a}$, sum of products in pairs $=\dfrac{c}{a}$, product $=-\dfrac{d}{a}$.
• Division algorithm: $p(x)=g(x)\cdot q(x)+r(x)$ with $\deg r<\deg g$ or $r=0$.
• To construct a quadratic with given zeros: $p(x)=k\bigl[x^{2}-(\text{sum})x+(\text{product})\bigr]$.

Previous Year Style Practice Set

Q1. If the sum of the zeros of the polynomial $p(x)=(k-1)x^{2}+kx+1$ is $-3$, find $k$. Sum of zeros $=-\dfrac{k}{k-1}=-3\Rightarrow \dfrac{k}{k-1}=3\Rightarrow k=3(k-1)=3k-3\Rightarrow -2k=-3\Rightarrow k=\dfrac{3}{2}$. Answer: $k=\dfrac{3}{2}$.

Q2. If $\alpha$ and $\beta$ are the zeros of $f(x)=2x^{2}-3x+1$, find a quadratic polynomial whose zeros are $\alpha+1$ and $\beta+1$. $\alpha+\beta=\dfrac{3}{2}$, $\alpha\beta=\dfrac{1}{2}$. New sum $=(\alpha+1)+(\beta+1)=\dfrac{3}{2}+2=\dfrac{7}{2}$. New product $=(\alpha+1)(\beta+1)=\alpha\beta+\alpha+\beta+1=\dfrac{1}{2}+\dfrac{3}{2}+1=3$. $p(x)=x^{2}-\dfrac{7}{2}x+3$, equivalently $2x^{2}-7x+6$. Answer: $2x^{2}-7x+6$.

Q3. If the sum of the zeros of $kx^{2}+2x+3k$ is equal to their product, find $k$. Sum $=-\dfrac{2}{k}$. Product $=\dfrac{3k}{k}=3$. $-\dfrac{2}{k}=3\Rightarrow k=-\dfrac{2}{3}$. Answer: $k=-\dfrac{2}{3}$.

Q4. Find the zeros of $\sqrt{2}x^{2}+7x+5\sqrt{2}$. $\sqrt{2}x^{2}+7x+5\sqrt{2}=\sqrt{2}x^{2}+5x+2x+5\sqrt{2}=x(\sqrt{2}x+5)+\sqrt{2}(\sqrt{2}x+5)=(\sqrt{2}x+5)(x+\sqrt{2})$. Zeros: $x=-\dfrac{5}{\sqrt{2}}=-\dfrac{5\sqrt{2}}{2}$ and $x=-\sqrt{2}$. Answer: $-\dfrac{5\sqrt{2}}{2}$ and $-\sqrt{2}$.

Q5. If $\alpha$ and $\beta$ are zeros of $p(x)=x^{2}-(k+6)x+2(2k-1)$, find $k$ given $\alpha+\beta=\dfrac{1}{2}\alpha\beta$. $\alpha+\beta=k+6,\;\alpha\beta=4k-2$. $k+6=\dfrac{1}{2}(4k-2)=2k-1\Rightarrow k=7$. Answer: $k=7$.

Q6. Apply the division algorithm to $p(x)=3x^{3}+x^{2}+2x+5$ when divided by $g(x)=1+2x+x^{2}$. Arrange $g(x)=x^{2}+2x+1$. $3x^{3}\div x^{2}=3x$. $3x(x^{2}+2x+1)=3x^{3}+6x^{2}+3x$. Subtract: $-5x^{2}-x+5$. $-5x^{2}\div x^{2}=-5$. $-5(x^{2}+2x+1)=-5x^{2}-10x-5$. Subtract: $9x+10$. Answer: Quotient $=3x-5$, Remainder $=9x+10$.

Q7. If two zeros of the polynomial $p(x)=2x^{4}-2x^{3}-7x^{2}+3x+6$ are $-\sqrt{\dfrac{3}{2}}$ and $\sqrt{\dfrac{3}{2}}$, find the other zeros. $\left(x-\sqrt{\dfrac{3}{2}}\right)\left(x+\sqrt{\dfrac{3}{2}}\right)=x^{2}-\dfrac{3}{2}$, equivalently $2x^{2}-3$ is a factor. Divide $2x^{4}-2x^{3}-7x^{2}+3x+6$ by $2x^{2}-3$: $2x^{4}\div 2x^{2}=x^{2}$. $x^{2}(2x^{2}-3)=2x^{4}-3x^{2}$. Subtract: $-2x^{3}-4x^{2}+3x+6$. $-2x^{3}\div 2x^{2}=-x$. $-x(2x^{2}-3)=-2x^{3}+3x$. Subtract: $-4x^{2}+6$. $-4x^{2}\div 2x^{2}=-2$. $-2(2x^{2}-3)=-4x^{2}+6$. Subtract: $0$. Quotient $=x^{2}-x-2=(x-2)(x+1)$. Other zeros: $2,-1$. Answer: $2$ and $-1$.

Q8. Find a quadratic polynomial whose zeros are $-2\sqrt{3}$ and $-\dfrac{9}{2\sqrt{3}}$. $-\dfrac{9}{2\sqrt{3}}=-\dfrac{9\sqrt{3}}{6}=-\dfrac{3\sqrt{3}}{2}$. Sum $=-2\sqrt{3}-\dfrac{3\sqrt{3}}{2}=\dfrac{-4\sqrt{3}-3\sqrt{3}}{2}=-\dfrac{7\sqrt{3}}{2}$. Product $=(-2\sqrt{3})\left(-\dfrac{3\sqrt{3}}{2}\right)=\dfrac{6\cdot 3}{2}=9$. $p(x)=x^{2}+\dfrac{7\sqrt{3}}{2}x+9$, or $2x^{2}+7\sqrt{3}x+18$. Answer: $2x^{2}+7\sqrt{3}x+18$.

Q9. If $\alpha$ and $\beta$ are zeros of the polynomial $f(x)=x^{2}-3x-2$, find a polynomial whose zeros are $\dfrac{1}{2\alpha+\beta}$ and $\dfrac{1}{2\beta+\alpha}$. $\alpha+\beta=3,\;\alpha\beta=-2$. $2\alpha+\beta=\alpha+3$ and $2\beta+\alpha=\beta+3$. Sum $=\dfrac{1}{\alpha+3}+\dfrac{1}{\beta+3}=\dfrac{(\alpha+3)+(\beta+3)}{(\alpha+3)(\beta+3)}=\dfrac{\alpha+\beta+6}{\alpha\beta+3(\alpha+\beta)+9}=\dfrac{9}{-2+9+9}=\dfrac{9}{16}$. Product $=\dfrac{1}{(\alpha+3)(\beta+3)}=\dfrac{1}{16}$. $p(x)=x^{2}-\dfrac{9}{16}x+\dfrac{1}{16}$, or $16x^{2}-9x+1$. Answer: $16x^{2}-9x+1$.

Q10. Show that $\dfrac{1}{2}$ and $-\dfrac{3}{2}$ are zeros of the polynomial $4x^{2}+4x-3$ and verify the relations. $p(\dfrac{1}{2})=4\cdot\dfrac{1}{4}+4\cdot\dfrac{1}{2}-3=1+2-3=0$. $p(-\dfrac{3}{2})=4\cdot\dfrac{9}{4}+4\cdot(-\dfrac{3}{2})-3=9-6-3=0$. $a=4,b=4,c=-3$. Sum $=\dfrac{1}{2}-\dfrac{3}{2}=-1=-\dfrac{4}{4}=-\dfrac{b}{a}$. Product $=\dfrac{1}{2}\cdot(-\dfrac{3}{2})=-\dfrac{3}{4}=\dfrac{-3}{4}=\dfrac{c}{a}$. ✓ Answer: Verified.

Glossary of Important Terms

Polynomial: An algebraic expression of the form $a_{n}x^{n}+a_{n-1}x^{n-1}+\dots+a_{0}$ where $a_{i}$ are real numbers and $n$ is a non-negative integer.

Degree: The highest power of the variable in a polynomial.

Linear polynomial: A polynomial of degree $1$, e.g. $ax+b$.

Quadratic polynomial: A polynomial of degree $2$, e.g. $ax^{2}+bx+c$.

Cubic polynomial: A polynomial of degree $3$, e.g. $ax^{3}+bx^{2}+cx+d$.

Zero of a polynomial: A value $k$ for which $p(k)=0$.

Geometric meaning of zero: The x-coordinate of a point where the graph $y=p(x)$ meets the x-axis.

Parabola: The U-shaped (or inverted U-shaped) graph of a quadratic polynomial.

Division algorithm: $p(x)=g(x)\cdot q(x)+r(x)$ where the degree of $r(x)$ is less than the degree of $g(x)$ or $r(x)=0$.

Quotient: The polynomial $q(x)$ obtained when $p(x)$ is divided by $g(x)$.

Remainder: The polynomial $r(x)$ left after division.

Factor: $g(x)$ is a factor of $p(x)$ if $p(x)=g(x)\cdot q(x)$, i.e., remainder is $0$.

Examination Tips

• Always identify the type of polynomial (linear, quadratic or cubic) before starting a question.
• For quadratic polynomial questions, the splitting-the-middle-term method is the fastest way to factorise. Practise it until it becomes automatic.
• When asked to verify the relationship between zeros and coefficients, write the values of $a$, $b$, $c$ (and $d$ for cubics) clearly first, then compute $-\dfrac{b}{a}$, $\dfrac{c}{a}$ and so on.
• While performing long division of polynomials, do not skip steps in the examination — examiners expect to see each subtraction.
• If a polynomial has irrational zeros like $\sqrt{p}$, then $-\sqrt{p}$ is also a zero (for polynomials with rational coefficients), and $(x^{2}-p)$ is a factor.
• For HOTS questions involving sums like $\alpha^{2}+\beta^{2}$ or $\dfrac{1}{\alpha}+\dfrac{1}{\beta}$, always express them in terms of $\alpha+\beta$ and $\alpha\beta$ first.
• Memorise the cubic identity $\alpha^{3}+\beta^{3}=(\alpha+\beta)^{3}-3\alpha\beta(\alpha+\beta)$ — it is asked frequently.

That completes the chapter Polynomials for Class 10 Mathematics under the ASSEB English Medium curriculum. Practise each exercise sincerely, draw rough graphs to visualise zeros, and remember the relationships between zeros and coefficients — they appear repeatedly in the HSLC examination. Best of luck from the team at HSLC Guru.