NCERT Polynomials Class 10: Complete Concepts, Formulas, and Practice Questions

Geometrical Meaning of the Zeroes of a Polynomial

In previous classes, you learned that a real number $k$ is a zero of a polynomial $p(x)$ if $p(k) = 0$. But what does this look like geometrically? Let's visualize this on a Cartesian plane.

For a linear polynomial $ax + b$ (where $a \neq 0$), the graph of $y = ax + b$ is always a straight line. This straight line intersects the x-axis at exactly one point, which is $(-b/a, 0)$. Therefore, a linear polynomial has exactly one zero.

Now, let's look at a quadratic polynomial $ax^2 + bx + c$ (where $a \neq 0$). The graph of the corresponding equation $y = ax^2 + bx + c$ is a symmetric U-shaped curve called a parabola. If $a > 0$, the parabola opens upwards (like a cup); if $a < 0$, it opens downwards. Depending on the values, the parabola can intersect the x-axis in three distinct ways:

1. Two intersecting points: The graph crosses the x-axis at two distinct points, meaning the polynomial has two distinct real zeroes.
2. One touching point: The graph touches the x-axis at exactly one point (which means it has two equal real zeroes).
3. No intersection: The graph lies entirely above or below the x-axis, meaning it has no real zeroes.

In general, a polynomial $p(x)$ of degree $n$ can cross the x-axis at most $n$ times. Hence, a polynomial of degree $n$ has at most $n$ zeroes.

How to Verify the Relationship Between Zeroes and Coefficients

1. Step 1: Identify the Coefficients — Write down the given quadratic polynomial in its standard form: $ax^2 + bx + c$. Identify the coefficients $a$, $b$, and the constant term $c$ along with their signs.
2. Step 2: Find the Zeroes using Splitting the Middle Term — Factorize the quadratic polynomial by splitting the middle term $bx$ into two terms whose product is equal to $ac$. Equate each factor to zero to obtain the two zeroes, let's call them $\alpha$ (alpha) and $\beta$ (beta).
3. Step 3: Verify the Sum of Zeroes — Calculate the sum of the zeroes: $\alpha + \beta$. Compare this value to $-b/a$ (i.e., $-\text{coefficient of } x / \text{coefficient of } x^2$). Both values must be equal.
4. Step 4: Verify the Product of Zeroes — Calculate the product of the zeroes: $\alpha \times \beta$. Compare this value to $c/a$ (i.e., $\text{constant term} / \text{coefficient of } x^2$). Both values must be equal.

Board Exam Tips & Common Student Mistakes

To score 100% in this chapter, pay attention to these highly common mistakes made by students during CBSE board exams:

• The Negative Sign Trap: When verifying the sum of zeroes $(\alpha + \beta = -b/a)$, students often forget the negative sign of $b$ itself. If $b = -5$, then $-b/a$ becomes $-(-5)/a = 5/a$. Write down steps explicitly to avoid silly sign errors.
• Using the Formula for Formulating a Polynomial: When you are given the sum ($S$) and product ($P$) of zeroes and asked to find the polynomial, use the standard formula $k[x^2 - Sx + P]$, where $k$ is a non-zero real constant. Do not write it as $x^2 + Sx + P$.
• Confusing Zeroes with the Number Zero: A 'zero' of a polynomial is a coordinate value of $x$ where the output $y$ becomes $0$. The zero itself does not have to be the number $0$; it can be any real number (positive, negative, fractional, or irrational).

Practice Questions with Solutions

• Q: Find the zeroes of the quadratic polynomial $x^2 - 2x - 8$ and verify the relationship between the zeroes and the coefficients. A: Step 1: Write down the polynomial: $p(x) = x^2 - 2x - 8$. Step 2: Factorize by splitting the middle term: we need two numbers whose sum is $-2$ and product is $-8$. These numbers are $-4$ and $+2$. $x^2 - 4x + 2x - 8 = 0$ $x(x - 4) + 2(x - 4) = 0$ $(x - 4)(x + 2) = 0$ So, $x = 4$ or $x = -2$. Therefore, the zeroes are $\alpha = 4$ and $\beta = -2$. Step 3: Identify coefficients from $x^2 - 2x - 8$: $a = 1, b = -2, c = -8$. Step 4: Verify Sum of zeroes: $\alpha + \beta = 4 + (-2) = 2$. From formula: $-b/a = -(-2)/1 = 2$. (Verified!) Step 5: Verify Product of zeroes: $\alpha \cdot \beta = 4 \times (-2) = -8$. From formula: $c/a = -8/1 = -8$. (Verified!) Final answer: Zeroes are $4$ and $-2$. The relationships are verified.
• Q: Find a quadratic polynomial, the sum and product of whose zeroes are $-3$ and $2$, respectively. A: Step 1: Let the zeroes of the required quadratic polynomial be $\alpha$ and $\beta$. Given, Sum of zeroes $(\alpha + \beta) = -3$ and Product of zeroes $(\alpha\beta) = 2$. Step 2: Use the standard formula of a quadratic polynomial: $p(x) = k[x^2 - (\alpha + \beta)x + \alpha\beta]$, where $k$ is a constant. Step 3: Substitute the values into the formula: $p(x) = k[x^2 - (-3)x + 2]$ $p(x) = k[x^2 + 3x + 2]$ Step 4: For simplicity, choosing $k = 1$, we get the polynomial as $x^2 + 3x + 2$. Final answer: The quadratic polynomial is $x^2 + 3x + 2$.
• Q: Find the zeroes of the quadratic polynomial $4s^2 - 4s + 1$ and verify the relationship between zeroes and coefficients. A: Step 1: Let $p(s) = 4s^2 - 4s + 1$. Step 2: Factorize using middle-term splitting: we need two numbers whose sum is $-4$ and product is $4 \times 1 = 4$. These are $-2$ and $-2$. $4s^2 - 2s - 2s + 1 = 0$ $2s(2s - 1) - 1(2s - 1) = 0$ $(2s - 1)(2s - 1) = 0$ Thus, $s = 1/2$ and $s = 1/2$. The zeroes are $\alpha = 1/2$ and $\beta = 1/2$. Step 3: Identify coefficients: $a = 4, b = -4, c = 1$. Step 4: Verify Sum: $\alpha + \beta = 1/2 + 1/2 = 1$. $-b/a = -(-4)/4 = 1$. (Verified!) Step 5: Verify Product: $\alpha \cdot \beta = 1/2 \times 1/2 = 1/4$. $c/a = 1/4$. (Verified!) Final answer: Zeroes are $1/2$ and $1/2$. The relationships are verified.
• Q: If $\alpha$ and $\beta$ are the zeroes of the quadratic polynomial $p(x) = 3x^2 - 5x + 2$, find the value of $\frac{1}{\alpha} + \frac{1}{\beta}$. A: Step 1: From the polynomial $3x^2 - 5x + 2$, identify the coefficients: $a = 3, b = -5, c = 2$. Step 2: Find the sum and product of the zeroes: Sum of zeroes $(\alpha + \beta) = -b/a = -(-5)/3 = 5/3$. Product of zeroes $(\alpha\beta) = c/a = 2/3$. Step 3: Simplify the expression we need to evaluate: $\frac{1}{\alpha} + \frac{1}{\beta} = \frac{\beta + \alpha}{\alpha\beta} = \frac{\alpha + \beta}{\alpha\beta}$. Step 4: Substitute the values of $(\alpha + \beta)$ and $(\alpha\beta)$: $\frac{\frac{5}{3}}{\frac{2}{3}} = \frac{5}{3} \times \frac{3}{2} = \frac{5}{2}$. Final answer: The value of $\frac{1}{\alpha} + \frac{1}{\beta}$ is $\frac{5}{2}$.

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