Case Study Based Questions for Class 10 Maths Chapter 2 Polynomials

The zeroes of a polynomial are the x-coordinates of the points where its graph intersects the x-axis. From the given options and the context of the subsequent questions, the graph intersects the x-axis at $x = -2$ and $x = 8$.

Given the zeroes $\alpha = -2$ and $\beta = 8$, we calculate the sum and product of the zeroes:

The general form of a quadratic polynomial is $k(x^2 – Sx + P)$. Because the tunnel forms a downward-opening parabola, $k$ must be negative. Taking $k = -1$:

Substitute $x = 4$ into the expression obtained in the previous question, $p(x) = -x^2 + 6x + 16$:

To find the zeroes, equate the polynomial to zero and solve for $x$ by splitting the middle term:

Thus, the zeroes are $x = 1$ and $x = 2$.

Given one zero $\alpha = 4$ and the sum of zeroes $S = -3$:

Calculate the product of zeroes ($P$):

The polynomial is $k(x^2 – Sx + P) = k(x^2 – (-3)x – 28) = k(x^2 + 3x – 28)$. Because it represents a tunnel, the parabola must open downwards, so $k = -1$.

The graph of any quadratic polynomial of the form $ax^2 + bx + c$ (where $a \neq 0$) is a continuous, U-shaped curve. In mathematics, this specific geometric shape is called a parabola.

To find the zeroes, equate the polynomial to zero and split the middle term:

Therefore, the zeroes are $x = -1$ and $x = 5$.

Substitute $x = 3$ directly into the polynomial expression:

Set the polynomial to zero and multiply by -1 to simplify the factoring process:

Split the middle term:

This gives the zeroes as $x = -2$ and $x = 4$.

A quadratic polynomial can be formed using the formula $k[x^2 – (\text{Sum of zeroes})x + (\text{Product of zeroes})]$.

Given the sum is $0$ and the product is $-9$: