# Case based 2:

While passing over an iron bridge, Hema noticed that the shape resembles a curve about which she had studied in mathematics.(Class 10 Case based problem of Chapter 2 Polynomials 2)

(A) What type of ploynomial does the shape of the bridge represents ?

(B) Find the zeroes of the polynomial xÂ² + 7x – 60

Solution: (A) The shape of the bridge represents quadratic polynomial

(B) To find the zeros of the polynomial xÂ² + 7x – 60, we need to factorize the polynomial.

xÂ² + 7x – 60 = xÂ²Â  + 12x – 5x – 60

=Â  x(x + 12) -5(x + 12)

=Â  (x + 12)(x – 5)

Therefore zeros of the polynomialÂ  = 5 , -12

# Case based 3:

Due to heavy storm, an electric wire got bent as shown in the figure. It followed a mathematical shape. Answer the following question below:

(A) Name the shape in which the wire is bent.

(a) SpiralÂ  Â  Â  Â  Â  Â  Â  Â  Â (b) Ellipse

(c) LinearÂ  Â  Â  Â  Â  Â  Â  Â  Â (d) Parabola

(B) How many zeros are there for the polynomaial(shape of the wire)

(a) 2Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â (b) 3

(c) 1Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â (d) 0

(C) The zeros of the polynomial are

(a) -1, 5Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  (b) -1, 3

(c) 3, 5Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  (d)Â  -4, 2

(D) What will be the expression of the polynomial?

(a) xÂ² + 2x – 3Â  Â  Â  Â  Â  Â  Â (b) xÂ² – 2x + 3

(c) xÂ² – 2x – 3Â  Â  Â  Â  Â  Â  Â  Â (d) xÂ² + 2x + 3

(E) What is the value of the polynomial if x = -1?

(a) 6Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  (b) -18

(c) 18Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  (d) 0

## Solution:

Explanation:

Graph of the polynomial intersect x-axis at two points

Hence, number of zeroesÂ  of the polynomialÂ  = 2

Explanation:

We know, points at which graph of a polynomial intersect the x- axis,are the zeros of the polynomial. Here, The graph intersects x-axis at points -1 and 3

Its zeros are = -1 and 3

(D) Answer (c) xÂ² – 2x – 3

Explanation:

We have the zeros of the polynomial are -1 and 3

= xÂ² – (Sum of zeroes) x + Product of zeroes

= xÂ² – (-1 + 3)x + (-1)(3)

= xÂ² – 2x – 3

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