**POLYNOMIALS**

NCERT Exercise 2.1 and other Q & A

**Q1: The graphs of**

*y = p(x)*are given in below, for some polynomials*p(x)*. Find the number of zeroes of p(x), in each case.
Answer:

(i) As shown in the graph, it does not intersect

*x-axis*. Hence it has no real zeroes.
(ii) From the graph, it intersects

*x-axis*at one point. ∴ it has one real zero.
(iii) As the graph shows, it intersects

*x-axis*at three points. ∴ the graph polynomial has three real zeroes.
(iv) From the graph, it is clear it intersects

*x-axis*at two points. Hence it has two zeroes.
(vi) As shown in the graph, it intersects

*x-axis*at three points. ∴ it has three zeroes.**Q2(CBSE 2011): In the figure below the number of zeroes of**

*y = f(x)*are:**(a) 1**

**(b) 2**

**(c) 3**

**(d) 4**

Answer: (c) 3. Since the graph intersects

*x-axis*at three points.**Q3(CBSE 2011): The number of zeroes for the polynomial y = p(x) from the given graph is :**

**(a) 3**

**(b) 1**

**(c) 2**

**(d) 0**

Answer: (b) 1

**Q4: Draw the graph of the quadratic polynomial p(x) = x**

^{2}+6x + 9 and find zeroes of it.
Answer: Given p(x) = x

^{2}+ 6x + 9
To find the zeroes of p(x), consider p(x) = 0 [Factor theorem]

∴ x

^{2}+ 6x + 9 = 0
⇒ (x + 3)

^{2}= 0
⇒ x = -3.

∴ -3 is the zero of the polynomial.

For the graph, we take the following different values of x and y = p(x) = x

^{2}+ 6x + 9x | -3 | -2 | -1 | -4 |

y=p(x) | 0 | 1 | 4 | 1 |

The graph plotted is shown below:

Comparing the quadratic polynomial as ax

^{2}+ bx + c, ∵ a =1 > 0, the graph parabola opens upwards(∪).
thanks for sharing the wonderful article

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