NCERT Solutions Class 10th Maths Chapter – 2 Polynomials Examples

Example 1. Look at the graphs in Fig. 2.9 given below. Each is the graph of y = p(x), where p(x) is a polynomial. For each of the graphs, find the number of zeroes of p(x).

Solution:
(i) The number of zeroes is 1 as the graph intersects the x-axis at one point only.
(ii) The number of zeroes is 2 as the graph intersects the x-axis at two points.
(iii) The number of zeroes is 3 as the graph intersects the x-axis at Three points.
(iv) The number of zeroes is 1 as the graph intersects the x-axis at one point only.
(v) The number of zeroes is 1 as the graph intersects the x-axis at one point only.
(vi) The number of zeroes is 4 as the graph intersects the x-axis at Four points.

Example 2. Find the zeroes of the quadratic polynomial x² + 7x + 10, and verify the relationship between the zeroes and the coefficients.

Solution: We have

x² + 7 x + 10 = (x + 2)(x + 5)

So, the value of x² + 7x + 10 is zero when x + 2 = 0 or x + 5 = 0, i.e., when x = -2 or

x = -5. Therefore, the zeroes of x² + 7x + 10 are – 2 and -5. Now,

sum of zeroes = −2 + (-5) = (7) = -(7)/1 = (Coefficient of x)/(Coefficient of x²)

product of zeroes = (-2) x (-5) = 10 = 10/1 = Constant term/Coefficient of x² – (7)

Example 3. Find the zeroes of the polynomial x² – 3 and verify the relationship between the zeroes and the coefficients.

Solution: Recall the identity a² – b² = (a – b)(a + b). Using it, we can write:

x² – 3 = (x − √3) (x + √3)

So, the value of x² – 3 is zero when x = √3 or x = -√3.

Therefore, the zeroes of x² – 3 are √3 and −√3. Now,

sum of zeroes = √3 – √3 = 0 = -(Coefficient of x)/(Coefficient of x²)

product of zeroes = (√3) – (√3) = -3 = -3/1 = Constant term/Cofficient of X²

Example 4. Find a quadratic polynomial, the sum and product of whose zeroes are -3 and 2, respectively.

Solution: Let the quadratic polynomial be ax² + bx + c, and its zeroes be a and B.
We have

α + β = – 3 = -b/a,
and αβ = 2 = C/a

If a = 1, then b = 3 and c = 2.

So, one quadratic polynomial which fits the given conditions is x² + 3x + 2

You can check that any other quadratic polynomial that fits these conditions will
be of the form k(x² + 3x + 2), where k is real.

Let us now look at cubic polynomials. Do you think a similar relation holds between the zeroes of a cubic polynomial and its coefficients?

Let us consider p(x) = 2x³ – 5x² – 14x + 8.

You can check that p(x) = 0 for x = 4, -2, 1/2.
Since p(x) can have atmost three zeroes, these are the zeores of 2x³ – 5x² – 14x + 8.
Now,

sum of the zeroes = 4 + (−2) + 1/2 = 5/2 = – (- 5)/2 = (Coefficient of – x²)/(Coefficient of x³),

product of the zeroes = 4 x (− 2) x 1/2 = – 4 = – 8/2 = – Constant Term/Coefficient of x³.

However, there is one more relationship here. Consider the sum of the products of the zeroes taken two at a time. We have

{4 x (-2)} + {(- 2) x 1/2} + {1/2 x 4}

= – 8 – 1 + 2 = – 7 = -14/2 = Coefficient of x/Coefficient of x³

In general, it can be proved that if a, ẞ, y are the zeroes of the cubic polynomial

ax³ + bx² + cx + d, then

α + B + y = -b/a,
αβ + βy + yα = c/a,
αβγ = -d/a.

Let us consider an example.

Example 5. Verify that 3, -1, -1/3 are the zeroes of the cubic polynomial p(x) = 3x³ – 5x² – 11x – 3, and then verify the relationship between the zeroes and the coefficients.

Solution: Comparing the given polynomial with ax³ + bx² + cx + d, we get

a = 3, b = – 5, c = -11, d = -3. Further
p(3) = 3 × 3³ – (5 × 3²) − (11 × 3) − 3 = 81 – 45 – 33 – 3 = 0,
p(-1) = 3 × (- 1)³ – 5 × (-1)² – 11 × (-1) – 3 = – 3 – 5 + 11 – 3 = 0,

p(-1/3) = 3 × (-1/3)³ – 5 × (-1/3) -3,
= -1/9 – 5/9 + 11/3 – 3 = – 2/3 + 2/3 = 0

Therefore, 3, -1 and -1/3 are the Zeroes of 3x³ – 5x² – 11x -3.

So, we take α = 3, β = -1 and y = -1/3
Now,

α + β + γ = 3 + (-1) + (-1/3) = 2 – 1/3 = 5/3 = -(-5)/3 = -b/a,

αβ + βγ + γα = 3 x (-1) + (-1) x (-1/3) + (-1/3) x 3 = -3 + 1/3 – 1 = -11/3 = c/a,

αβγ = 3 x (-1) x (-1/3) = 1 = -(-3)/3 = -d/d