NCERT Solutions Class 10th Maths Chapter – 2 Polynomials Exercise – 2.4

1. Verify that the numbers given alongside of the cubic polynomials below are their zeroes. Also verify the relationship between the zeroes and the coefficients in each case:

(i) 2x³ + x²– 5x + 2; – 1/2, 1, – 2

Solution: Given,
p(x) = 2x³+ x²– 5x + 2
And zeroes for p(x) are = 1/2, 1, – 2
∴ p(1/2) = 2(1/2)³ + (1/2)²– 5(1/2) + 2 = (1/4) +(1/4) – (5/2) + 2 = 0
p(1) = 2(1)³+ (1)²– 5(1) + 2 = 0
p(- 2) = 2(- 2)³+ (- 2)²– 5(- 2) + 2 = 0
Hence, proved 1/2, 1, -2 are the zeroes of 2x³ + x²– 5x + 2.
Now, comparing the given polynomial with general expression, we get;
∴ ax³+ bx² + cx + d = 2x³ + x²– 5x + 2
a = 2, b = 1, c = – 5 and d = 2
As we know, if α, β, γ are the zeroes of the cubic polynomial ax³+ bx² + cx + d , then;
α + β + γ = – b/a
αβ + βγ + γα = c/a
α βγ = – d/a.
Therefore, putting the values of zeroes of the polynomial,
α + β + γ = ½ + 1 + (- 2) = -1/2 = – b/a
αβ + βγ + γα = (1/2 × 1) + (1 × – 2) + (- 2 × 1/2) = – 5/2 = c/a
α β γ = ½ × 1 × (- 2) = – 2/2 = – d/a
Hence, the relationship between the zeroes and the coefficients are satisfied.

(ii) x³– 4x² + 5x – 2 ;2, 1, 1

solution: Given,
p(x) = x³– 4x² + 5x – 2
And zeroes for p(x) are 2,1,1.
∴ p(2)= 2³– 4(2)² + 5(2) – 2 = 0
p(1) = 1³– (4 × 1² ) + (5 × 1) – 2 = 0
Hence proved, 2, 1, 1 are the zeroes of x³– 4x²+ 5x – 2
Now, comparing the given polynomial with general expression, we get;
∴ ax³+ bx² + cx + d = x³– 4x² + 5x – 2
a = 1, b = – 4, c = 5 and d = – 2
As we know, if α, β, γ are the zeroes of the cubic polynomial ax³+ bx² + cx + d , then;
α + β + γ = – b/a
αβ + βγ + γα = c/as
α β γ = – d/a.
Therefore, putting the values of zeroes of the polynomial,
α + β + γ = 2 + 1 + 1 = 4 = – (- 4)/1 = – b/a
αβ + βγ + γα = 2 × 1 + 1× 1 + 1 × 2 = 5 = 5/1= c/a
αβγ = 2 × 1 × 1 = 2 = -(- 2)/1 = – d/a
Hence, the relationship between the zeroes and the coefficients are satisfied.

2. Find a cubic polynomial with the sum, sum of the product of its zeroes taken two at a time, and the product of its zeroes as 2, –7, –14 respectively.

Solution: Let us consider the cubic polynomial is ax³ + bx² + cx + d and the values of the zeroes of the polynomials be α, β, γ.
As per the given question,
α + β + γ = – b/a = 2/1
αβ + βγ + γα = c/a = -7/1
αβγ = – d/a = -14/1
Thus, from above three expressions we get the values of coefficient of polynomial.
a = 1, b = -2, c = -7, d = 14
Hence, the cubic polynomial is x³– 2x² – 7x + 14

3. If the zeroes of the polynomial x³-3x²+x+1 are a – b, a, a + b, find a and b.

Solution: We are given with the polynomial here,
p(x) = x³– 3x² + x + 1
And zeroes are given as a – b, a, a + b
Now, comparing the given polynomial with general expression, we get;
∴px³+ qx² + rx + s = x³– 3x² + x + 1
p = 1, q = -3, r = 1 and s = 1
Sum of zeroes = a – b + a + a + b
-s/p = 1 – b²
Putting the values q and p.
-(- 3)/1 = 3a
a = 1
Thus, the zeroes are 1- b, 1, 1 + b.
Now, product of zeroes = 1(1 – b)(1 + b)
-s/p = 1 – b²
-1/1 = 1 – b²
b² = 1 + 1 = 2
b = ±√2
Hence,1 – √2, 1 ,1 + √2 are the zeroes of x³– 3x² + x + 1.

4. If two zeroes of the polynomial x⁴– 6x³– 26x² +138x – 35 are 2 ±√3, find other zeroes.

Solution: Since this is a polynomial equation of degree 4, hence there will be total 4 roots.
Let f(x) = x⁴– 6x³– 26x² + 138x – 35
Since 2 + √3 and 2 – √3 are zeroes of given polynomial f(x).
∴ [x − (2 + √3)] [x − (2 – √3)] = 0
(x − 2 − √3)(x − 2 + √3) = 0
On multiplying the above equation we get,
x²– 4x + 1, this is a factor of a given polynomial f(x).
Now, if we will divide f(x) by g(x), the quotient will also be a factor of f(x) and the remainder will be 0.

So, x⁴ – 6x³– 26x² + 138x – 35 = (x²– 4x + 1)(x² – 2x − 35)
Now, on further factorizing (x² – 2x − 35) we get,
x²– (7 − 5)x − 35 = x²– 7x + 5x + 35 = 0
x(x −7) + 5(x − 7) = 0
(x + 5)(x − 7) = 0
So, its zeroes are given by:
x= − 5 and x = 7.
Therefore, all four zeroes of given polynomial equation are: 2 + √3 , 2 – √3, −5 and 7.