Question

 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
(ii) x3 - 4x2 + 5x - 2; 2, 1, 1

Answer 1

Thus, 1/2, 1 and -2 are the zeroes of the given polynomial.

Comparing the given polynomial with ax3 + bx2 + cx + d, we get a=2, b=1, c=-5, d=2
Also, α=1/2, β=1 and γ=-2
Now,
-b/a = α+β+γ
⇒ 1/2 = 1/2 + 1 - 2
⇒ 1/2 = 1/2

c/a = αβ+βγ+γα
⇒ -5/2 = (1/2 × 1) + (1 × -2) + (-2 × 1/2)
⇒ -5/2 = 1/2 - 2 - 1
⇒ -5/2 = -5/2

-d/a = αβγ
⇒ -2/2 = (1/2 × 1 × -2)
⇒ -1 = 1

Thus, the relationship between zeroes and the coefficients are verified.

(ii)  p(x) = x3 - 4x2 + 5x - 2
Now for zeroes, putting the given value in x.

p(2) = 23 - 4(2)2 + 5(2) - 2
= 8 - 16 + 10 - 2
= 0

p(1) = 13 - 4(1)2 + 5(1) - 2
= 1 - 4 + 5 - 2
= 0

p(1) = 13 - 4(1)2 + 5(1) - 2
= 1 - 4 + 5 - 2
= 0

Thus, 2, 1 and 1 are the zeroes of the given polynomial.

Comparing the given polynomial with ax3 + bx2 + cx + d, we get a=1, b=-4, c=5, d=-2
Also, α=2, β=1 and γ=1
Now,
-b/a = α+β+γ
⇒ 4/1 = 2 + 1 + 1
⇒ 4 = 4

c/a = αβ+βγ+γα
⇒ 5/1 = (2 × 1) + (1 × 1) + (1 × 2)
⇒ 5 = 2 + 1 + 2
⇒ 5 = 5

-d/a = αβγ
⇒ 2/1 = (2 × 1 × 1)
⇒ 2 = 2

Thus, the relationship between zeroes and the coefficients are verified.

Answer 2

(1) 2x³ + x² -5x + 2
= 2x³ -2x² + 3x² -3x - 2x + 2
= 2x²( x -1) + 3x(x -1) -2(x -1)
= (x -1)(2x² + 3x -2)
= (x -1) { 2x² + 4x - x -2 }
=(x -1) { 2x(x +2) -(x +2) }
=(x -1)(x +2)(2x -1)
hence,
x = 1, -2 , 1/2 are the zeros of 2x³ + x² -5x + 2 .
given numbers are same as evaluate numbers . hence, verified .

now,
sum of roots = -(coefficient of x²)/(coefficient of x³)
=> - (1)/2 = 1/2 + 1 -2
=> -1/2 = -1/2

products of roots = -( constant )/coefficient of x³
=> - (2)/2 = 1/2 × 1 × -2
=> -1 = -1

sum of products of two consecutive roots = ( coefficient of x)/coefficient of x³
=> -5/2 = 1/2 × 1 + 1 × (-2) + (-2) × 1/2
=> -5/2 = -5/2

hence, verified /


(ii) x³ - 4x² + 5x -2
= x³ - x² -3x² + 3x +2x -2
= x²(x -1) -3x(x -1) +2(x -1)
= (x² -3x +2 )(x -1)
={x² -2x - x +2 }(x -1)
=(x -2)(x -1)(x -1)

hence, 2, 1 , 1 are the zeros of x³ - 4x² + 5x -2 .
given numbers are same as evaluate numbers so, verified .

now,
sum of roots :
=> -(-4)/1 = 2 + 1 + 1
=> 4 = 4

product of roots :
=> -(-2)/1 = 2 × 1 × 1
=> 2 = 2

sum of products of two consecutive roots :
=> 5/1 = 2 × 1 + 1 × 1 + 1 × 2
=> 5 = 5
hence, verified /