Question

Given that the zeroes of the cubic polynomial x³ – 6x² + 3x + 10 are of the form a, a + b, a + 2b for some real numbers a and b, find the values of a and b as well as the zeroes of the given polynomial.

Answer 1

Given that a, a + b, a + 2b are roots of given polynomial x³ – 6x² + 3x + 10

Sum of the roots

⇒ a + 2b + a + a + b = `(-("coefficient of"  x^2))/("coefficient of"  x^3)`

⇒ 3a + 3b = `(-(-6))/1` = 6

⇒ 3(a + b) = 6

⇒ a + b = 2   .........(1)

b = 2 – a

Product of roots

⇒ (a + 2b)(a + b)a = `(-("constant"))/("coefficient of"  x^3)`

⇒ (a + b + b)(a + b)a = `-10/1`

Substituting the value of a + b = 2 in it

⇒ (2 + b)(2)a = –10

⇒ (2 + b)2a = –10

⇒ (2 + 2 – a)2a = –10

⇒ (4 – a)2a = –10

⇒ 4a – a² = –5

⇒ a² – 4a – 5 = 0

⇒ a² – 5a + a – 5 = 0

⇒ (a – 5)(a + 1) = 0

a – 5 = 0 or a + 1 = 0

a = 5 a = –1

a = 5, –1 in (1) a + b = 2

When a = 5,

5 + b = 2

⇒ b = –3

a = –1, –1 + b = 2

⇒ b = 3

∴ If a = 5 then b = –3

or

If a = –1 then b = 3