Show that 3 is a zero of the polynomial 2x³ – x² – 13x – 6. Find all the other zeroes of this polynomial.

p(x) = 2x³ – x² – 13x – 6

p(x) is cubic hence it will have 3 zeroes

‘a’ is said to be zero of p(x) if p(a) = 0

which means that we substitute x by a in p(x) and if we get zero then we say that ‘a’ is zero of p(x)

we have to show here that 3 is the zero of p(x)

here a = 3

⇒ p(3) = 2(3)³ – 3² – 13(3) – 6

⇒ p(3) = 2(27) – 9 – 39 – 6

⇒ p(3) = 54 – 54

⇒ p(3) = 0

Hence 3 is zero of polynomial p(x)

As 3 is a zero of p(x) (x – 3) will be a factor of p(x)

Divide p(x) by x – 3

Using dividend = divisor × quotient + remainder

⇒ p(x) = (x – 3)(2x² + 5x + 2) + 0

Remaining zeroes would be found when 2x² + 5x + 2 = 0

⇒ 2x² + 5x + 2 = 0

⇒ 2x² + 4x + x + 2 = 0

⇒ 2x(x + 2) + 1(x + 2) = 0

⇒ (2x + 1)(x + 2) = 0

⇒ 2x + 1 = 0 and x + 2 = 0

⇒ x = –1/2 and x = –2

All zeroes of 2x³ – x² – 13x – 6 are x = 3, x = –1/2 and x = –2