Question

If –1 /2 is a zero of the polynomial 2x³ + x² –6x – 3, find the sum and product of its other two zeroes.

Answer 1

GIVEN that -½ is a zero of the polynomial 2x³ +  x²  –6x  – 3.

Let f(x) = 2x³ +  x²  –6x  – 3.
Since -1/2 is a zero of f(x) so (x-½) is a factor of f(x)
On dividing f(x) by (x-½),

x-½)  2x³ +  x²  –6x  – 3 ( 2x² -6
 2x³ + x²
    (-)    (-)
 ------------------------
  –6x  – 3
     –6x  –3
 (+)      (+)
 ------------------
0

f(x) = 2x³ +  x²  –6x  – 3.
=(x-½) ( 2x² -6)
= (x-½)  2(x²-3)
f(x) = 0
(x-½)  2(x²-3) = 0
 (x-½)  or x²=3
(x-½)  or x=±√3

The other two zeroes of f(x) are √3 and -√3

Sum of two zeroes = √3+(-√3)= √3-√3)=0
Sum of two zeroes = 0

Product of two zeroes = (√3)(-√3)= -3
Product of two zeroes = -3
Hence the Sum of two zeroes= 0 & Product of two zeroes = -3

HOPE THIS WILL HELP YOU...

Answer 2

Answer:

Step-by-step explanation:

GIVEN that -½ is a zero of the polynomial 2x³ + x² –6x – 3.

Let f(x) = 2x³ + x² –6x – 3.

Since -1/2 is a zero of f(x) so (x-½) is a factor of f(x)

On dividing f(x) by (x-½),

x-½) 2x³ + x² –6x – 3 ( 2x² -6

2x³ + x²

(-) (-)

------------------------

–6x – 3

–6x –3

(+) (+)

------------------

0

f(x) = 2x³ + x² –6x – 3.

=(x-½) ( 2x² -6)

= (x-½) 2(x²-3)

f(x) = 0

(x-½) 2(x²-3) = 0

(x-½) or x²=3

(x-½) or x=±√3

The other two zeroes of f(x) are √3 and -√3

Sum of two zeroes = √3+(-√3)= √3-√3)=0

Sum of two zeroes = 0

Product of two zeroes = (√3)(-√3)= -3

Product of two zeroes = -3

Hence the Sum of two zeroes= 0 & Product of two zeroes = -3