Question

If two of the zeroes of the polynomial P(x)=2x^(3)-x^(2)-7x+6 are (3)/(2) and -2 the other zero will be

Answer 1

Explanation:

The given equation is p(x)=
`2x^3-x^2-7x+6` of the form
`ax^3+bx^2+cx+d`

and the zeroes are:

α=3/2 and β= -2

We have to find the third zero let say "γ".

p(x)=
`2x^3-x^2-7x+6`

∴ a=2, b=1, c=-7 and d=6

Then αβγ=
`\cfrac{-d}{a}`

⇒
`(\cfrac{3}{2}) (-2)`·γ=
`\cfrac{-6}{2}`

⇒ -3γ= -3

⇒γ=1

Therefore, the third zero of
`2x^3-x^2-7x+6` is 1.