two zeroes of the cubic polynomial …

Let {tex}p(x) =ax^3 + 3x^2 - bx - 6{/tex}
{tex}\because{/tex} - 1 is a zero
{tex}\therefore{/tex} {tex}p(-1) =0{/tex}
{tex}\Rightarrow{/tex} {tex}a(-1)^3 + 3(-1)^2 - b(-1) - 6 = 0{/tex}
{tex}\Rightarrow{/tex} {tex}-a + 3 + b - 6 = 0{/tex}
{tex}\Rightarrow{/tex} {tex}-a + b = 3.{/tex}.........(i)
Also, -2 is another zero
{tex}\therefore{/tex} {tex}p(-2) =0{/tex}
{tex}\Rightarrow{/tex} {tex}a(-2)^3 + 3(-2)^2 - b(-2) - 6 = 0{/tex}
{tex}\Rightarrow{/tex} {tex}- 8a + 12 + 2b - 6 = 0{/tex}
{tex}\Rightarrow{/tex} {tex}- 8a + 2b = -6{/tex}
{tex}\Rightarrow{/tex} {tex}4a - b = 3{/tex}.........(ii)
Solving equation (i) and (ii), we get
a = 2 and b = 5
{tex}\therefore{/tex} p(x) = 2x³ + 3x² - 5x - 6
Let third zero = k
Sum of zeroes = {tex}\frac { - 3 } { 2 }{/tex}
{tex}\Rightarrow{/tex} -1 + (-2) + k = {tex}\frac { - 3 } { 2 } \Rightarrow k = \frac { - 3 } { 2 } + 3 = \frac { 3 } { 2 }{/tex}
Therefore, third zero = {tex}\frac { 3 } { 2 }{/tex}