Two zeroes of a cubic polynomial ax3+3x2-bx-6are -1and-2.find the third zero and values of a and b.

Two zeroes of a cubic polynomial ax3+3x2-bx-6are -1and-2.find the thir

1.	Two zeroes of a cubic polynomial ax3+3x2-bx-6are -1and-2.find the third zero and values of a and b.
Answer» Let {tex}p(x) =ax^3 + 3x^2\xa0- bx\xa0- 6{/tex}{tex}\\because{/tex}\xa0- 1 is a zero{tex}\\therefore{/tex}\xa0{tex}p(-1) =0{/tex}{tex}\\Rightarrow{/tex}\xa0{tex}a(-1)^3\xa0+ 3(-1)^2\xa0- b(-1) - 6 = 0{/tex}{tex}\\Rightarrow{/tex}\xa0{tex}-a + 3 + b - 6 = 0{/tex}{tex}\\Rightarrow{/tex}\xa0{tex}-a + b = 3.{/tex}.........(i)Also, -2 is another zero{tex}\\therefore{/tex}\xa0{tex}p(-2) =0{/tex}{tex}\\Rightarrow{/tex}\xa0{tex}a(-2)^3\xa0+ 3(-2)^2\xa0- b(-2) - 6 = 0{/tex}{tex}\\Rightarrow{/tex}\xa0{tex}- 8a + 12 + 2b - 6 = 0{/tex}{tex}\\Rightarrow{/tex}\xa0{tex}- 8a + 2b = -6{/tex}{tex}\\Rightarrow{/tex}\xa0{tex}4a - b = 3{/tex}.........(ii)Solving equation (i) and (ii), we geta = 2 and b = 5{tex}\\therefore{/tex} p(x) = 2x3 + 3x2\xa0- 5x\xa0- 6Let third zero = kSum of zeroes =\xa0{tex}\\frac { - 3 } { 2 }{/tex}{tex}\\Rightarrow{/tex}\xa0-1 + (-2) + k =\xa0{tex}\\frac { - 3 } { 2 } \\Rightarrow k = \\frac { - 3 } { 2 } + 3 = \\frac { 3 } { 2 }{/tex}Therefore, third zero =\xa0{tex}\\frac { 3 } { 2 }{/tex}