A polynomial ppp has zeros when x=-2x=βˆ’2x, equals, minus, 2, x=\dfrac13x= 3 1 ​ x, equals, start fraction, 1, divided by, 3, end fraction, and x=3x=3x, equals, 3. What could be the equation of ppp?

Accepted Solution

Answer:Polynomial equation is, [tex]p=(x+2)(3x-1)(x-3)=3x^3-4x^2-17x+6[/tex]Step-by-step explanation:Given:The zeros of the polynomial, [tex]p[/tex], are Β [tex] x=-2,\frac{1}{3}, 3[/tex].Therefore, [tex]x= -2\\(x+2)=0[/tex]So, [tex](x+2)[/tex] is a factor of the polynomial.[tex]x=\frac{1}{3}\\3x=1\\3x-1=0[/tex]So, [tex](3x-1)[/tex] is a factor of the polynomial.[tex]x=3\\x-3=0[/tex]So, [tex](x-3)[/tex] is also a factor of the polynomial.Therefore, a polynomial can be expressed in terms of the product of its factors.Thus, [tex]p=(x+2)(3x-1)(x-3)[/tex]Expanding [tex](x+2)\textrm{ and }(3x-1)[/tex], we get[tex](x+2)(3x-1)=3x^2-x+6x-2=3x^2+5x-2[/tex]Now, expanding [tex](3x^2+5x-2)\textrm{ and }(x-3)[/tex], we get[tex](3x^2+5x-2)(x-3)=3x^3-9x^2+5x^2-15x-2x+6=3x^3-4x^2-17x+2[/tex]Therefore, the polynomial could be [tex]p=(x+2)(3x-1)(x-3)=3x^3-4x^2-17x+6[/tex]