Find a polynomial of degree 3 with real coefficients and zeros of minus​3,minus​1, and​ 4, for which ​f(minus​2)equals30.

Answer:[tex]f(x)=5x^{3}-65x-60[/tex]Step-by-step explanation:we know thatThe roots arex=-3,x=-4 and x=1soThe equation of a polynomial of degree 3 with real coefficients and zeros of minus​3,minus​1, and​ 4 is equal to[tex]f(x)=a(x+3)(x+1)(x-4)[/tex]Remember that[tex]f(-2)=30[/tex] ----> given valueFor x=-2, f(x)=30substitute and solve for the coefficient a[tex]30=a(-2+3)(-2+1)(-2-4)[/tex][tex]30=a(1)(-1)(-6)[/tex][tex]30=6a[/tex][tex]a=5[/tex]soThe polynomial is[tex]f(x)=5(x+3)(x+1)(x-4)[/tex]Apply distributive property[tex]f(x)=5(x+3)(x+1)(x-4)\\f(x)=5(x+3)(x^{2}-4x+x-4)\\f(x)=5(x+3)(x^{2}-3x-4)\\f(x)=5(x^{3}-3x^{2}-4x+3x^{2}-9x-12)\\f(x)=5(x^{3}-13x-12)\\ f(x)=5x^{3}-65x-60[/tex]