Find a polynomial of degree 3 with real coefficients that satisfies the given conditions.
-Zeros: $3 , \mathrm { i } , - \mathrm { i } ; \mathrm { f } ( 2 ) = 15$
A) $- 3 x ^ { 3 } + 9 x ^ { 2 } - 3 x + 9$
B) $3 x ^ { 3 } - 9 x ^ { 2 } - 3 x + 9$
C) $- 3 x ^ { 3 } + 9 x ^ { 2 } + 3 x - 9$
D) $3 x ^ { 3 } - 9 x ^ { 2 } + 3 x - 9$

Question

Quizplus · Accepted Answer

Since the zeros are $3 , \mathrm { i } , - \mathrm { i } $, the polynomial can be written as $$f(x)=a(x-3)(x^2+1)$$$$f(x)=a(x^3-3x^2+x-3x^2+3x-3)$$$$f(x)=a(x^3-6x^2+4x-3)$$To find the value of $a$, we use the given condition $f(2)=15$.$$f(2)=a(2^3-6(2)^2+4(2)-3)=15$$$$a(8-24+8-3)=15$$$$a(-9)=15$$$$a=-15/9=-5/3$$Therefore, the polynomial is $$f(x)=-\frac{5}{3}(x^3-6x^2+4x-3)$$$$f(x)=-5x^3+10x^2-\frac{20}{3}x+5$$

Find a Polynomial of Degree 3 with Real Coefficients That 3,i,−i;f(2)=153 , \mathrm { i } , - \mathrm { i } ; \mathrm { f } ( 2 ) = 153,i,−i;f(2)=15

Find a Polynomial of Degree 3 with Real Coefficients That $3 , \mathrm { i } , - \mathrm { i } ; \mathrm { f } ( 2 ) = 15$