Use Synthetic Division to Factor the following polynomials completely by given factors or roots and Find all the zeros; When you reach quadratic equation, performance regular factoring or Quadratic Formula. [tex]x^{5} -9x^{3} -x^{2} +9;(x-3),(x-1)[/tex], as factors and -3 as a root

Answer:The factors are (x - (1 - i√3)/2) , (x - (1 + i√3)/2) , (x - 3) , (x + 3) , (x - 1)The zeroes are 1 , 3 , -3 , (1 - i√3)/2 , (1 + i√3)/2Step-by-step explanation:[tex](x^{5}+0-9x^{3} -x^{2}+0+9)[/tex] ÷ (x - 3) =[tex](x^{4}-9x^{3}-x^{2}+0+9)[/tex] ÷ (x - 3) =[tex]x^{4}+3x^{3}+(-x^{2}+0+9)[/tex] ÷ (x - 3) =[tex]x^{4}+3x^{3}-x+(-3x+9)[/tex] ÷ (x - 3) = [tex](x^{4}+3x^{3}-x-3)[/tex][tex](x^{4}+3x^{3}+0-x-3)[/tex] ÷ (x - 1) =x³ + (4x³ + 0 - x - 3) ÷ (x - 1) =x³ + 4x² + (4x² - x - 3) ÷ (x - 1) =x³ + 4x² + 4x + (3x - 3) ÷ (x - 1) =x³ + 4x² + 4x + 3∵ -3 is a root ⇒ (x + 3) is a factor(x³ + 4x² + 4x + 3) ÷ (x + 3) = x² + (x² + 4x + 3) ÷ (x + 3) =x² + x + (x + 3) ÷ (x + 3) =x² + x + 1 ⇒ use the formula to find the factors of this quadratic∵ a = 1 , b = 1 and c = 1∴ [tex]x=\frac{-1+\sqrt{(1)^{2}-4(1)(1)} }{2(1)}=\frac{-1+\sqrt{-3}}{2}=\frac{-1+i\sqrt{3}}{2}[/tex]∴ [tex]x=\frac{-1-i\sqrt{3} }{2}[/tex]∴ The factors are (x - (1 - i√3)/2) , (x - (1 + i√3)/2) , (x - 3) , (x + 3) , (x - 1)The zeroes:x - 3 = 0 ⇒ x = 3x + 3 = 0 ⇒ x = -3x - 1 = 0 ⇒ x = 1x - (1 - i√3)/2 = 0 ⇒ x = (1 - i√3)/2x - (1 + i√3)/2 = 0 ⇒ x = (1 + i√3)/2The zeroes are 1 , 3 , -3 , (1 - i√3)/2 , (1 + i√3)/2