Question 3. The polynomial function that has roots 1, 2, -3 and -1 has factors (x-1), (x-2), (x+3) and (x+1), respectively. Then the form of this polynomial function is
[tex]f(x)=(x-1)(x-2)(x+3)(x+1)=((x-1)(x+1))((x-2)(x+3))=(x^2-1)(x^2+x-6)=x^4+x^3-7x^2-x+6.[/tex]
Answer: correct choice is A
Question 4. Factor the polynomial function [tex]f(x)=3x^3+15x^2+18x.[/tex] All three terms of the polynomial function have common factor [tex]3x,[/tex] then
[tex]f(x)=3x(x^2+5x+6).[/tex]
Factor the quadratic function [tex]g(x)=x^2+5x+6.[/tex]
[tex]D=5^2-4\cdot 6=25-24=1,\\ \\\sqrt{D}=1,\\ \\x_{1,2}=\dfrac{-5\pm1}{2}=-3,\ -2,[/tex]
then
[tex]g(x)=(x+3)(x+2),\\ \\f(x)=3x(x+3)(x+2).[/tex]
This means that roots are -3, -2 and 0 of multiplicity 1.
Answer: correct choice is C
Question 5. The equation [tex]x^3-64=0[/tex] can be rewritten as
[tex](x-4)(x^2+4x+16)=0.[/tex]
The quadratic equation [tex]x^2+4x+16=0[/tex] has roots
[tex]D=4^2-4\cdot 16=16-64=-48,\\ \\\sqrt{D}=4\sqrt{3}i,\\ \\x_{1,2}=\dfrac{-4\pm4\sqrt{3}i}{2}=-2\pm2\sqrt{3}i.[/tex]
The roots of the equation [tex]x^3-64=0[/tex] are [tex]4,\ -2-2\sqrt{3}i,\ -2+2\sqrt{3}i.[/tex]
Answer: correct choice is C
