Answer:
4. [tex]x^4+\dfrac{13}{5}x^3-\dfrac{86}{5}x^2-\dfrac{208}{5}x+\dfrac{96}{5}[/tex]
5. [tex]x^4+6x^3-3x^2+24x-28[/tex]
6. [tex]5, -3, \dfrac{7}{2}[/tex]
Step-by-step explanation:
4. If numbers [tex]-4 , 4 , -3 , \dfrac{2}{5}[/tex] are roots of the equation, then
[tex]x-(-4)=x+4\\ \\x-4\\ \\x-(-3)=x+3\\ \\x-\dfrac{2}{5}[/tex]
are factors of the equation. The equation, therefore, can be written in the following way:
[tex](x+4)(x-4)(x+3)\left(x-\dfrac{2}{5}\right)\\ \\=(x^2-16)\left(x^2-\dfrac{2}{5}x+3x-\dfrac{6}{5}\right)\\ \\=(x^2-16)\left(x^2+\dfrac{13}{5}x-\dfrac{6}{5}\right)\\ \\=x^4+\dfrac{13}{5}x^3-\dfrac{6}{5}x^2-16x^2-\dfrac{208}{5}x+\dfrac{96}{5}\\ \\=x^4+\dfrac{13}{5}x^3-\dfrac{86}{5}x^2-\dfrac{208}{5}x+\dfrac{96}{5}[/tex]
5. If numbers [tex]\pm 2i, -7, 1[/tex] are roots of the equation, then
[tex]x-2i\\ \\x+2i\\ \\x-(-7)=x+7\\ \\x-1[/tex]
are factors of the equation. The equation, therefore, can be written in the following way:
[tex](x-2i)(x+2i)(x+7)(x-1)\\ \\=(x^2+4)(x^2-x+7x-7)\\ \\=(x^2+4)(x^2+6x-7)\\ \\=x^4+6x^3-7x^2+4x^2+24x-28\\ \\=x^4+6x^3-3x^2+24x-28[/tex]
6. Given polynomial
[tex]2x^3 - 11x^2 - 16x + 105[/tex]
and its zero 5, then
[tex]2x^3 - 11x^2 - 16x + 105\\ \\=2x^3 -10x^2 -x^2 +5x-21x+105\\ \\=2x^2 (x-5)-x(x-5)-21(x-5)\\ \\=(x-5)(2x^2 -x-21)\\ \\=(x-5)(2x^2 +6x-7x-21)\\ \\=(x-5)(2x(x+3)-7(x+3))\\ \\=(x-5)(x+3)(2x-7)[/tex]
Therefore, zeros are
[tex]5, -3, \dfrac{7}{2}[/tex]
