Question

4) create an equation with the following roots: -4 , 4 , -3 , 2/5 (fraction)

5) create an equation with the following roots: ± 2i , -7 , 1

6) find all the zeros on the following polynomial: 2x^3 - 11x^2 - 16x + 105 Given zero: 5

Answer 1

4.
`x^4+(13)/(5)x^3-(86)/(5)x^2-(208)/(5)x+(96)/(5)`

5.
`x^4+6x^3-3x^2+24x-28`

6.
`5, -3, (7)/(2)`

Explanation:

4. If numbers
`-4 , 4 , -3 , (2)/(5)` are roots of the equation, then

`x-(-4)=x+4\\ \\x-4\\ \\x-(-3)=x+3\\ \\x-(2)/(5)`

are factors of the equation. The equation, therefore, can be written in the following way:

`(x+4)(x-4)(x+3)\left(x-(2)/(5)\right)\\ \\=(x^2-16)\left(x^2-(2)/(5)x+3x-(6)/(5)\right)\\ \\=(x^2-16)\left(x^2+(13)/(5)x-(6)/(5)\right)\\ \\=x^4+(13)/(5)x^3-(6)/(5)x^2-16x^2-(208)/(5)x+(96)/(5)\\ \\=x^4+(13)/(5)x^3-(86)/(5)x^2-(208)/(5)x+(96)/(5)`

5. If numbers
`\pm 2i, -7, 1` are roots of the equation, then

`x-2i\\ \\x+2i\\ \\x-(-7)=x+7\\ \\x-1`

are factors of the equation. The equation, therefore, can be written in the following way:

`(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`

6. Given polynomial

`2x^3 - 11x^2 - 16x + 105`

and its zero 5, then

`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)`

Therefore, zeros are

`5, -3, (7)/(2)`