Question

List all possible rational roots and then find all real and complex roots (zeros) of the polynomial

Answer 1

We are given the following polynomial

`2x^3+12x^2+29x+33`

and we want to find all possible rational roots. To do so, we first identify the coefficient 2 as the coefficient of the leading term and 33 as the independent term. To find all possible rational roots of the polynomial, we will first find the list of all divisors of the leading coefficient and the independent term.

leading coefficient(2): it can be divided by the following numbers: 2, -2, 1, -1.

independent term(33): Note that 33=11*3. So, it can be divided by 11, -11, 3, -3, 1, -1, 33, -33.

Now, to list all possible rational roots of the polynomial, we take all possibly combinations of one divisor of the independent term and divide it by one divisor of the leading coefficient. So, for example, one possible combination would be

`(11)/(-2)`

If we do this, we would get the following list of all possible combinations:

11/2, 11/-2, 11/1, 11/-1, 3/2, 3/-2, 3/1, 3/-1, 1/2, 1/-2, 1/1, 1/-1, 33/2, 33/-2, 33/1, 33/-1.

the proper way to find the rational roots would be to take x= one root of the list and check if the polynomial is 0 at that value. Let us take x=-3. We can check that

`2\cdot(-3)^3+12\cdot(-3)^2+29\cdot(-3)+33=-2\cdot27+12\cdot9-3\cdot29+33=-54+108-87+33=0`

since the polynomial is 0 at x=-3, we have that it can be factored as

`(x+3)\cdot\text{polynomial of degre}e\text{ 2}`

So if

`2x^3+12x^2+29x+33=(x+3)\cdot(ax^2+bx+c)`

now, we want to find the values of a,b,c so the equation remains true

If we distribute the product on the rigt, we have

`ax^3+(b+3a)x^2+(c+3b)x+3c`

since this two polynomials should be equal, their coefficients should be equal. Then, by comparing the coefficient of x^3 we get that

`a=2`

If we compare the coefficients of x^2 we get

`b+3a=12`

since a=2 we get

`b+3\cdot2=12=b+6`

by subtracting 6 on both sides, we get

`b=12-6=6`

Finally if we compare the independent term we can check that

`3c=33`

So by dividing both sides by 3 we get

`c=(33)/(3)=11`

Finally, we have that

`2x^3+12x^2+29x+33=(x+3)\cdot(2x^2+6x+11)`

Now, to find all roots, we should factor the quadractic factor

`2x^2+6x+11`

Recall that given a polynomial of the form

`ax^2+bx+c`

The roots are given by the expression

`x=\frac{-b\pm\sqrt[]{b^2-4ac}}{2a}`

In our case a=2, b=6 and c=11. So we have

`x=\frac{-6\pm\sqrt[]{6^2-4\cdot2\cdot11}}{2\cdot2}=\frac{-6\pm\sqrt[]{36-88}}{4}=\frac{-6\pm\sqrt[]{52}i}{4}=\frac{-6\pm\sqrt[]{13\cdot4}i}{4}=\frac{-6\pm2\sqrt[]{13}i}{4}=\frac{-3\pm\sqrt[]{13}i}{2}`

So the roots of the polynomial are

`x_1=-3`
`x_2=\frac{-3+\sqrt[]{13}i}{2}`

and

`x_3=\frac{-3-\sqrt[]{13}i}{2}`