solve 3x^4+3x^3-17x^2+x-6=0?

Answer 1

x=2

Answer 2

Hi,

We have

3·x^4+3·x^3-17·x^2+x-6=0

I'm going to assume you have a passing familiarity with the remainder theorem, the factor theorem, and synthetic division.

Looking at the last term, we notice that it is 6. So, some possible roots of our equation are ±6, ±3, ±2, and ±1. We can use synthetic division and the factor theorem to solve the equation. Now, since testing each number and typing the results out is sort of a drag, and doesn't really show anything, I'll just cut to the chase

Using synthetic division and dividing by -3, we get

3 3 -17 1 - 6 | -3
..-9 18 -3 ··6
----------------------------
3 -6 1 -2 ···0

So, we note that the remainder is zero, and -3 is one root of our equation.

We are left with the depressed equation

3·x³·-6·x² + x - 2 = 0

We note that the last term is 2, and the possible integer roots are ±1 and ±2. Again, we have to test all the values until the remainder is zero, so I'll just reproduce the division that gives that result. Testing the value 2, and using synthetic divison, we get

3 -6 1 -2 | 2
····6 0 · 2
____________
3 · 0· 1 · 0

We see that the remainder is zero, and our second depressed equation is

3·x² + 1 = 0

We can simply use the quadratic formula to find that the roots are -i / √3 and i / √3. Or, you could simply solve the quadratic eqation by inspection.

So, the solutions to the given equation are

-i / √3 , i / √3, -3, and 2


James :-)

Answer 3

a element is any form expression used in a multiplication difficulty. so there are 2 element expressions in this equation a million] x - 10, and 2] x + 10. considering the fact that any factor more desirable by ability of 0 = 0 {a million x 0, 0 x 1000, a million/2 x 0, 0.034 x 0, ... = 0. So if the two of those expressions = 0, the entire factor equals 0. So set up 2 seperate effortless equtions applying the expressions: x - 10 = 0, and x + 6 = 0, resolve each and every seperately, the 1st by ability of including 10 to the two area, and getting x = 10, the 2d by ability of subtracting 6 from the two area and getting x = -6. isn't math exciting?:) sturdy success!

Answer 4

The rational root theorem suggests 1,2,3,6 etc as possible roots. Synthetic division shows 2 works, and the depressed equation is clearly factorable:

2 ....... 3 .. 3 .. -17 .. 1 .. -6
................ 6 ... 18 .. 2 ... 6
------------------------------------
........... 3 .. 9 .... 1 ... 3 ... 0

[3x³ + 9x² + x + 3](x - 2) = 0
[3x²(x+3) + 1(x+3)]x - 2) = 0
(3x² + 1)(x + 3)(x - 2) = 0
x = 2, x = -3, or
x² = -1/3
x = ±(i√3)/3

Answer 5

3x^4 + 3x^3 -17x^2 + x -6

= (x-2)(3x^3 + 9x^2 +x +3)

= (x-2)(x+3)(3x^2 +1)

x=2
x=-3
x=+/- i/√3

Answer 6

the solutions are

i*sqrt(1/3) (about .577*i)
-i*sqrt(1/3) (about -.577*i)
2
and
-3.

where i is the sqrt(-1)

You have to use the quartic formula.