Solving a 4th degree polynomial

Question

Does the equation

x⁴ + 3x³ + x + 3 = 0

have any real solutions? If so, what are they?

Answer 1

Hi, I just wanted to add another method to the mix called synthetic division.

We can first find the possible roots with the rational roots test. For x⁴ + 3x³ + +0x² +x + 3 = 0 we take the ± of prime factors of the lowest degree coefficient (in this case x⁰ is 3,1) and divide by the ± prime factors of the highest degree coefficient (in this case x⁴ is 1) , ± 3,1/1 gives ± 3 and ±1 as possible rational roots. I'll start with 1 as the first possible root.

1 | 1 3 0 1 3 To set this up the chosen root 1 is to the far left, then draw an L shape (as I tried).

| 1 4 4 5 The 1 3 0 1 3 to the right of possible root 1 are the coefficients. then bring down

|___________ below the bottom line the first coefficient 1. multiply the root 1 times the lower

1 4 4 5 8 coefficient 1, that is the 1 on the second line. Add second coefficient on the first

line 3 to the 1 on the second to get 4 at the bottom. Continue until done. If the last

addition yields a zero you have a root. Which in this case we don't.

-1 | 1 3 0 1 3 In this case we found a root of -1. So (x+1) is a solution and we can take the bottom

| -1 -2 2 -3 numbers to write our factored equation (x+1)(x³+2x²-2x+3). Again our rational roots

|___________ for our new equation are ±3,1/1 or ±1,3.

1 2 -2 3 0

3 | 1 2 -2 3 This time I start with 3 but no luck.

| 3 15 39

|__________

1 5 13 42

-3 | 1 2 -2 3 Another root. We now have (x+1)(x+3)(x²-x+1). Using the quadratic formula we get

| -3 3 -3 1±√1-4*1*1|/2 or 1±√-3/2 and the are no real roots left.

|__________

1 -1 1 0

I hope this helps, Joe.

Answer 2

x⁴ + 3x³ + x + 3 = 0

x³(x + 3) + 1(x + 3) = 0

(x + 3)(x³ + 1) = 0

(x + 3)(x + 1)(x² - x + 1) = 0

If x + 3 = 0, then x = -3

If x + 1 = 0, then x = -1

The remaining factor has non-real roots (use the quadratic formula)

So, the only real roots are -3 and -1.

Answer 3

There are 2 real roots...... -1 & -3