Question

Find the roots of the polynomial equation. x^3-x^2+x+39

1+3i, 1-3i, -3
1+3i, 1-3i, 3
-2+3i, -2-3i, -3
2+3i, 2-3i, -3

Answer 1

it will be +- factors of 39, plug each in to see which of them work.

Answer 2

the factors would be 1, 3, 13, 39 right?

Answer 3

yes, the factors, yes, but the roots can be negative or positive.

Answer 4

Plug in x = 3 and x = -3 into the polynomial.
Whichever of the two gives you zero is a root. Call it k.
Then using long division or synthetic division, divide the original polynomial by x - k to get a quadratic equation.
Then use the quadratic formula to solve the quadratic equation.

Answer 5

I plugged in 3 and -3 both don't equal 0 @mathstudent55

Answer 6

Try again. One of the two definitely makes it zero.

Answer 7

One is 60 and the other is 18

Answer 8

\(x^3-x^2+x+39 = 0\)

\( (3)^3 - (3)^2 + (3) + 39 \)

\( (-3)^3 - (-3)^2 + (-3) + 39 \)

Answer 9

Let's do each one.

(3)^3 - (3)^2 + (3) + 39
= 27 - 9 + 3 + 39
= 60
Your 60 is correct.

Answer 10

(-3)^3 - (-3)^2 + (-3) + 39
= -27 - 9 - 3 + 39
= 0
Your 18 is incorrect.

Remember that (-3)^3 = -27 and
-(-3)^2 = - 9

Answer 11

Oh, now I divide the equation by x-3?

Answer 12

Be careful.
When a polynomial equation has roots, a, b, c, d, etc.
The binomial factors are (x - a), (x - b), (x - c), (x - d), etc.
Since the root is -3, you need x - (-3) = x + 3.
Divide the polynomial by x + 3.

Answer 13

Oh I see.

Answer 14

After synthetic division I got 1 4 13 0

Answer 15

@mathstudent55

Answer 16

I get \(x^2 - 4x + 13\) with long division. Can you check on your synthetic division?