How do you find the roots of x^3+x+10=0x3+x+10=0?

2 Answers
Dec 5, 2016

1 + 2i1+2i, 1 - 2i12i and -22

Explanation:

We can use the remainder and factor theorems to effectively factor this trinomial.

If y = ax^n + bx^(n - 1) + ... + p, then the possible factors are "factors of p"/"factors of a".

This gives us (+-10, +-5, +- 2, +-1)/(+- 1). Usually, when I do these sorts of problems, I start at +1, then work my way up. If you evaluate these factors within the equation and they make the equation true, you have a zero. You will find x = -2 is one of these values.

We now use synthetic division to factor the trinomial.

We want to divide x^3 + x + 10 by x + 2.

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The equation then becomes (x + 2)(x^2 - 2x + 5) = 0

x^2 - 2x + 5 can be solved using the quadratic formula.

x= (-(-2) +- sqrt((-2)^2- 4 xx 1 xx 5))/(2 xx 1)

x = (2 +- sqrt(-16))/2

x = (2+- 4i)/2

x = 1 +- 2i

Putting all of this together, the equation x^3 + x + 10 = 0 has roots of 1 + 2i, 1 - 2i and -2.

Hopefully this helps!

Dec 5, 2016

-2 and 1+-i2

Explanation:

The number of changes in signs of the coefficients is 0.

So, there are no positive roots.

Checking signs of the cubic, for #x = -1 and -2, we

find that -2 is a zero of the cubic.

So, it is of the form

(x+2)(x^2+ax+b).

Comparing like coefficients,

a =-2 and and b = 5. Now, the zeros of the quadratic factor

x^2-2x+5 are 1+-i2