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

2 Answers
Dec 5, 2016

Answer:

#1 + 2i#, #1 - 2i# and #-2#

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

Answer:

#-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#