What are the roots of the equation x^3 +4x^2-4x- 16=0x3+4x2−4x−16=0?

2 Answers
George C.
Dec 18, 2016

The roots are:

x = 2x=2, x = -2x=−2 and x=-4x=−4

Explanation:

The difference of squares identity can be written:

a^2-b^2 = (a-b)(a+b)a2−b2=(a−b)(a+b)

We use this with a=xa=x and b=2b=2 later.

Given:

x^3+4x^2-4x-16 = 0x3+4x2−4x−16=0

Note that the ratio between the first and second terms is the same as the ratio between the third and fourth terms, so this cubic factors by grouping:

0 = x^3+4x^2-4x-160=x3+4x2−4x−16

color(white)(0) = (x^3+4x^2)-(4x+16)0=(x3+4x2)−(4x+16)

color(white)(0) = x^2(x+4)-4(x+4)0=x2(x+4)−4(x+4)

color(white)(0) = (x^2-4)(x+4)0=(x2−4)(x+4)

color(white)(0) = (x^2-2^2)(x+4)0=(x2−22)(x+4)

color(white)(0) = (x-2)(x+2)(x+4)0=(x−2)(x+2)(x+4)

Hence the roots are:

x = 2x=2, x = -2x=−2 and x=-4x=−4

Alan P.
Dec 18, 2016

x in {2,-4,-2}x∈{2,−4,−2}

Explanation:

x^3+4x^2-4x-16=0x3+4x2−4x−16=0

rArr x^3+4x^2=4x+16⇒x3+4x2=4x+16

rarr x * x^2 +4 * x^2 = x * 2^2 + 4 * 2^2→x⋅x2+4⋅x2=x⋅22+4⋅22

rarr →(at least one possible solution is) x=2x=2
(i.e. (x-2)(x−2) is a factor)

Dividing x^3+4x^2-4x-16x3+4x2−4x−16 by (x-2)(x−2) (using either synthetic or long division) gives:
color(white)("XXX")x^2+6x+8XXXx2+6x+8
which factors using standard operations as:
color(white)("XXX")(x+4)(x+2)XXX(x+4)(x+2)

Therefore
x^3+4x^2-4x-16=0x3+4x2−4x−16=0
rArr (x-2)(x+4)(x+2)=0⇒(x−2)(x+4)(x+2)=0

rArr x in {2,-4,-2}⇒x∈{2,−4,−2}

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