How do you find the zeros of x^3-x^2-7x+15=0x3−x2−7x+15=0?

1 Answer
George C.
May 19, 2016

x=-3x=−3 and x=2+-ix=2±i

Explanation:

f(x) = x^3-x^2-7x+15f(x)=x3−x2−7x+15

By the rational root theorem, any rational zeros of f(x)f(x) must be expressible in the form p/qpq for integers pp, qq with pp a divisor of the constant term 1515 and qq a divisor of the coefficient 11 of the leading term.

So the only possible rational zeros are:

+-1±1, +-3±3, +-5±5, +-15±15

Trying each of these in turn, we find:

f(-3) = -27-9+21+15 = 0f(−3)=−27−9+21+15=0

So x=-3x=−3 is a zero and (x+3)(x+3) a factor:

x^3-x^2-7x+15 = (x+3)(x^2-4x+5)x3−x2−7x+15=(x+3)(x2−4x+5)

We can factor x^2-4x+5x2−4x+5 by completing the square and using the difference of squares identity:

a^2-b^2=(a-b)(a+b)a2−b2=(a−b)(a+b)

with a=(x-2)a=(x−2) and b=ib=i as follows:

x^2-4x+5 = x^2-4x+4+1 = (x-2)^2-i^2 = (x-2-i)(x-2+i)x2−4x+5=x2−4x+4+1=(x−2)2−i2=(x−2−i)(x−2+i)

Hence zeros:

x = 2+-ix=2±i

Related questions
See all questions in Zeros
Impact of this question
4024 views around the world
You can reuse this answer
Creative Commons License