How do you factor #x^3+x^2-7x-3#?

1 Answer
Aug 18, 2016

Answer:

#x^3+x^2-7x-3 = (x+3)(x-1-sqrt(2))(x-1+sqrt(2))#

Explanation:

#f(x) = x^3+x^2-7x-3#

By the rational roots theorem, any rational zeros of #f(x)# are expressible in the form #p/q# for integers #p, q# with #p# a divisor of the constant term #-3# and #q# a divisor of the coefficient #1# of the leading term.

That means tha the only possible rational zeros are:

#+-1, +-3#

We find:

#f(-3) = -27+9+21-3 = 0#

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

#x^3+x^2-7x-3 = (x+3)(x^2-2x-1)#

We can factor the remaining quadratic by completing the square and using the difference of squares identity:

#a^2-b^2=(a-b)(a+b)#

with #a=(x-1)# and #b=sqrt(2)# as follows:

#x^2-2x-1#

#=x^2-2x+1-2#

#=(x-1)^2-(sqrt(2))^2#

#=(x-1-sqrt(2))(x-1+sqrt(2))#

Putting it all together:

#x^3+x^2-7x-3 = (x+3)(x-1-sqrt(2))(x-1+sqrt(2))#