How do you factor #x^3 + 7x^2 + 9x + 3#?

1 Answer
Jul 26, 2015

Answer:

Use shortcut concerning coefficients, then quadratic formula to find:

#x^3+7x^2+9x+3 = (x+1)(x+3+sqrt(6))(x+3-sqrt(6))#

Explanation:

Let #f(x) = x^3+7x^2+9x+3#

First notice that the sum of coefficients of the terms of odd powers of #x# is equal to the sum of the coefficients of the terms of even powers of #x#. That is #1 + 9 = 10 = 7 + 3#.

As a result, #x=-1# is a root of #f(x) = 0#

#f(-1) = -1+7-9+3 = 0#

So #(x+1)# is a factor of #f(x)#

#x^3+7x^2+9x+3 = (x+1)(x^2+6x+3)#

Next #x^2+6x+3# is of the form #ax^2+bx+c#, with #a=1#, #b=6# and #c=3#. This has discriminant #Delta# given by the formula:

#Delta = b^2-4ac = 6^2-(4xx1xx3) = 36-12 = 24#

Since this is positive, but not a perfect square #x^2+6x+3=0# has irrational roots given by the quadratic formula:

#x = (-b+-sqrt(b^2-4ac))/(2a) = (-b+-sqrt(Delta))/(2a)#

#=(-6+-sqrt(24))/2 = -3+-sqrt(6)#

Hence #x^2+6x+3 = (x+3+sqrt(6))(x+3-sqrt(6))#

Putting it all together:

#x^3+7x^2+9x+3 = (x+1)(x+3+sqrt(6))(x+3-sqrt(6))#