How do you factor #y= x^3+x^2+2x-4#

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Question

How do you factor #y=  x^3+x^2+2x-4#

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Savannah Anderson · Accepted Answer

#x^3+x^2+2x-4#

#= (x-1)(x^2+2x+4)#
#= (x-1)(x+1-sqrt(3)i)(x+1+sqrt(3)i)#

First note that the sum of the coefficients is zero. That is #1+1+2-4 = 0#. So #x=1# is a zero and #(x-1)# a factor: #x^3+x^2+2x-4 = (x-1)(x^2+2x+4)# The remaining quadratic factor is of the form #ax^2+bx+c# with #a=1#, #b=2# and #c=4#. This has discriminant #Delta# given by the formula: #Delta = b^2-4ac = 2^2-(4*1*4) = -12# Since this is negative, the quadratic has no linear factors with Real coefficients. If we want to proceed further, we can use the quadratic formula to find the Complex roots and derive factors: #x = (-b+-sqrt(b^2-4ac))/(2a)# #= (-b+-sqrt(Delta))/(2a)# #= (-2+-sqrt(-12))/2# #= (-2+-sqrt(12)i)/2# #= (-2+-2sqrt(3)i)/2# #=-1+-sqrt(3)i# Hence: #x^2+2x+4 = (x+1-sqrt(3)i)(x+1+sqrt(3)i)#

Ethan Adkins · Answer

undefined To factor the polynomial $ y = x^3 + x^2 + 2x - 4 $, you can use polynomial factorization techniques such as grouping or factoring by grouping.

First, group the terms:

$ y = (x^3 + x^2) + (2x - 4) $

Now, factor out the greatest common factor from each group:

$ y = x^2(x + 1) + 2(x - 2) $

Finally, factor out the common binomial factor:

$ y = (x^2 + 2)(x + 1) $