How do you factor #3x^4-x^3-3x^2+1#?

We can see clearly that all elements that contain the variable #x# are multiplying #x^2#, so we can factor it out as follows:

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

#x^2[x(3x-1)-3]+1#

Answer: #3 x^4 - x^3 - 3 x^2 + 1# #=# #(x - 1)(3 x^3 + 2 x^2 - x - 1)#

Explanation:

Let's denote our polynomial #3 x^4 - x^3 - 3 x^2 + 1# by #P(x)#.

We use the Remainder Theorem: when dividing a polynomial #P(x)# by #(x-c)#, where #c# is a constant, the remainder equals #P(c)#.

Hence, for #x - c# to be a divisor of the polynomial #P(x)#, the remainder #P(c)# needs to be equal to zero.

Therefore, we need to find a constant #c# such that #P(c) = 0#.

#P(c) = 0# #<=># #3 c^4 - c^3 - 3 c^2 + 1 ##=# #0# #<=># #3 c^2 (c^2 - 1) - (c^3 - 1)##=##0# #<=># #3 c^2(c+1)(c-1) - (c - 1)(c^2 + c + 1)##=##0# #<=># #(c - 1)(3 c^3 + 3 c^2 - c^2 - c - 1)##=##0# #<=># #(c - 1)(3 c^3 + 2 c^2 - c - 1)##=##0#

So, we found that, if #c##=##1#, then #P(c) = 0#. Therefore, #x -1# is a divisor of the polynomial #P(x)#.

Verification : divide #P(x)##=##3 x^4 - x^3 - 3 x^2 + 1# by #x - 1# and obtain
#P(x)##/##(x - 1)##=##3 x^3 + 2 x^2 - x - 1#