To find the error the Ron-Jon made, we will the given by on our own first. Notice that the first two terms share a common factor of
x2 and the of the last two terms is
1.
3x3+x2+3x+1
x2(3x+1)+3x+1
x2(3x+1)+1(3x+1)
Midway through, we obtained a different partial result than Ron-Jon. He made a mistake when factoring the second group. Let's continue to correctly factor the polynomial.
x2(3x+1)+1(3x+1)
(3x+1)(x2+1)