Let p(x)=x2+3x+2.
Let factors be (x−a) and (x−b)
So, p(x)=(x–a)(x–b)
⇒p(x)=x2–bx–ax+ab
On comparing the constants, we get ab = 2.
The factors of 2 are +1,-1,-2 and 2.
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Now, p(−2)=(−2)2+(3×(−2))+2=4−6+2=0.
So, (x + 2) is a factor of p(x).
Also, p(−1)=(−1)2+(3×(−1))+2=1−3+2=0
So, (x +1) is also a factor of p(x).
Therefore, x2+3x+2=(x+1)(x+2)
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