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Quiz 8 (take-home factoring quiz)
Factor the following polynomials using integers. If it is not possible, state so.

1.) x2+4x+3
Solution: Must find p and q so that pq=3 and p+q=4. Thus p=3 and q=1 work. Now rewrite the polynomial and factor by grouping:
x2+4x+3=x2+3x+x+3=(x2+3x)+(x+3)=x(x+3)+(x+3)=(x+1)(x+3).

2.) x2+x13
Solution: Must find p and q so that pq=13 and p+q=1. No such p and q exist, so this is not factorable (using integers).

3.) x2+2x+1
Solution: This could be factored as a perfect square trinomial, but we will proceed with the p and q method. Must find p and q so that pq=1 and p+q=1. This means taking p=1 and q=1 will work. Therefore x2+2x+1=x2+x+x+1=(x2+x)+(x+1)=x(x+1)+(x+1)=(x+1)(x+1)=(x+1)2.

4.) x24
Solution: This could be factored as a difference of squares, but we will proceed with the p and q method, though. We must find p and q so that pq=4 and p+q=0. Therefore p=2 and q=2 works. Now x24=x2+2x+(2)x+(4)=(x2+2x)+[(2)x+(4)]=x(x+2)+(2)(x+2)=(x2)(x+2).

5.) 6x2+3x3
Solution: We must find p and q so that pq=18 and p+q=3. Therefore p=6 and q=3 works. Now 6x2+3x3=6x2+6x+(3)x+(3)=(6x2+6x)+[(3)x+(3)]=6x(x+1)+(3)(x+1)=(6x3)(x+1)=3(2x1)(x+1)

6.) x281
Solution: This could be factored as a difference of squares, but we will use the p and q method. We must find p and q so that pq=81 and p+q=0. Therefore p=9 and q=9 works. Therefore x281=x29x+9x81=(x29x)+(9x81)=x(x9)+9(x9)=(x+9)(x9).

7.) x2+3x+2
Solution: We must find p and q so that pq=2 and p+q=3. Therefore p=2 and q=1 works. Now x2+3x+2=x2+2x+x+2=(x2+2x)+(x+2)=x(x+2)+(x+2)=(x+1)(x+2).