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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+x−13
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.) x2−4
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
x2−4=x2+2x+(−2)x+(−4)=(x2+2x)+[(−2)x+(−4)]=x(x+2)+(−2)(x+2)=(x−2)(x+2).
5.) 6x2+3x−3
Solution: We must find p and q so that pq=−18 and p+q=3. Therefore p=6 and q=−3 works. Now
6x2+3x−3=6x2+6x+(−3)x+(−3)=(6x2+6x)+[(−3)x+(−3)]=6x(x+1)+(−3)(x+1)=(6x−3)(x+1)=3(2x−1)(x+1)
6.) x2−81
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
x2−81=x2−9x+9x−81=(x2−9x)+(9x−81)=x(x−9)+9(x−9)=(x+9)(x−9).
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).