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

1.) $x^2+4x+3$
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:
$$\begin{array}{ll} x^2+4x+3 &= x^2+3x+x+3 \\ &= (x^2+3x) + (x+3) \\ &= x(x+3) + (x+3) \\ &=(x+1)(x+3). \end{array}$$

2.) $x^2+x-13$
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.) $x^2+2x+1$
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 $$\begin{array}{ll} x^2+2x+1 &= x^2+x+x+1 \\ &= (x^2+x) + (x+1) \\ &= x(x+1) + (x+1) \\ &= (x+1)(x+1) \\ &= (x+1)^2. \end{array}$$

4.) $x^2-4$
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 $$\begin{array}{ll} x^2-4 = x^2 + 2x + (-2)x + (-4) \\ &= (x^2+2x) + [(-2)x+(-4)] \\ &= x(x+2) + (-2)(x+2) \\ &= (x-2)(x+2). \end{array}$$

5.) $6x^2+3x-3$
Solution: We must find $p$ and $q$ so that $pq=-18$ and $p+q=3$. Therefore $p=6$ and $q=-3$ works. Now $$\begin{array}{ll} 6x^2+3x-3 &= 6x^2 +6x+(-3)x +(-3) \\ &= (6x^2+6x) + [(-3)x+(-3)] \\ &= 6x(x+1) + (-3)(x+1) \\ &= (6x-3)(x+1) \\ &= 3(2x-1)(x+1) \end{array}$$

6.) $x^2-81$
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 $$\begin{array}{ll} x^2-81 &= x^2 -9x + 9x-81 \\ &= (x^2-9x) + (9x-81) \\ &= x(x-9) + 9(x-9) \\ &= (x+9)(x-9). \end{array}$$

7.) $x^2+3x+2$
Solution: We must find $p$ and $q$ so that $pq=2$ and $p+q=3$. Therefore $p=2$ and $q=1$ works. Now $$\begin{array}{ll} x^2+3x+2 &= x^2 + 2x + x + 2 \\ &= (x^2+2x) + (x+2) \\ &= x(x+2) + (x+2) \\ &= (x+1)(x+2). \end{array}$$