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