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Quiz 9
Factor the following polynomials using integers. If it is not possible, state so.

note: all these can be solved with the quadratic formula; I chose not to use it when possible but it will always work!
1.) Solve $x^2+1=0$.
Solution: Subtract $1$ to get $$x^2=-1.$$ Now take square roots of each side to get $$x = \pm \sqrt{-1} = \pm i.$$

2.) Solve $x^2+2x+1=0$.
Solution: The left-hand side factors to give us $$(x+1)(x+1)=0.$$ This yields two equations (which happen to be the same): $x+1=0$ and $x+1=0$. Hence the only solution is $x=-1$ (which appears twice).

3.) Solve $x^2-16=0$.
Solution: Add $16$ to both sides to get $$x^2=16.$$ Now take the square root of each side to get $$x= \pm \sqrt{16} = \pm 4.$$

4.) Solve $5x^2-3x+2=0$.
Solution: For this we use the quadratic formula with $a=5$, $b=-3$, and $c=2$ to get $$\begin{array}{ll} x&= \dfrac{-(-3) \pm \sqrt{(-3)^2-4(5)(2)}}{2 \cdot 5} \\ &=\dfrac{3 \pm \sqrt{9-40}}{10} \\ &=\dfrac{3 \pm \sqrt{-31}}{10} \\ &=\dfrac{3 \pm \sqrt{31}i}{10} \\ &= \dfrac{3}{10} \pm \dfrac{\sqrt{31}}{10} i. \end{array}$$

5.)Solve $-3x^2+x+4=0$.
Solution: The left-hand side factors and we get $$-(x+1)(3x-4)=0,$$ hence the solution is $x=-1, \dfrac{4}{3}$.