bc46dc…: “Nearer the quick angle”

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1 : Complete the square on the following quadratic: $\sage {p1f2} = \answer {\sage {p1c2}}(\answer {\sage {x-p1c1}})^2 + \answer {\sage {p1c3}}$

Don’t forget to factor out any leading term, like the $\sage {p1c2}$ (ignore if this is 1, since this is randomly generated). Once you have factored out the leading term (if needed), you need to divide the resulting middle term ( $\sage {p1c4}$ ) and then square it to get $\sage {(p1c4)^2}$ , which you need to add and subtract to the expression.

2 : Complete the square on the following quadratic: $\sage {p2f2} = \answer {\sage {p2c2}}(\answer {\sage {x-p2c1}})^2 + \answer {\sage {p2c3}}$

Don’t forget to factor out any leading term, like the $\sage {p2c2}$ (ignore if this is 1, since this is randomly generated). Once you have factored out the leading term (if needed), you need to divide the resulting middle term ( $\sage {p2c4}$ ) and then square it to get $\sage {(p2c4)^2}$ , which you need to add and subtract to the expression.

3 : Complete the square on the following quadratic: $\sage {p3f2} = \answer {\sage {p3c2}}(\answer {\sage {x-p3c1}})^2 + \answer {\sage {p3c3}}$

Don’t forget to factor out any leading term, like the $\sage {p3c2}$ (ignore if this is 1, since this is randomly generated). Once you have factored out the leading term (if needed), you need to divide the resulting middle term ( $\sage {p3c4}$ ) and then square it to get $\sage {(p3c4)^2}$ , which you need to add and subtract to the expression.

4 : Complete the square on the following quadratic: $\sage {p4f2} = \answer {\sage {p4c2}}(\answer {\sage {x-p4c1}})^2 + \answer {\sage {p4c3}}$

Don’t forget to factor out any leading term, like the $\sage {p4c2}$ (ignore if this is 1, since this is randomly generated). Once you have factored out the leading term (if needed), you need to divide the resulting middle term ( $\sage {p4c4}$ ) and then square it to get $\sage {(p4c4)^2}$ , which you need to add and subtract to the expression.