Find the value of \(c\) that makes it a perfect square trinomial. Then factor the perfect square trinomial.

\(x^2 + 6x + c\)

\(c = 9,\ (x + 3)^2\)

\(x^2 - 9x + c\)

\(c = \dfrac{81}{4},\ \left(x - \dfrac{9}{2}\right)^2\)

\(x^2 + 9x + c\)

\(c = \dfrac{81}{4},\ \left(x + \dfrac{9}{2}\right)^2\)

The following problem is harder, because it requires you to square 25 and 26, which are not easy to square mentally!

\(x^2 - \dfrac{25}{13}x + c\)

\(c = \dfrac{625}{676},\ \left(x - \dfrac{25}{26}\right)^2\)

\(x^2 - 14x + c\)

\(c = 49,\ (x - 7)^2\)

\(y^2 + 12y + c\)

\(c = 36,\ (y + 6)^2\)

\(y^2 - 3y + c\)

\(c = \dfrac{9}{4},\ \left(y - \dfrac{3}{2}\right)^2\)

\(x^2 + 8x + c\)

\(c = 16,\ (x + 4)^2\)

\(x^2 - 4x + c\)

\(c = 4,\ (x - 2)^2\)

\(t^2 + \dfrac{5}{6}t + c\)

\(c = \dfrac{25}{144},\ \left(t + \dfrac{5}{12}\right)^2\)

\(u^2 - \dfrac{u}{4} + c\)

\(c = \dfrac{1}{64},\ \left(u - \dfrac{1}{8}\right)^2\)

\(b^2 - \dfrac{5}{3}b + c\)

\(c = \dfrac{25}{36},\ \left(b - \dfrac{5}{6}\right)^2\)

\(x^2 + 17x + c\)

\(c = \dfrac{289}{4},\ \left(x + \dfrac{17}{2}\right)^2\)

\(a^2 + 12a + c\)

\(c = 36,\ (a + 6)^2\)

\(h^2 - 20h + c\)

\(c = 100,\ (h - 10)^2\)

\(p^2 - p + c\)

\(c = \dfrac{1}{4},\ \left(p - \dfrac{1}{2}\right)^2\)

\(m^2 + 11m + c\)

\(c = \dfrac{121}{4},\ \left(m + \dfrac{11}{2}\right)^2\)