Find the value of n so that the expression is a perfect square trinomial and then factor the trinomial.
-$$x ^ { 2 } + 20 x + n$$
A) $n = 100 ; ( x + 10 ) ^ { 2 }$
B) $n = 400 ; ( x + 20 ) ^ { 2 }$
C) $n = 100 ; ( x + 10 ) ( x - 10 )$
D) $n = 100 ; ( x - 10 ) ^ { 2 }$

Question

Quizplus · Accepted Answer

For a trinomial $x^2 + bx + c$ to be a perfect square, $c$ must be $(\frac{b}{2})^2$. Here, $b = 20$, so $c = (\frac{20}{2})^2 = 100$, making the trinomial $x^2 + 20x + 100$, which factors to $(x + 10)^2$.

Find the Value of N So That the Expression Is x2+20x+nx ^ { 2 } + 20 x + nx2+20x+n

Find the Value of N So That the Expression Is $x ^ { 2 } + 20 x + n$