Question

find the perfect square trinomial whose first two terms are n² - 6n. the perfect square trinomial is n² - 6n + . (type an integer or a simplified fraction.)

Explanation:

Step1: Recall the formula for perfect - square trinomial

The general form of a perfect - square trinomial is $(a - b)^2=a^{2}-2ab + b^{2}$. In the given expression $n^{2}-6n$, we have $a = n$ and $-2ab=-6n$.

Step2: Find the value of $b$

Since $a = n$ and $-2ab=-6n$, substitute $a = n$ into $-2ab=-6n$. We get $-2nb=-6n$. Divide both sides of the equation by $-2n$ (assuming $n
eq0$; when $n = 0$, the form still holds). So, $b = 3$.

Step3: Calculate the last term

The last term of the perfect - square trinomial $a^{2}-2ab + b^{2}$ is $b^{2}$. Since $b = 3$, $b^{2}=9$.

Answer:

9