Factor the expression by grouping. First, the expression needs to be rewritten as $y^{2}+ay+by+81$. To find $a$ and $b$, set up a system to be solved.
$$a+b=-18$$ $$ab=1\times 81=81$$
Since $ab$ is positive, $a$ and $b$ have the same sign. Since $a+b$ is negative, $a$ and $b$ are both negative. List all such integer pairs that give product $81$.
$$-1,-81$$ $$-3,-27$$ $$-9,-9$$
Calculate the sum for each pair.
$$-1-81=-82$$ $$-3-27=-30$$ $$-9-9=-18$$
The solution is the pair that gives sum $-18$.
$$a=-9$$ $$b=-9$$
Rewrite $y^{2}-18y+81$ as $\left(y^{2}-9y\right)+\left(-9y+81\right)$.
$$\left(y^{2}-9y\right)+\left(-9y+81\right)$$
Factor out $y$ in the first and $-9$ in the second group.
$$y\left(y-9\right)-9\left(y-9\right)$$
Factor out common term $y-9$ by using distributive property.
$$\left(y-9\right)\left(y-9\right)$$
Rewrite as a binomial square.
$$\left(y-9\right)^{2}$$
Steps Using Square Of Binomial
This trinomial has the form of a trinomial square, perhaps multiplied by a common factor. Trinomial squares can be factored by finding the square roots of the leading and trailing terms.
$$factor(y^{2}-18y+81)$$
Find the square root of the trailing term, $81$.
$$\sqrt{81}=9$$
The trinomial square is the square of the binomial that is the sum or difference of the square roots of the leading and trailing terms, with the sign determined by the sign of the middle term of the trinomial square.
$$\left(y-9\right)^{2}$$
Steps Using the Quadratic Formula
Quadratic polynomial can be factored using the transformation $ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right)$, where $x_{1}$ and $x_{2}$ are the solutions of the quadratic equation $ax^{2}+bx+c=0$.
$$y^{2}-18y+81=0$$
All equations of the form $ax^{2}+bx+c=0$ can be solved using the quadratic formula: $\frac{-b±\sqrt{b^{2}-4ac}}{2a}$. The quadratic formula gives two solutions, one when $±$ is addition and one when it is subtraction.
