Question

Is $81 + 18y + {y^2}$ a perfect square trinomial and how do you factor it?

Answer 1

Hint: Let us consider the coefficient of ${y^2}$, the coefficient of $y$ and the constant term as $a,b$ and $c$ respectively. Now, we calculate the discriminant $D$ for the given question which is given by $D = {b^2} - 4ac$. If the value of discriminant is zero, then we can say that the given quadratic polynomial is a perfect square trinomial. In order to factor the given equation, we must first write it in the form of ${a^2} + 2ab + {b^2}$ and then use the conversion ${\left( {a + b} \right)^2}$ to get the required answer.

Complete step by step answer:
Here in this question, we are given the quadratic polynomial: $81 + 18y + {y^2}$ and we asked to check if it is a perfect square trinomial or not. Next, we are asked to factorize it. The given quadratic polynomial can be said to be a perfect square trinomial if it can be written in the form ${\left( {a + b} \right)^2}$. As we can clearly see that both the roots of the equation will be the same, if such a condition arises, then the value of discriminant must be equal to zero. Now, let's consider the coefficient of ${y^2}$, the coefficient of $y$ and the constant terms as $a,b$ and $c$ respectively.
For $81 + 18y + {y^2}$,
$a = 1,b = 18,c = 81$
Applying the discriminant formula, we get,
$D = {b^2} - 4ac \\
\Rightarrow D = {\left( {18} \right)^2} - 4 \times 1 \times 81 \\
\Rightarrow D = 324 - 324 \\
\Rightarrow D = 0 \\ $
Therefore, the value of discriminant is zero, so we can say that $81 + 18y + {y^2}$ is a perfect square trinomial. Now, we factor the given polynomial. The given expression can be rewritten as:
$81 + 18y + {y^2} = {y^2} + 18y + 81 \\
\Rightarrow {y^2} + 18y + 81 = {\left( y \right)^2} + 2 \times y \times 9 + {\left( 9 \right)^2}$
Clearly, we can see that R.H.S. of the given expression is in the form of ${a^2} + 2ab + {b^2}$ whose factored form is ${\left( {a + b} \right)^2}$, so we can write it as,
${y^2} + 18y + 81 = {\left( {y + 9} \right)^2}$

Hence, ${\left( {y + 9} \right)^2}$ is the factored form.

Note:Students should note that if the value of discriminant would have been greater than zero then we would have used either the middle term split method or completing the square method to factorize the quadratic polynomial. If the value of discriminant would have been negative, then we won’t be able to factorize the polynomial as in that case, the roots wouldn’t have been real.