Chapter 8.7, Problem 53HP

To determine

To find: The trinomial which does not belong.

Answer to Problem 53HP

  $4x^2+10x+4$

Does not belong.

Explanation of Solution

Given

First trinomial is

  $4x^2-36x+81$

On simplifying

We get

  $\begin{array}{l}\Rightarrow 4x^2-36x+81={\left(2x\right)}^2-2\left(2x\right)\left(9\right)+{\left(9\right)}^2\\ \Rightarrow 4x^2-36x+81={\left(2x-9\right)}^2\end{array}$

So the given $4x^2-36x+81$ is a perfect square trinomial.

Taking second trinomial

  $25x^2+10x+1$

On simplifying

We get

  $\begin{array}{l}\Rightarrow 25x^2+10x+1={\left(5x\right)}^2+2\left(5x\right)\left(1\right)+{\left(1\right)}^2\\ \Rightarrow 25x^2+10x+1={\left(5x+1\right)}^2\end{array}$

So the given $25x^2+10x+1$ is a perfect square trinomial.

Taking given fourth polynomial

  $9x^2-24x+16$

On solving

We get

  $\begin{array}{l}\Rightarrow 9x^2-24x+16={\left(3x\right)}^2-2\left(3x\right)\left(4\right)+{\left(4\right)}^2\\ \Rightarrow 9x^2-24x+16={\left(3x-2\right)}^2\end{array}$

So the given $9x^2-24x+16$ is a perfect square trinomial.

Taking third polynomial

  $4x^2+10x+4$

This cannot be written as a perfect square trinomial

So,

  $4x^2+10x+4$

does not belong.