Chapter 8.4, Problem 57HP

To determine

To determine which expression does not belong from the given options.

Answer to Problem 57HP

  $(2c+d)(2c-d)$

  $$

Explanation of Solution

Given:

  $(2c-d)(2c-d)$ ; $(2c+d)(2c-d)$ ; $(2c+d)(2c+d)$ ; $(c+d)(c+d)$

Calculation:

A perfect square binomial is a trinomial that when factored gives the square of a binomial.

In the first expression,

Expression 1: $(2c-d)(2c-d)$

  $\Rightarrow {(2c-d)}^2$

So, ${(2c-d)}^2$ is a perfect square binomial.

Expression 2: $(2c+d)(2c-d)$

But it cannot be represented as perfect square binomial as the terms are unlike.

So, $(2c-d)(2c+d)$ is a not a perfect square binomial.

Expression 3: $(2c+d)(2c+d)$

  $\Rightarrow {(2c+d)}^2$

So, ${(2c+d)}^2$ is a perfect square binomial.

Expression 4: $(c+d)(c+d)$

  $\Rightarrow {(c+d)}^2$

So, ${(c+d)}^2$ is a perfect square binomial.

Conclusion:

Therefore, the expression that does not belong here is $(2c+d)(2c-d)$ as it is a not a perfect square binomial.