Chapter A.3, Problem 137AYU

To determine

To find: The expression $x^2+4$ is prime.

Answer to Problem 137AYU

The expression $x^2+4$ must be prime.

Explanation of Solution

Given information:

The expression is $x^2+4$ .

Calculation:

Calculate the polynomial to be prime, it should be able to written as a product of the form $\left(x+a\right)\left(x+b\right)$ . The value of $a$ and $b$ indicate the values of $x$ that cause the original quadratic equation to be equal to $0$ .

If,

  $\begin{array}{l}{x}^2+4=0\\ x^2=-4\end{array}$

We see that the square of a number is equal to negative number, which can never be possible for real values of $x$ . The square of a real number is always positive. This shows that $x^2+4$ is prime.

Consider when $a$ and $b$ are 1 and 4

  $\left(x+1\right)\left(x+4\right)=x^2+5x+4$

Since this doesn’t equal $x^2+4$ when multiply out, these factors are not an option.

Consider when $a$ and $b$ are -1and -4

Since $\left(x-1\right)\left(x-4\right)=x^2-5x+4$

Since this doesn’t equal $x^2+4$ when multiplied out, these factors are not option either.

Consider the positive for $a,b=2,2$

  $\left(x+2\right)\left(x+2\right)=x^2+4x+4$

Since doesn’t equal $x^2+4$ when multiplied out these factors are also not an option consider the possibilities for $a,b=2,2$

  $\left(x-2\right)\left(x-2\right)=x^2-4x+4$

Therefore this doesn’t equal $x^2+4$ when multiplied out, these are also not an option. this means that $x^2+4$ must be prime.