183k views
5 votes
Factor completely x² + 64

Factor completely x² + 64-example-1
User Benten
by
5.2k points

2 Answers

7 votes

Answer:

(b) Prime

Step-by-step explanation:

The difference of squares can be factored as ...

a² -b² = (a -b)(a +b)

This polynomial expression can be considered to be the difference of squares.

__

For the squares x² and (-64), the linear factors will have constants equal to the root of -64, which is an imaginary number.

√(-64) = 8i

x² +64 = (x -8i)(x +8i)

This is the complete factorization.

However, the constants in this factorization are not integers. Hence we say this polynomial cannot be factored using integers.

When it cannot be factored using integers, it is considered prime.

User INeal
by
4.5k points
9 votes

Answer:

B) Prime

Step-by-step explanation:


\quad \large{\boxed{\sf x^2 +64}}

The polynomial (x² + 64) is prime because it has only two factors.

Factors: 1 and (x² + 64)

Note: Prime polynomials are those whose factors are 1 and itself. Prime polynomials cannot be factored.

User David Mann
by
4.7k points