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Chapter 6: Problem 31

Factor completely, or state that the polynomial is prime. $$x^{2}+36$$

Short Answer

Expert verified
The polynomial \(x^{2} + 36\) is a prime polynomial as it can't be factored using integer coefficients.

Step by step solution

01

Step 1

Identify the polynomial and its components: we have \(x^{2} + 36\)
02

Step 2

Try to factor the polynomial: In our case, since there is no integral root which when squared gives 36, and the polynomial can't be expressed as difference of squares. It doesn't factor into two binomials with integral coefficients.
03

Step 3

Conclude: As a result, the original polynomial \(x^{2} + 36\) can't be factored using non-integer coefficients, so it is a prime polynomial.

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Most popular questions from this chapter

Contain polynomials in several variables. Factor each polynomial completely and check using multiplication. $$24 a^{4} b+60 a^{3} b^{2}+150 a^{2} b^{3}$$

Now let's move on to factorizations that may require two or more techniques. Factor completely, or state that the polynomial is prime. Check factorizations using multiplication or a graphing utility. $$100 y^{2}+49$$

Now let's move on to factorizations that may require two or more techniques. Factor completely, or state that the polynomial is prime. Check factorizations using multiplication or a graphing utility. $$9 x^{4}+18 x^{3}+6 x^{2}$$

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Factor completely. $$(x-7)^{2}-4 a^{2}$$
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