# I need help factoring this x^2 + 5x + 8

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### 5 Answers

This just goes to show that "factoring `x^2+5x+8`" is ambiguous, and we need to know which factors are allowed. For example,

`x^2-2`

can't be factored over the integers because we can't write

`x^2-2=(ax-b)(cx-d)` where `a,b,c,d` are integers. However, we can factor the same expression over the real numbers as

`x^2-2=(x-sqrt(2))(x+sqrt(2)).`

Likewise, `x^2+5x+8` can't be factored over the real numbers, but it can be factored over the complex numbers. You're probably the best judge on what's meant by the question (if you've covered complex numbers, then you probably want aruv's factorization, otherwise the first two answers are probably what you want).

Factor `x^2 + 5x + 8`

In order to factor this expression, we need to find 2 factors of 8 that have a sum of 5.

`(x+a)(x+b) = x^2 + ax+bx + ab`

There are no factors of 8 that will make a sum of 5

`2*4=8 , 2+4=6`

`1*8=8, 1+8 = 9`

**This expression can not be factored.**

The expression x^2 + 5x + 8 has to be factored. It is not possible to express 5 as a sum of two numbers a and b such that a*b = 8.

**As a result, the expression x^2 + 5x + 8 cannot be factored.**

`x^2 + 5x + 8`

`a=1` `b=5` `c=8`

First you multiply a by c

`1xx8=8`

now find factors of 8 that add to 5, there is no such thing so this problem can't be factored by real numbers

`x^2+5x+8=x^2+2(5/2)x+(5/2)^2-(5/2)^2+8`

`=(x+5/2)^2+7/4`

`=(x+5/2)^2-(i^2 7)/4`

`=(x+5/2)^2-((isqrt(7))/2)^2`

`=(x+(5+isqrt(7))/2)(x+(5-isqrt(7))/2)`