How do you factor x^2-3x+8?

Question

campbell_st · Answer

you can't ... there are no factors of 8 that add to -3... both factors would be negative

anonymous · Answer

use completing the square

anonymous · Answer

you can still factor it...

anonymous · Answer

don't listen to cambell...it may not factor nicely but it is factorable

campbell_st · Answer

well the interesting thing with this question is there are no real roots ... ... just an observation

anonymous · Answer

still factorable, even if the roots are complex

anonymous · Answer

With the quadratics formula, it's possible, but I don't understand how to use it.

anonymous · Answer

just plug and chug.

anonymous · Answer

You can always add or subtract numbers to make it factor nicely. But if you do this then you will not be solve for =0 but instead some number,

anonymous · Answer

which is completing the square

alexwee123 · Answer

wouldn't the roots be imaginary?

campbell_st · Answer

the formula is 

$$x=\frac{-b \pm \sqrt{b^2 -4ac}}{2a}$$

in your question a = 1 b = -3 c = 8

substitute them and evaluate 

the problem will be inside the square root... you will get -23

this can be handled by the use of complex numbers...

anonymous · Answer

For example you can subtract 19 to both sides. 
x^2-3x+8=0
x^2-3x+8-19=-19
x^2-3x-10    = -19
(x-5)(x+2)     = -19

anonymous · Answer

an easier number is half the coefficient of b, squared

anonymous · Answer

Can we keep the topic on math? Thanks. @Romero

anonymous · Answer

YOLO!!! ╭∩╮ - _ - ╭∩╮