Question

How would you factor 8x3+27

Answer 1

Rewrite it like this:
\[(2x)^3+(3)^3\]
And then use this pattern:
\[a^3+b^3=(a+b)(a^2-ab+b^2)\]

Answer 2

do you mean \[8x ^{2}+3x+27\] or do you mean \[8x ^{3}+27\]

Answer 3

How do you guys write like that >.< lol i can't figure that out!

Answer 4

Use the equation button at the bottom of this box.

Answer 5

Alternatively learn LaTeX. :P

Answer 6

How i would do the problem is i would automatically plug in the 8x, so it would be \[(8x \pm x) and (x \pm x)\]

Answer 7

Oh then the equation i have to factor is \[8x ^{3}+27\]

Answer 8

Actually im going to have to correct myself because i cant seem to figure out how i used to do this

Answer 9

Jocelyn, good job using the equation editor - makes it much easier to read!

Here's how I would factor it.

\[8x^3+27\] has two cubes for coefficients. I'll take the cube roots as the coefficients of my first term:
\[(8x^3+27)=(2x+3)(ax^2+bx+c)\]
Now, let's multiply that out
\[(8x^3+27)=2ax^3+2bx^2+2cx+3ax^2+3bx+3c\]\[=2ax^3+(2b+3a)x^2+(2c+3b)x+3c\]
Let's figure out the values of a,b,c that make that fit our original.
\[2a=8\]\[3c=27\]giving us\[a=4,c=9\]and
\[8x^3+(2b+3(4))x^2+(2(9)+3b)x+27\]
But we need those x^2 and x terms to vanish, so we solve \[2b+12=0\] giving us b=-6.
Our factored equation is therefore \[(8x^3+27)=(2x+3)(4x^2-6x+9)\]

Answer 10

You could also do polynomial division after choosing the first term. Harder to illustrate that :-)