Question

2x^2 - 8

How would I factor this? I mean, I CAN get it factored, but I spend far too much time hunting for an answer. I know there are more optimal methods out there of factoring polynomials like this, so I figured I'd ask here.

Answer 1

take 2 common and then use \(\huge a^2-b^2=(a+b)(a-b)\)

Answer 2

both terms are multiples of 2

then use difference of squares

Answer 3

I generally start this kind with
(2x )(x )
But I have a hard time with the ones that have anything other than 1 in the leading term.

Answer 4

2(x^2-4) = 2(x^2-2^2)
i hope u can take it further.

Answer 5

I would apply what hartnn suggested. It's the right approach.

Answer 6

@halimabegum
you cannot solve it, its an expression not an equation

Answer 7

2x^2 -8 = 2 ( x^2 - 4)

Answer 8

2[(x^2)-(2^2)] =2((x-2)(x+2)) difference of two squares or 2(x-2)(x+2)

Answer 9

So I can just factor out the 2?
\[2(x^{2}-4)\] and then use the difference of square from there?
2 ((x-2)(x+2))
When would I be able to tell that this method (factoring something out and then using the difference of squares) is applicable? Unless I would look for something to factor out right off the bat... would I?

Answer 10

2x^2-8
2( x^2 -4)
2 ( x + 2) ( x- 2)

Answer 11

take some other example of your own....with leading co-efficient not 1,. and try this method.

Answer 12

Okay, @hartnn
\[-5x ^{2}+5\]
I would factor out the 5:
\[5(-x ^{2}+1)\]
5(-x+1)(x+1) ?

Answer 13

nopes., take -5 out.

Answer 14

make leading co=efficient = 1, not -1

Answer 15

What would happen to the 5? -1?

Answer 16

Okay, so I would convert it to 5x^2 - 5?

Answer 17

-1(5x^2-5) = -5 (x^2-1)

Answer 18

-5(x+1)(x-1)

Answer 19

yup.

Answer 20

Sweet, thanks bro!

Answer 21

welcome :)

Answer 22

@melbel what you did earlier 5(1-x^2) = 5(-x+1)(x+1) <---- this is fine too. it's 5(1+x)(1-x)