Please help I dont remember how to factor and have an exam tomorrow. Please show/explain all the steps?

Question

Factor x^2 + 2x − 8

Please show all the steps..the book shows this but Im not sure how to get the -32? Please show me!
=8x(x+2) - 32(x+2)
=(8x -32)(x +2)
=2(x-4)(x+2)
= (x-4)(x+2)

Answer 1

FOIL
first outer inner last
multiply the first in each set of () then the outer most then the inner most and finaly the last

(a+b)(c-d)
=ac+-ad+bc+-bd

or in the case of
a(b+d)
=ab+ad
just multiply though

Answer 2

Hey man, That was a good try but when you don't have a constant in front of the x^2 all you have to do is find out what two numbers when multiplied equals -8 and when added equals +2
The first two number that pop into my mind are 4 and 2.
-2(4)=-8
-2+4=2
So those two numbers are what you need to factor x^2+2x-8:
x^2+2x-8
=(x+4)(x-2)

The way the book showed you is over complicating things...and actually I think it's wrong (but who knows). In the third step it looks like they factored out a 2 when they should have factored out an 8..as in:
(8x-32)=8(x-4)

Lets say you had something like 8x^2-12x-8..then you should factor it in a manner similar to the way the book shows. To factor 8x^2-12x-8, you would want two numbers who when multiplied together equal 8(-8)=-64 but when added together equals -12. I guess something like 16 and 4 would do the trick:
-16(4)=-64
-16+4=-12
So we rewrite 8x^2-12x-8 as:
8x^2-16x+4x-8
Then we group the first two terms and the last two terms together and factor them if we can:
8x(x-2)+4(x-2)
so what ever is in the brackets is one factor and whatever is outside of the brackets is your other factor:
(8x+4)(x-2) is 8x^2-12x-8 factored.

I suppose we could of done the same thing with x^2+2x-8:

we would have to find what two numbers when multiplied together equals 1(-8)=-8
and when added together equals +2
remember it's 4 and -2 so we can rewrite x^2+2x-8 as:
x^2+4x-2x-8 and group the two first terms together and factor them if we can, then group the last two terms together and factor them if we can:
x(x+4)-2(x+4)
so the outer numbers make up the first factor and the numbers in the brackets make up the second factor (the numbers in the left brackets have to be exactly the same as the numbers in the right brackets, that's how you know you factored it right).
(x-2)(x+4) is x^2+2x-8 factored.

Hopefully this helped you understand what's going on, it's tricky at first but once you get the hang of it (with practice) you will get it. Practice is the most important part...

Good luck on your exam!!

Answer 3

The steps you show seem to be for a completely different problem. Here's how I'd do it:

The ending term is negative, so we have one plus and one minus:
(x + ?)(x - ?)
Now we need to find two numbers whose product is 8 (the last number) and whose difference is 2 (the middle number). 4 and 2 would work here. We have to end up with the 2 being positive, so we want 4 - 2 = 2 and not 2 - 4 = -2. So the 4 goes after the positive and the 2 goes after the negative:
(x + 4)(x - 2)

Answer 4

x² + 2x + 8 = 0
You now require two numbers that multiply to give 8 and add or subtract to give 2
These numbers are 4 and 2 with appropriate + or - sign as shown:-
(x + 4).(x - 2)

Check
x.(x - 2) + 4.(x - 2)
= x² - 2x + 4x - 8
= x² + 2x - 8 as required.

Answer 5

you ought to locate components of -8 that upload as much as +2. achievable pairs of components are... -a million, 8 a million, -8 -2, 4 2, -4 The pair that upload as much as +2 are -2, 4, so your quadratic components to (x-2)(x+4)