Are these two equations factorable?

Question

If so please factor them.

x^4 - 3x^2 - 2x + 4

x^3 -2x^2 - x +2

Answer 1

Good morning!
There's a theorem that says if you plug a number into a polynomial expression and the expression then calculates out to zero, THEN X - THE NUMBER IS A FACTOR. I think it's called the Remainder theorem, but I can't be certain

For x^4 - 3x^2 -2x +4, I pick the number 1 to try.
I get (1)^4 -3(1)^2 -2(1) +4, =1-3-2+4=0 Bingo! That means x-1 is a factor. I now have to divide the expression by x-1

I write x^4 -3x^2 -2x +4 as follows:
X^4 +0x^3 -3x^2 -2x +4 to get neat descending powers of x. What comes next is going to drive us both crazy!
___________________
x-1 | x^4 + 0x^3 - 3x^2 -2x +4

x^3
______________________
x-1| x^4 + 0x^3 -3x^2 -2x +4
.......x^4 - 1x^3
----------------------
....0 + x^3 I changed signs and added

--- x^3 +x^2
--- ___________________
x-1|-----x^3 -3x^2 -2x +4
----------x^3 -1x^2
----------------------
---------0 -2x^2 (I changed the signs and added)

.....x^3 + x^2 -2x
.... ______________
x-1|-2x^2 -2x +4
-2x^2 +2x
....________________
...............-4x +4 ( Changed the signs and added)

.....x^3 +x^2 -2x -4
....._____________
x-1| -4x +4
......-4x +4
......_________
.........0
TA DAH! I'm done.I've successfully factored to get
(x-1)(x^3+x^2 -2x -4).
Now alI I have to do is repeat the process of pulling a number out of thin air and see if I can get zero for
x^3 +x^2 -2x -4. I can't find one, so I can go no further. Maybe it exists- I can't find it

The second question is done the same way. I get 1 to give me an equation value of zero, so x-1 is a factor. After division I get (x-1)(x^2 -x -2). The
(x^2 -x -2 can be further factored to give (x-2)(x+1).

My final answer is (x-1)(x-2)(x+1).

I sure hope this is understandable by you.

Answer 2

These are not equations as there is not an = sign.
They are expressions.

Question 1
f (x) = x^4 - 3x² - 2x + 4
f(1) = 1 - 3 - 2 + 4 = 0
Thus x - 1 is a factor of f(x).
To find other factors , use synthetic division:-

**** |**1***0***-3***-2****4
***1|**0***1****1****-2***-4
******1****1***-2*****0***0

f(x) = (x - 1) (x³ + x²- 2x)
f(x) = (x - 1) (x) (x² + x - 2)
f(x) = (x - 1) (x) (x + 2) (x - 1)
f(x) = x (x + 2) (x - 1)²

Question 2
f(x) = x³ - 2x² - x + 2
f(1) = 1 - 2 - 1 + 2 = 0
Thus x - 1 is a factor
****|1****-2****-1*****2
**1 |******1*****-1****-2
*****1****-1****-2****0
f(x) = (x - 1) (x ² - x - 2)
f(x) = (x - 1) (x - 2 ) (x + 1)

Answer 3

Try groupings.

a) (x^4 - 3 x^2) - (2x-4)

x^2(x^2 - 3) - (2x-4). I think not.

or x(x^3 -3x-1) + 4. Nope.

b) (x^3 - 2x^2) - (x-2); take out common factor, x^2
x^2(x-2) - (x-2), take out common factor, (x-2)
(x-2)(x^2 - 1), factor difference of squares
(x-2)(x+1)(x-1)

Answer 4

quit trying get your homework done on Yahoo answers. Thats cheating.