Factor completely:?

Question

x3 – 2x2 + 7x – 14

Answer 1

first move - 14 to left side of equation by using additive inverse.

x^3 - 2x^2 +7x - 14 + 14 = 0 +14
X^3 - 3x^2 +7x = 14
now factor out x.
x(x^2 -2x +7) =14
seven is a prime number so only 1 and 7 can be multiplied to give you seven. and it will have to be - 1 and -7. That will not come out to a - 2 when added. so as far as you go that way.

Next we try this was use additive inverse to mov x^3 to the other side of the equation.

x^3 - x^3 -2x^2 +7x -14 = 0 -x^3

-2x^ +7x -14 = - x^3

now use additive inverse to get - 2x^2 to other side of equation

7x -14 = -x^3 +2x^2

7(x-2) = x^2 ( 2 -x) now divide both sides by x^2(2 -x)

7(x-2) / x^2( 2 - x) = 1
that will give you.

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

You can also divide going the other way:

7(x -2) = x^2( 2 -x)

1 = x^2(2 -x) / 7 ( x -2)
which gives you.

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

Fractions are not good when factoring or simplifying an equation so try it this way:

7(x - 2) = x^2(2 - x)
use associative law to get x^2(2 -x) inside paranthesies

7(x - 2) = (x^2(2 - x))
now use additive inverse:

7(x -2) -(x^2(2 - x)) = (x^2( 2 - x)) - (x^2(2 - x))
7(x -2) -(x^2(2 -x)) = 0
now clear as many parathesis as possible:

Remember when you have some thing in paranthesis like this : (x +y) that a 1 is understood to be in front of it like this:
1*(x+y)

7(x-2) + x^2( -(2 - x)) =0
7 (x -2) + x^2(-2 +x) =0
7(x- 2) + x^2 (x-2) = 0
now factor out x - 2

(x -2) (7 +x^2) = 0 this is your answer:

now check your work to make sure: :
x - 2
7 +x^2
--------
7x -14
...........x^3 - 2x^
7x - 14 +x^3 - 2x^2
rearrange

x^3 - 2x^2 - +7x - 14 =0 came out same must be right.

Answer 2

Hai love alwayz,

x^3 – 2x^2 + 7x – 14 = x^2(x-2) + 7(x-2) = (x^2+7)(x-2)