Math Factorization Help (Easy)?

2x^3 - x^2 - 2x + 1
= x^2 (2x - 1) - 1(2x -1)
= (x^2 -1)(2x -1)
= (x + 1)(x -1) (2x -1)

f (1) = 2 - 1 - 2 + 1 = 0
Thus x - 1 is a factor
Use synthetic division to find other factors:-

1 | 2_____-1_____-2_____1
_ |_______2______1____ -1
_ | 2_____ 1_____ -1_____0

(x - 1)(2x² + x - 1) = 0

(x - 1)(2x - 1)(x + 1) = 0

a^2 - b^2 = (a + b)(a - b)

2x^3 - x^2 - 2x + 1
= (2x^3 - x^2) - (2x - 1)
= x^2(2x - 1) - 1(2x - 1)
= (2x - 1)(x^2 - 1)
= (2x - 1)(x^2 - 1^2)
= (2x - 1)(x + 1)(x - 1)
= (x + 1)(2x - 1)(x - 1)

take your constant term (one without an X in it) and find its factors
this one is 1 and its factors are +/- 1
yes you have to use positive AND negative
anyways you substitute the X in the equation with one or the other
what we are trying to do is trial and error until we arrive at a solution with 0 in it
so in this case we'll try both

so lets try (-1)

2(-1)^3 - (-1)^2 - 2(-1) +1
-2-1-2+1= -4

So this isnt the one we want because it comes out with an answer not equal to zero... let's try positive 1

2(1)^3 - (1)^2 - 2(1) +1
2-1-2+1=0

and there we have our answer to 0

and since +1 was our answer we use (x-1) as one of our factors thats where they got that factor

now you take the given co-efficients and you use them for Horner's method

so we found +1 to be our solution so we go

1......2...-1....-2......1
_______.2___1___-1_
........2....1....-1......0

so the first term (2) goes down on the bottom and then you move it over and the add the top and bottom together and move your answer over like we did with the 2 and add them together and the last number should come out as a 0
from there you have 2 1 -1
thats the new co efficients for you new trinome

2x^2 + 1x -1 =0
and you just solve from there i multiply the 2 with the -1
x^2 +1x -2 =0
factorise
(x-1) (x+2)
then you have to divide by 2 because you multiplied it by 2 before so...
(x-1/2) (x+1)
(x-1/2) can also be written (2x-1)
and all together giving you the three factors (x+1)(x-1)(2x-1)
hope this helps

the constant term, in this case +1, when divided by the coefficient of the highest term, in this case 2, is the product of the roots. You can guess that 1 or -1 or -1/2 or 1/2 is a root. So try one of them.

Easiest way to do it without the annoying long division is by grouping. Group the first two terms and the last two terms :

(2x³ - x²) - (2x - 1)

GCF of the first two terms is x² :

x² (2x - 1) - (2x - 1)

Now since we grouped you may have wondered why the last term changed from (2x + 1) to (2x -1). This is because of the minus on the outside being multiplied through.

Now we have (2x - 1) in common (remember there's an imaginary 1 in front of the (2x - 1) :

(2x - 1) (x² - 1)

(x² - 1) is a difference of two squares so that leaves us with :

(2x - 1) (x + 1) (x - 1)

2x³ - x² - 2x + 1 =

Group.
(2x³ - 2x) + (-x² + 1) =

Factor.
2x(x² - 1) - 1(x² - 1) =

Due to distributive properties, you can rewrite this as:
(2x - 1)(x² - 1) =

x² - 1 is the difference of two squares.
x² = x * x = (x)²
1 = 1 * 1 = (1)²
x² - 1 = (x)² - (1)²

Remember how to factor the difference of two squares.
a² - b² = (a - b)(a + b)

Given: x² - 1 = (x)² - (1)²
Means: a = x, b = 1

Apply the factored form.
(2x - 1)(x² - 1) =
(2x - 1)(x - 1)(x + 1) =

Rearrange.
(x + 1)(x - 1)(2x - 1)

using synthetic division and the rational root theorem.

dividing by 1 gives 2x^2 + x - 1.
so one factor is (x-1) and the other is 2x^2 +x -1
factoring 2x^2 +x -1 to (2x - 1)(x + 1) gives the other two.