Can anyone please factorize x^4-x^3-x+1 ? ?

Question

Can anyone please factorize x^4-x^3-x+1 ? ?

Answer 1

x^4 - x^3 - x + 1

= (x^4 - x^3) - (x - 1)

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

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

Answer 2

Sure. You can either use the factor theorem, or you can eyeball it.

The factor theorem tells us that if we substitute x = a into the polynomial and we get 0, (x - a) will be a factor. We can see that, x = 1 makes the polynomial 0, so (x - 1) is a factor. Then we can use polynomial division to take the (x - 1) factor out. Or you can just see that:

x^3(x - 1) - (x - 1)

Now we can factorise out the (x - 1) easily:

(x - 1)(x^3 - 1)

(x^3 - 1) can easily be factorised using difference of two cubes:

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

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

A property of difference of two cubes is that the quadratic is unfactorable over real numbers, so that's fully factorised.

Answer 3

x⁴ - x³ - x + 1

= (x⁴ - x³) - (x - 1)

= x³(x - 1) - 1(x - 1)

= (x - 1)(x³ - 1)

= (x - 1)(x - 1)(x² + x + 1)

= (x - 1)²(x² + x + 1)

Answer 4

x^4 - x^3 - x  = -1

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

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

......No sorry

Answer 5

by inspectiion we know

f(1) = 1-1-1+1 - 0

so x-1 is a factor

x^4-x^3-x+1

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

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

Answer 6

factor by grouping...

rewrite (I put some parentheses in)...

(x^4 - x^3) - (x - 1)

pull out x^3 from the first two terms...

x^3(x - 1) - (x - 1)

factor out the common (x - 1) term...

(x - 1)(x^3 - 1)

your teacher may accept that. But just in case, you can take it a little further.

(x^3 - 1) factors into (x - 1)(x^2 + x +1) ← (that's memorized, difference of cubes)

so...

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

(x - 1)² (x² + x + 1) ← final answer

Answer 7

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

                  = (x-1)(x^3 -1)

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