How do you factor $x^5 -4x^4 + 8x^3 - 14x^2 +15x -6$ ?

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Answer 1 · 2015-11-12T07:38:05

Find factor $(x-1)$ by examining the sum of the coefficients, divide and repeat, then factor by grouping to find:

$x^5-4x^4+8x^3-14x^2-14x^2+15x-6$

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

$f(x) = x^5-4x^4+8x^3-14x^2-14x^2+15x-6$

First notice that $f(1) = 0$ , since the sum of the coefficients is $0$ .

So $(x-1)$ is a factor of $f(x)$ .

$f(x) = (x-1)(x^4-3x^3+5x^2-9x+6)$

Again, the remaining quartic factor is divisible by $(x-1)$ since its coefficients also add up to $0$ .

$x^4-3x^3+5x^2-9x+6 = (x-1)(x^3-2x^2+3x-6)$

The remaining cubic factor is factorizable by grouping:

$x^3-2x^2+3x-6 = (x^3-2x^2)+(3x-6) = x^2(x-2)+3(x-2)$

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

The remaining quadratic factor has no linear factors with Real coefficients since $x^2+3 >= 3 > 0$ for all $x in RR$

Putting this together we get:

$x^5-4x^4+8x^3-14x^2-14x^2+15x-6 = (x-1)(x-1)(x-2)(x^2+3)$

If you allow Complex coefficients then you can factor further using:

$x^2+3 = (x-sqrt(3)i)(x+sqrt(3)i)$

So:

$x^5-4x^4+8x^3-14x^2-14x^2+15x-6$

$= (x-1)(x-1)(x-2)(x-sqrt(3)i)(x+sqrt(3)i)$

How do you factor x^5 -4x^4 + 8x^3 - 14x^2 +15x -6 x^5 -4x^4 + 8x^3 - 14x^2 +15x -6 ?