How to factor this?

Question

 

-x^3+6x^2-11x+6=0

Answer 1

When asked to factor a cubic polynomial, it is always a good idea to first look for simple, integer roots, such as 0, 1, or -1. When you find one such root x₀, you can divide the polynomial by the factor (x-x₀) and are left with a quadratic polynomial.

 

In this case, you see that x₀=1 is a root since -1+6-11+6=0, so you can divide the polynomial by (x-1) through the process of long (or synthetic) division:

 

-x³+6x²-11x+6:x-1=-x²+5x-6

 

This resulting quadratic polynomial can be easily factored as -(x-3)(x-2), so the answer is

 

-x³+6x²-11x+6=-(x-3)(x-2)(x-1)=0.

Answer 2

Darren has used Gorner's schema known in algebra to factor polynomials of any degree" It says the polynomial (all coefficients are multiplued by -1)

 

    x³ -6x² +11x - 6 = (x-a)(x-b)(x-c)       (a,b,c are roots of the polynomial)      (1)

 

The expression on the right side can be rewritten as

 

   x³ - (a+b+c)x² = (ab = ac + bc) x - abc                                                    (2)

 

Both sides in  (1) will be equal if the coefficients for equal powers of x are equal. Thus we can write

 

a +b +c = 6

ab + bc + ac + -11

abc = 6

 

You can easily conclude tha  a + b +c = abc = 6 whish is true if a=1, b=2, c=3

*sum of the first tree consequitive numbers  equals to their product). Thjus your polynomial can be factored as

                                - (x-1)(x-2(x-3)

 

Robert has applied more rational way to factor the polynomial, but the Gorner's schema has more universality and is applicable to more complex polynomials.

Answer 3

Hey Sun: a "root test" ==> (+ or - factors of last term) / (+ or - factors of 1st term) ...

 

candidates are +/- (1,2,3) / (+/- 1) ... always try +/- 1 first, then try others +/- (2,3) ... 

x=1 works (12-12) ... you may factor out (x-1)(-x^2 +5x -6) ==> (x-1)(x-2)(3-x) ...

OR you may test the other candidate roots ... x= -1 nogo (23-1) ,,, x=2 works (30-30) ...

x= -2 nogo (52-8) ... x=3 works (60-60)  Best wishes, sir :)

 

 

 

Answer 4

(As noted below, my answer used to reflect a separate approach. I have modified it for clarity's sake.)

To make things simpler, you can look for where you might be able to fit in a squared polynomial, which basically operates like so:

 

x² ± bx + c

If the square root of c is equal to half of b, then you've got something that can be written like this:

 

(x ± m)²                                //where m is half of b, as described above

 

So. Let's do that!

 

-x³ + 6x² - 11x + 6 = 0          // your formula

-x³ + 6x² - 9x - 2x + 6 = 0    // We've turned -11x into -9x and -2x.

-x(x² + 6x - 9) - 2x + 6 = 0   // We've factored out x

-x(x + 3)² - 2x + 6 = 0          // Now we see if we can find (x - 3) again.

-x(x + 3)² - 2(x + 3) = 0       // We can! Now we'll factor that out.

(x + 3)(-x(x + 3) - 2) = 0      //

(x + 3)(-x² - 3x - 2) = 0        // Let's get that negative out of there too.

-(x + 3)(x² + 3x + 2) = 0      // We can factor the rest out by sight.

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

 

And you can take it from there!

Answer 5

Regrouping,

 

-x^3+6x^2-11x+6

= -x(x^2-6x+9) - 2x+6

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

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

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