Factoring, still lost: x^4-x^3-7x^2+x+6?

To come out cleanly, it will have to factor into something along the lines of ...

(x + a)(x + b)(x + c)(x + d)

... and abcd = 6

6 doesn't have too many factors, and at least two of them in this case are likely to equal 1 or -1.

Therefore I'd try dividing the expression by x + 1 and seeing what washes up.

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<later that day - lol>

Firstly, divide x^4 - x^3 - 7x^2 + x + 6 by x + 1:

= x^3 - 2x^2 - 5x + 6

Divide this by x - 1:

= x^2 - x - 6

... and this equals (x + 2)(x - 3)

Therefore x^4 - x^3 - 7x^2 + x + 6 = (x + 1)(x - 1)(x + 2)(x - 3)

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First group this long equation to two groups because only in grouping can we apply existing factoring rules.

(x^4 - x^3) + (-7x^2 + x + 6) <<Take note here... I used addition "+" to divide the two groups. And the negative sign "-" is placed in the 2nd group. negative 7x squared "-7x^2">>

For the first group (x^4 - x^3) factor out x^3. There exist a factoring rule for this. After factoring it will look like this...

x^3 (x - 1)

For the second group (-7x^2 + x + 6) factor it in accordance to the rules found in "Factoring Trinomials." as the second group is actually a TRInomial or having 3 terms. After factoring it will look like this...

(7x + 6) (-x + 1)

Combine the two groups...

x^3 (x-1) + (7x + 6) (-x + 1)

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I think this is the final answer...

Please verify...

God bless!

I........................................................

have NO idea