- Thread starter oldblue
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the more practical way to factorise this is to find f(x)=0.

f(0)=1

f(1)=0

therefore (x-1) is a factor.

To find the second factor... (x-1)(ax[sup:rqdp27i3]2[/sup:rqdp27i3]+bx+c)=3x[sup:rqdp27i3]3[/sup:rqdp27i3]-3x[sup:rqdp27i3]2[/sup:rqdp27i3]-x+1

gives a=3, b=0 and c=-1.

(x-1)(3x[sup:rqdp27i3]2[/sup:rqdp27i3]-1) = 3x[sup:rqdp27i3]3[/sup:rqdp27i3]-3x[sup:rqdp27i3]2[/sup:rqdp27i3]-x+1.

As you have three "x" terms and two "3" terms, you can start grouping terms using the "3".

3(x[sup:rqdp27i3]3[/sup:rqdp27i3]-x[sup:rqdp27i3]2[/sup:rqdp27i3])-x+1 and you see that x[sup:rqdp27i3]2[/sup:rqdp27i3] can be taken out also.

Or, nesting.... x(3x[sup:rqdp27i3]2[/sup:rqdp27i3]-3x-1)+1 = x{3x(x-1)-1}+1

=3x[sup:rqdp27i3]2[/sup:rqdp27i3](x-1)-x+1=-3x[sup:rqdp27i3]2[/sup:rqdp27i3](1-x)+(1-x) = (1-x)(1-3x[sup:rqdp27i3]2[/sup:rqdp27i3]).

When we write quadratics, cubics etc

the answers are embedded in the factors themselves.

x=5 can be written x-5=0...

2(x-5)=0 then x=5 since 2 is not zero....

(x-2)(x-3)=0.... x is 2 or 3 and so on.

I should have added earlier that for ax[sup:3uhuggyh]2[/sup:3uhuggyh]+bx+c, both "a" and "c" are immediately obvious,

because x(ax[sup:3uhuggyh]2[/sup:3uhuggyh]) is 3x[sup:3uhuggyh]3[/sup:3uhuggyh] and (-1)c = 1.

"b" can be discovered in 2 ways by adding together the "x" parts or the "x[sup:3uhuggyh]2[/sup:3uhuggyh]" parts,

....x(bx)-3x[sup:3uhuggyh]2[/sup:3uhuggyh]=-3x[sup:3uhuggyh]2[/sup:3uhuggyh] so b=0, or -bx-x = -x so b=0.

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When you have four terms and nothing comes out of all of them, you should check foroldblue said:Factor the below by grouping.

3x^3-3x^2-x+1

. . . . .(3x[sup:3c3fn4x8]3[/sup:3c3fn4x8] - 3x[sup:3c3fn4x8]2[/sup:3c3fn4x8]) - (x - 1)

. . . . .3x[sup:3c3fn4x8]2[/sup:3c3fn4x8](x - 1) - 1(x - 1)

Take the common factor out front:

. . . . .(x - 1)(3x[sup:3c3fn4x8]2[/sup:3c3fn4x8] - 1)

Hope that helps! :wink: