oldblue

12-13-2009, 05:59 PM

Just looking for a little help for this problem. I'm sure I'm making more out of it than it is.

Factor the below by grouping.

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

Factor the below by grouping.

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

View Full Version : Factoring help

oldblue

12-13-2009, 05:59 PM

Just looking for a little help for this problem. I'm sure I'm making more out of it than it is.

Factor the below by grouping.

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

Factor the below by grouping.

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

chrisr

12-13-2009, 06:35 PM

You can write "nested" factors.

3 terms have a common value.

Grouping those and examining what you've got in brackets,

you find common terms a second time.

Think it through, what is your first common term, given that x is some number?

3 terms have a common value.

Grouping those and examining what you've got in brackets,

you find common terms a second time.

Think it through, what is your first common term, given that x is some number?

oldblue

12-13-2009, 06:39 PM

Other than x, 1?

chrisr

12-13-2009, 06:49 PM

Yes,

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]).

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]).

oldblue

12-13-2009, 06:53 PM

Thank you. I wasn't sure of myself, but I will practice more!

chrisr

12-13-2009, 07:01 PM

The factors are based on the fact that "zero multiplied by anything equals zero".

When we write quadratics, cubics etc as multiplications, using the values that cause it to be zero,

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.

When we write quadratics, cubics etc as multiplications, using the values that cause it to be zero,

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.

stapel

12-14-2009, 11:59 AM

Factor the below by grouping.

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

When you have four terms and nothing comes out of all of them, you should check for factoring pairs (http://www.purplemath.com/modules/simpfact2.htm). In your case, you have:

. . . . .(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:

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

When you have four terms and nothing comes out of all of them, you should check for factoring pairs (http://www.purplemath.com/modules/simpfact2.htm). In your case, you have:

. . . . .(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:

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