1. ## division

Hi all - me again,

I have a math exam in the morning. Lucky for me, I have some past papers, and only the numbers, not the questions, change each year.

I was wondering if someone could help me with the solution I have here:

1
a) Fully factorize:

3x $^3$-7x $^2$-3x+7

The solution which I have been given is to first of all do this:

1 I...3...............-7............-3..........7
....I
....I_________3_____-4_____7_____
.......3............... -4............-7..........0

Which then goes onto this:

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

so (x-1)(3x-7)(x+1)

Can someone please explain all of the above to me? I have no idea what the grid thing does, how the 4 gets in there? Basically, help? Please?

Cheers,

Martin

2. Originally Posted by Mathy
Hi all - me again,

I have a math exam in the morning. Lucky for me, I have some past papers, and only the numbers, not the questions, change each year.

I was wondering if someone could help me with the solution I have here:

1
a) Fully factorize:

3x $^3$-7x $^2$-3x+7

The solution which I have been given is to first of all do this:

1 I...3...............-7............-3..........7
....I
....I_________3_____-4_____7_____
.......3............... -4............-7..........0

Which then goes onto this:

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

so (x-1)(3x-7)(x+1)

Can someone please explain all of the above to me? I have no idea what the grid thing does, how the 4 gets in there? Basically, help? Please?

Cheers,

Martin
i don't see what you are doing here. it seems you were trying some sort of long division. that was completely unnecessary. this is a "nice" polynomial. it factors easily by grouping

$3x^3 - 7x^2 - 3x + 7$

factor out $x^2$ from the first two terms and -1 from the last two, we get:

$x^2 (3x - 7) - (3x - 7)$

now the 3x - 7 is common, so pull that out

$(3x - 7)(x^2 - 1)$

the last part is the difference of two squares, so we have:

$(3x - 7)(x + 1)(x - 1)$