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iruleqwerty1234 · Apr 17, 2009, 12:17 PM

What's 9k power of 2 minus 12k plus 4

iruleqwerty1234 · Apr 17, 2009, 12:19 PM

12a(2)b(2)-27a(2)

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Unknown008 · Apr 17, 2009, 08:11 PM

Ok, here's your question:

$9k^2 -12k +4$

Give the factors of +4 (this is like trial and error)
They are +1, +2, +4 but also -1, -2, -4.

Now, factors of 9, which are +1, +3, +9 (no need to find negative factors here)

Write them in a certain form, like for example this one;

$9k^2$ OOOOOO $4$
$k$ OOOOOOO $+1$ OOOOOO $+9k$
$9k$ OOOOOO $+4$ OOOOOO $+4k$ OOOOOO
OOOOOOOOOOOOOOOO $+13k$

You obtain +13k when you add the products of 9k and +1, and k and +4. You need to obtain -12k.

By trying different numbers, you'll see that the answer is from;

$9k^2$ OOOOOO $4$
$3k$ OOOOOO $-2$ OOOOOO $-6k$
$3k$ OOOOOO $-2$ OOOOOO $-6k$ OOOOOO
OOOOOOOOOOOOOOOO $-12k$

And you factors will be $(3k-2)(3k-2)$

I know this may be a little confusing, there are several methods, and that's the one I was taught. Post back if you don't understand.

Unknown008 · Apr 17, 2009, 08:23 PM

Originally Posted by iruleqwerty1234

12a(2)b(2)-27a(2)

IN THE BRACKETS ARE THE POWERS

Here's your question;

$12a^2b^2 - 27a^2$

That's more simple than the previous one;

what are the common factors in $12a^2b^2$ and $27a^2$ ?

You'll find them be $3a^2$ , so factorise by that term and put what remains into brackets;

$3a^2(4b^2 - 9)$

You'll find your original question if you expand. Now, it happens that the terms in brackets are perfect squares, 4, $b^2$ and 9. From your formula of $(a-b)(a+b) = a^2 - b^2$ , further factorise to give;

$3a^2(2b+3)(2b-3)$