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akotoh · Aug 2, 2009, 07:44 AM

Please.. Anyone.. Help me with my problem.. I tried my very best to answer this problems and yet, I end up with nothing. I've answered already 75% of my problem.. And it's the only problem I can't solve.. Please help. :((

Here it is: factoring..
1.) 4a^4 + 4a^3 - 8a^2 + a + 2
2.) x^2 - 2xy + y^2 - 12x + 12y + 36
3.) x^4 + 8x^2 + 144
4.) 49x^4 + 90x^2 + 121
5.) x^2 y^2 - 4x^2 y - 9y^2 + 36y - 36

anyone's help will be much appreciated. Thanks.

galactus · Aug 2, 2009, 08:32 AM

Originally Posted by akotoh

Please.. Anyone.. Help me with my problem.. I tried my very best to answer this problems and yet, i end up with nothing. I've answered already 75% of my problem.. And it's the only problem i can't solve.. Please help. :((

1.) $4a^{4} + 4a^{3} - 8a^{2} + a + 2$

By breaking up $4a^{3}=8a^{3}-4a^{3}$

Rewrite as $(4a^{4}-4a^{3}+a)+(8a^{3}-8a^{2}+2)$

factor: $a(4a^{3}-4a^{2}+1)+2(4a^{3}-4a^{2}+1)$

See? What's inside the parentheses is the same.

$(a+2)(4a^{3}-4a^{2}+1)$

Unknown008 · Aug 2, 2009, 08:57 AM

$1.) 4a^4 + 4a^3 - 8a^2 + a + 2$

You can also use the factor theorem here.

Let f(a) = 4a^4 + 4a^3 - 8a^2 + a + 2

Then, look for a factor of two, which will make f(a)=0. You'll find that -2 does the job.

Therefore, putting a=-2 will give f(a)=0. The factor is thus (a+2). [since a+2=0 from a=-2.]

Then, you have to do long division, to give you $(a+2)(4a^33-4a^2+1)$

galactus · Aug 2, 2009, 09:24 AM

5.) $x^{2} y^{2} - 4x^{2} y - 9y^{2} + 36y - 36$

Factor out an $x^{2}y$ and a 9

$x^{2}y(y-4)-9(y^{2}-4y+4)$

But $(y-2)^{2}=y^{2}-4y+4$

We get $x^{2}(y-4)-9(y-2)^{2}$