Addition and Subtraction of suitable terms

[Knuckler]

Hello, I am going through Schaum's College algebra and am on the chapter of factoring. I am having problems with this technique when dealing with trinoials.

In one example 36x^4 + 15x^2 + 4
9x^2 is both added and subtracted, yet it is not explained how 9x^2 is derived. I thought you were suppose to add and subtract 2 * (product of the square root of the two terms)

In another example
u^8 - 14u^4 + 25
4u^4 is added and subtracted. Again I have no idea how 4 is derived as it is neither a factor of 25 or 14. Can anyone help please?

 

[tkhunny]

No, you need to CREATE 2xsquare root of the two items).

In the first, you are looking for 24x^2, but you've already 15, You need only 9 more.

The second it a hair trickie due to the sign of the middle term. You need -10 and you've already -14. Only 4 to go.

 

[galactus]

The general form of the given expression is $\displaystyle a^{4}+a^{2}b^{2}+b^{4}$

With your problem, this means $\displaystyle a=\sqrt{6}x$ and $\displaystyle b=\sqrt{2}$

Because $\displaystyle (\sqrt{6}x)^{4}=36x^{4}$ and $\displaystyle (\sqrt{2})^{4}=4$

But, the middle term $\displaystyle a^{2}b^{2}=(\sqrt{6}x)^{2}(\sqrt{2})^{2}=12x^{2}$

Adding $\displaystyle a^{2}b^{2}=12x^{2}$ gives $\displaystyle 24x^{2}$.

We now have a perfect square trinomial. $\displaystyle a^{4}+2a^{2}b^{2}+b^{4}=(a^{2}+b^{2})^{2}$

But, the problem has a $\displaystyle 15x^{2}$, but we need a $\displaystyle 24x^{2}$.

This is where the $\displaystyle 9x^{2}$ comes in.

By adding the $\displaystyle 9x^{2}$, it becomes $\displaystyle 36x^{4}+24x^{2}+4=(6x^{2}+2)^{2}$

Subtract off the $\displaystyle 9x^{2}$ and we get $\displaystyle (6x^{2}+2)^{2}-9x^{2}$

Now, apply the difference of two squares factorization.

 

[Knuckler]

thank you all. So basically were just forcing the polynomial into a perfect square trinomial by getting 2 * (a * c) and replacing b with it, then taking the difference of the old term and new term and using that as to do the difference of 2 squares factorization?