Need urgent help on factorising!

**holmesb** · June 7th 2007, 01:00 AM

Just a few questions I am having trouble with here before my exam:
Factorize
(1) p^6-64
(2) 25e^2-30ef-16f^2
(3) 2a^2-2b^2+5a+2am-5b-2bm

Thanks for the help.

**tukeywilliams** · June 7th 2007, 01:35 AM

1. $p^{6} - 64 = (p-2)(p+2)(p^{2}+2p+4)(p^{2}-2p+4)$ .

Try the rest. For the first one I used the rational roots theorem and synthetic division to find the roots. Then used polynomial long division.

So: $\frac{\pm 1,2,4,8,16,32,64}{\pm 1}$ are canidates for possible rational roots.

2. For this one, it is just like factoring a regular quadratic equation like $x^{2} + 2x + 1$ .

3. I am not sure that this can be factored.

**topsquark** · June 7th 2007, 02:21 AM

Originally Posted by holmesb

(1) p^6-64

We can be a bit more analytic than using the rational roots theorem. There are two ways of looking at this one:
First:
$p^4 - 64 = (p^3)^2 - 8^2$

$= (p^3 +8)(p^3 - 8)$

$= (p^3 + 2^3)(p^3 - 2^3)$

Now,
$a^3 \pm b^3 = (a \pm b)(a^2 \mp ab + b^2)$

So
$p^3 + 2^3 = (p + 2)(p^2 - 2p + 4)$
$p^3 - 2^3 = (p - 2)(p^2 + 2p + 4)$

Thus
$p^4 - 64 = (p + 2)(p^2 - 2p + 4)(p - 2)(p^2 + 2p + 4)$

Second:
$p^6 - 64 = (p^2)^3 - 4^3$

$= (p^2 - 4)(p^4 + 4p^2 + 16)$

$= (p + 2)(p - 2)(p^4 + 4p^2 + 16)$

Now, $p^4 + 4p^2 + 16$ is not particularly easy to factor, but note that:
$p^4 + 4p^2 + 16 = (p^4 + 4p^2 + 16) + (4p^2 - 4p^2)$ (Just adding 0 to the expression)

$= (p^4 + 8p^2 + 16) - 4p^2$

$= (p^2 + 4)^2 - (2p)^2$ <-- The difference of two squares

$= ([p^2 + 4] + 2p)([p^2 + 4) - 2p)$

$= (p^2 + 2p + 4)(p^2 - 2p + 4)$

So again we get:
$p^6 - 64 = (p + 2)(p - 2)(p^2 + 2p + 4)(p^2 - 2p + 4)$

-Dan

**topsquark** · June 7th 2007, 02:35 AM

Originally Posted by holmesb

(3) 2a^2-2b^2+5a+2am-5b-2bm

We can try a variety of things here:
One way is:
$2a^2 - 2b^2 + 5a + 2am - 5b - 2bm$

$= (2a^2 + 5a) - (2b^2 + 5b) + 2am - 2bm$

$= 2 \left ( a^2 + \frac{5}{2}a \right ) - 2 \left ( b^2 + \frac{5}{2}b \right ) + 2am - 2bm$

Now complete the square on the first two terms. We have to add $\left ( \frac{5}{4} \right ) ^2 = \frac{25}{16}$ to each term so:
$= 2 \left ( a^2 + \frac{5}{2}a + \frac{25}{16} \right ) - 2 \left ( b^2 + \frac{5}{2}b + \frac{25}{16} \right ) + 2am - 2bm - 2 \cdot 2 \cdot \frac{25}{16}$
(The second factor of 2 comes from the fact that each of the first two factors is multiplied by 2. If you doubt me, expand it all out. You'll see that I'm just adding a total of 0 to the expression.)

$= \left ( a + \frac{5}{4} \right ) ^2 - \left ( b + \frac{5}{4} \right ) ^2 + 2am - 2bm - \frac{25}{4}$

$= \left ( a + \frac{5}{4} \right ) ^2 - \left ( b + \frac{5}{4} \right ) ^2 + 2m(a - b) - \frac{25}{4}$

Another way is:
$2a^2 - 2b^2 + 5a + 2am - 5b - 2bm$

$(2a^2 + 2am) - (2b^2 + 2bm) + 5a - 5b$

$2(a^2 + am) - 2(b^2 + bm) + 5a - 5b$ <-- Again complete the square:

$= 2 \left ( a^2 + am + \frac{m^2}{4} \right ) - 2 \left ( b + bm + \frac{m^2}{4} \right ) + 5a - 5b - 2 \cdot 2 \cdot \frac{m^2}{4}$

$= 2 \left ( a + \frac{m}{2} \right ) ^2 - 2 \left ( b + \frac{m}{2} \right ) ^2 + 5(a - b) - m^2$

-Dan

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