1. ## Help with factorizing.

Hi, this is probably easy as hell for you guys but I'm just not the kind of guy who's supposed to be doing this ****, but i would really appreciate it if someone would help me with these because otherwise I'll get an F in the report card, this is higher level maths we're talking about, i'll change to lower level but need to do these first.

factorize (if possible) as far as possible.

a) 4a^2 - 2b^2/32

b) 4x^2 + 12x^2 - 9

c) 4x^3 - 3x^2 + 8x - 6

2. Originally Posted by richardd
Hi, this is probably easy as hell for you guys but I'm just not the kind of guy who's supposed to be doing this ****, but i would really appreciate it if someone would help me with these because otherwise I'll get an F in the report card, this is higher level maths we're talking about, i'll change to lower level but need to do these first.

factorize (if possible) as far as possible.

a) 4a^2 - 2b^2/32

b) 4x^2 + 12x^2 - 9

c) 4x^3 - 3x^2 + 8x - 6
$\displaystyle a)\ 4a^2-\frac{2b^2}{32}$

$\displaystyle = 4a^2-\frac{b^2}{16}$

$\displaystyle = (2a)^2-(\frac{b}{4})^2$

$\displaystyle = (2a+\frac{b}{4})(2a-\frac{b}{4})$

3. Originally Posted by richardd
Hi, this is probably easy as hell for you guys but I'm just not the kind of guy who's supposed to be doing this ****, but i would really appreciate it if someone would help me with these because otherwise I'll get an F in the report card, this is higher level maths we're talking about, i'll change to lower level but need to do these first.

factorize (if possible) as far as possible.

a) 4a^2 - 2b^2/32

b) 4x^2 + 12x^2 - 9

c) 4x^3 - 3x^2 + 8x - 6
$\displaystyle 4a^2 - \frac{2b^2}{32}$

$\displaystyle 4a^2 - \frac{b^2}{16} = \left(2a-\frac{b}{4}\right)\left(2a+\frac{b}{4} \right)$

from $\displaystyle (x^2-y^2) = (x-y)(x+y)$

$\displaystyle 4x^2+12x^2-9$

$\displaystyle 16x^2-9 = (4x-3)(4x+3)$

$\displaystyle 4x^3-3x^2+8x-6$

$\displaystyle 4x^3+8x -3x^2 -6$

$\displaystyle 4x(x^2+2) -3(x^2+2) = (x^2+2)(4x-3)$

Hi, this is probably easy as hell for you guys
hell is not easy for all

4. Originally Posted by richardd
a) 4a^2 - 2b^2/32

b) 4x^2 + 12x^2 - 9

c) 4x^3 - 3x^2 + 8x - 6
a)
$\displaystyle 4a^2 - \frac{2b^2}{32}$
= $\displaystyle (2a)^2 - \frac{b^2}{16}$
= $\displaystyle (2a)^2 - (\frac{b}{4})^2$
= $\displaystyle (2a - \frac{b}{4})(2a + \frac{b}{4})$

b)
$\displaystyle 4x^2 + 12x^2 - 9$
= $\displaystyle 16x^2 - 3^2$
= $\displaystyle (4x)^2 - 3^2$
= (4x - 3)(4x + 3)

c)
$\displaystyle 4x^3 - 3x^2 + 8x - 6$
= $\displaystyle 4x^3 + 8x - 3x^2 - 6$
= $\displaystyle (4x)(x^2 + 2) - (3)(x^2 + 2)$
= $\displaystyle (4x - 3)(x^2 + 2)$

...a piece of cake.