# simplifying expressions

• Sep 29th 2007, 06:39 AM
raza88sab
simplifying expressions
...and if possible please show working out so i can understand better thanks

(a) (x+1)^3 -3x (x+1)

(b) y(2y+3) +2y^2 (y+1)(y-1)

(c) 14x^3 (3-2x) / x^2(x-1)(4x-6)
• Sep 29th 2007, 06:46 AM
topsquark
Quote:

Originally Posted by raza88sab
(a) (x+1)^3 -3x (x+1)

This is similar in concept to
$a^3 - ba = a(a^2 - b)$

Here $a = x + 1$ and $b = 3x$

So
$(x + 1)^3 - 3x(x + 1) = (x + 1) ( (x + 1)^2 - 3x)$

$= (x + 1)(x^2 + 2x + 1 - 3x) = (x + 1)(x^2 -x + 1)$

-Dan
• Sep 29th 2007, 06:48 AM
topsquark
Quote:

Originally Posted by raza88sab
(b) y(2y+3) +2y^2 (y+1)(y-1)

The only thing I can think of to do here is expand:
$y(2y + 3) + 2y^2(y + 1)(y - 1)$

$= 2y^2 + 3y + 2y^2(y^2 - 1)$

$= 2y^2 + 3y + 2y^4 - 2y^2$

$= 2y^4 + 3y$

$= y(2y^3 + 3)$

-Dan
• Sep 29th 2007, 06:52 AM
topsquark
Quote:

Originally Posted by raza88sab
(c) 14x^3 (3-2x) / x^2(x-1)(4x-6)

$\frac{14x^3(3 - 2x)}{x^2(x - 1)(4x - 6)}$

Note that $4x - 6 = 2(2x - 3) = -2(3 - 2x)$

So
$\frac{14x^3(3 - 2x)}{x^2(x - 1)(4x - 6)} = -\frac{14x^3(3 - 2x)}{2x^2(x - 1)(3 - 2x)}$

Now we may cancel the common $3 - 2x$ and the common 2x^2, provided we remember that $x \neq 0, \frac{3}{2}$.
$\frac{14x^3(3 - 2x)}{x^2(x - 1)(4x - 6)} = -\frac{14x^3(3 - 2x)}{2x^2(x - 1)(3 - 2x)} = -\frac{7x}{x - 1};~x \neq 0, \frac{3}{2}$

-Dan
• Sep 29th 2007, 06:58 AM
raza88sab
thanks
thanks dan mate....im just going to see if i can do this my self and lets see if it sticks in my head somehere lol.....much appriciated:)