can you guys please help me with the following questions??!!!

1. Factorise:

a. $\displaystyle 8x^2 - 2x - 15$

b. $\displaystyle 2y^3 + y^2 - 2y - 1$

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- Dec 26th 2008, 10:48 PMsudha_14factorize
can you guys please help me with the following questions??!!!

1. Factorise:

a. $\displaystyle 8x^2 - 2x - 15$

b. $\displaystyle 2y^3 + y^2 - 2y - 1$ - Dec 26th 2008, 11:33 PMnzmathman
By inspection,

$\displaystyle 8x^2 - 2x - 15 = (2x - 3)(4x + 5) $

You could have used the quadratic formula here to find the 2 roots and thus factorise the equation.

$\displaystyle \text{Let} \,\,f(y) = 2y^3 + y^2 - 2y - 1$

Using the remainder theorem, we can see $\displaystyle f(1) = 2 + 1 - 2 - 1 = 0 $

And therefore $\displaystyle y = 1$ is a root, therefore $\displaystyle (y - 1)$ is a factor of the polynomial.

Factorising: $\displaystyle 2y^3 + y^2 - 2y - 1 = (y - 1)(2y^2 + 3y + 1) = (y - 1)(2y + 1)(y + 1)$ - Dec 26th 2008, 11:48 PMsudha_14Quote:

Originally Posted by**nzmathman**

thanks in adv...(Happy) - Dec 27th 2008, 12:01 AMnzmathman
Hmmm I know how to do the long division but cannot present it here nicely for you to see...I am not that advanced in the math coding system. I'm just able to factorise such expressions straight away...

- Dec 27th 2008, 12:03 AMsudha_14
hmmm...sure .. no probz..ill try the long division again and if i still have a doubt..i'll tell u...

- Dec 27th 2008, 12:16 AMMathstud28
- Dec 27th 2008, 12:16 AMred_dog
b) Another approach, without the long division:

$\displaystyle 2y^3+y^2-2y-1=y^2(2y+1)-(2y+1)=(2y+1)(y^2-1)=(2y+1)(y-1)(y+1)$ - Dec 27th 2008, 12:26 AMsudha_14
thanks A LOT guys....i had completely forgotten about that method...

anyways...thanks again!!!!! alot!!