1. factorize

1. Factorise:
a. $\displaystyle 8x^2 - 2x - 15$
b. $\displaystyle 2y^3 + y^2 - 2y - 1$

2. 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: $\displaystyle 2y^3 + y^2 - 2y - 1 = (y - 1)(2y^2 + 3y + 1) = (y - 1)(2y + 1)(y + 1)$

3. Originally Posted by nzmathman
By inspection,

You could have used the quadratic formula here to find the 2 roots and thus factorise the equation.
thanks A LOT! for the solution of the first equation....but for the second one...can you please show the long division..cuz even i understood HOW to do it....but i go wrong in the procedure....ALWAYS!!!! so i kinda needed help with that....

4. 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...

5. hmmm...sure .. no probz..ill try the long division again and if i still have a doubt..i'll tell u...

6. Originally Posted by sudha_14

1. Factorise:
a. $\displaystyle 8x^2 - 2x - 15$
b. $\displaystyle 2y^3 + y^2 - 2y - 1$
There is no need for using polynomial division

\displaystyle \begin{aligned}2y^3+y^3-2y-1&=(2y^3+y^2)-(2y+1)\\ &=y^2(2y+1)-(2y+1)\\ &=(y^2-1)(2y+1)\\ &=(y-1)(y+1)(2y+1)\end{aligned}

7. b) Another approach, without the long division:
$\displaystyle 2y^3+y^2-2y-1=y^2(2y+1)-(2y+1)=(2y+1)(y^2-1)=(2y+1)(y-1)(y+1)$