1. ## Polynomial Division

I divided (-4x^3 + x^2 - 4)/(x - 1) and got a totally different answer on three attempts than the book.

-4x^2 - 3x - 3 remainder -7

How can this be?

Can you show me step by step why the above answer is correct?

2. Hello,
Originally Posted by magentarita
I divided (-4x^3 + x^2 - 4)/(x - 1) and got a totally different answer on three attempts than the book.

-4x^2 - 3x - 3 remainder -7

How can this be?

Can you show me step by step why the above answer is correct?
My method may not be like yours, but I can't manage to change it ~~

$-4x^3+x^2-4$ and x-1.

First, I see the first term of the polynomial -4x^3 and the first term of x-1 : x.
$-4x^3={\color{green}-4x^2}(x)$.

So I'll write :
$-4x^3+x^2-4={\color{green}-4x^2}(x-\underbrace{{\color{red}1}){\color{red}-4x^2}}_{\text{this is 0}} \quad +x^2-4=-4x^2(x-1)-3x^2-4$

And so on... :

$-4x^2(x-1)-3x^2-4=-4x^2(x-1)-3x(x-1)-3x-4=(x-1)(-4x^2-3x)-3x-4$

$(x-1)(-4x^2-3x)-3x-4=(x-1)(-4x^2-3x)-3(x-1)-3-4$ $=\boxed{(x-1)(-4x^2-3x-3)\textbf{-7}}$

If you want to know where your mistake(s) is/are, just show your work ^^

3. ## Perfect!

Originally Posted by Moo
Hello,

My method may not be like yours, but I can't manage to change it ~~

$-4x^3+x^2-4$ and x-1.

First, I see the first term of the polynomial -4x^3 and the first term of x-1 : x.
$-4x^3={\color{green}-4x^2}(x)$.

So I'll write :
$-4x^3+x^2-4={\color{green}-4x^2}(x-\underbrace{{\color{red}1}){\color{red}-4x^2}}_{\text{this is 0}} \quad +x^2-4=-4x^2(x-1)-3x^2-4$

And so on... :

$-4x^2(x-1)-3x^2-4=-4x^2(x-1)-3x(x-1)-3x-4=(x-1)(-4x^2-3x)-3x-4$

$(x-1)(-4x^2-3x)-3x-4=(x-1)(-4x^2-3x)-3(x-1)-3-4$ $=\boxed{(x-1)(-4x^2-3x-3)\textbf{-7}}$

If you want to know where your mistake(s) is/are, just show your work ^^
Perfectly done!