algebra troubles :( (1 Viewer)

coyazayo · Jan 18, 2016

2 Qs that i need help with. Thank you in advance!

Q1. Factorise:x^2+4y^2+x+2y-5xy-2

Q2. Factorise
ab(a-b)+bc(b-c)+ca(c-a)

leehuan · Jan 18, 2016

Green Yoda · Jan 18, 2016

leehuan said:
$ab(a-b)+bc(b-c)+ca(c-a)\\ ={ a }^{ 2 }b-a{ b }^{ 2 }+{ b }^{ 2 }c-bc^{ 2 }+abc+{ a }^{ 2 }c-abc-a{ c }^{ 2 }\\ =\left( { a }^{ 2 }b-a{ b }^{ 2 }+{ a }^{ 2 }c-abc \right) +\left( { b }^{ 2 }c-bc^{ 2 }+abc-a{ c }^{ 2 } \right) \\ =a\left( ab-{ b }^{ 2 }+ac-bc \right) -c\left( { bc-{ b }^{ 2 }+ac-ab } \right) \\ =a\left( a-b \right) \left( b-c \right) -c\left( c-b \right) \left( b-a \right) \\ =a(a-b)(b-c)-c(b-c)(a-b)\\ =(a-c)(a-b)(b-c)$

in your second line where did you get "abc" from?

InteGrand · Jan 18, 2016

Rathin said:
in your second line where did you get "abc" from?

The classic "add and subtract".

InteGrand · Jan 18, 2016

coyazayo said:
2 Qs that i need help with. Thank you in advance! Q1. Factorise:x^2+4y^2+x+2y-5xy-2

Q2. Factorise
ab(a-b)+bc(b-c)+ca(c-a)

$\noindent For Question 1, we start off doing it as though it's a usual quadratic in $x$, so treat $y$ as a constant. Calling the expression $E$, we have$

$\begin{align*}E\equiv x^2+4y^2+x+2y-5xy-2 &= x^2 +\left(1-5y\right)x + \left(4y^2 + 2y-2 \right).\end{align*}$

$\noindent First we'll factorise the ``constant term'' here (the quadratic in $y$). We have$

$\begin{align*}4y^2 + 2y-2 &= \frac{\left(4y-2 \right)\left(4y+4 \right)}{4} \\ &= \left(4y-2 \right)\left(y+1 \right). \end{align*}$

$\noindent Thus$

$\begin{align*}E&= x^2 +\left(1-5y\right)x + \left(4y-2 \right)\left(y+1 \right).\end{align*}$

$\noindent Now, let's factorise this as a quadratic in $x$, in the form $(x+\alpha)(x+\beta)$. We want to find expressions $\alpha,\beta$ such that $\alpha \beta = \left(4y-2 \right)\left(y+1 \right)$ and $\alpha + \beta = 1-5y$. Clearly we take $\alpha = -\left(4y-2 \right)$ and $\beta = -\left(y+1 \right)$ (check that this works). Then our final answer will be$

$\begin{align*}E&= \left(x- \left(4y-2 \right)\right)\left(x- \left(y+1 \right)\right) \\ &=\left(x-4y+2\right)\left(x- y-1\right). \end{align*}$

Green Yoda · Jan 18, 2016

InteGrand said:
The classic "add and subtract".

I still dont seem to understand what is happening.. :/

InteGrand · Jan 18, 2016

Rathin said:
I still dont seem to understand what is happening.. :/

$\noindent He basically expanded it normally and added and subtracted the term $abc$ in order to allow it to be factorised nicely. It is valid to add and subtract $abc$ like this because it is equivalent to adding $0$.$

leehuan · Jan 18, 2016

Basically this

InteGrand said:
$\noindent He basically expanded it normally and added and subtracted the term $abc$ in order to allow it to be factorised nicely. It is valid to add and subtract $abc$ like this because it is equivalent to adding $0$.$

What does +abc -abc equal to?

It's like saying decompose x/(x+1)

+1 -1:
(x+1-1)/(x+1)

Seperate the fractions
(x+1)/(x+1) - 1/(x+1)

= 1 - 1/(x+1)

leehuan · Jan 18, 2016

InteGrand said:
$\noindent For Question 1, we start off doing it as though it's a usual quadratic in $x$, so treat $y$ as a constant. Calling the expression $E$, we have$

$\begin{align*}E\equiv x^2+4y^2+x+2y-5xy-2 &= x^2 +\left(1-5y\right)x + \left(4y^2 + 2y-2 \right).\end{align*}$

$\noindent First we'll factorise the ``constant term'' here (the quadratic in $y$). We have$

$\begin{align*}4y^2 + 2y-2 &= \frac{\left(4y-2 \right)\left(4y+4 \right)}{4} \\ &= \left(4y-2 \right)\left(y+1 \right). \end{align*}$

$\noindent Thus$

$\begin{align*}E&= x^2 +\left(1-5y\right)x + \left(4y-2 \right)\left(y+1 \right).\end{align*}$

$\noindent Now, let's factorise this as a quadratic in $x$, in the form $(x+\alpha)(x+\beta)$. We want to find expressions $\alpha,\beta$ such that $\alpha \beta = \left(4y-2 \right)\left(y+1 \right)$ and $\alpha + \beta = 1-5y$. Clearly we take $\alpha = -\left(4y-2 \right)$ and $\beta = -\left(y+1 \right)$ (check that this works). Then our final answer will be$

$\begin{align*}E&= \left(x- \left(4y-2 \right)\right)\left(x- \left(y+1 \right)\right) \\ &=\left(x-4y+2\right)\left(x- y-1\right). \end{align*}$

This made no sense on WolframAlpha until you explained it lol

Awkward method though.

algebra troubles :( (1 Viewer)

coyazayo

Member

leehuan

Well-Known Member

Green Yoda

Hi Φ

InteGrand

Well-Known Member

InteGrand

Well-Known Member

Green Yoda

Hi Φ

InteGrand

Well-Known Member

leehuan

Well-Known Member

leehuan

Well-Known Member

Users Who Are Viewing This Thread (Users: 0, Guests: 1)