Factorise a six-termed quadratic in ##a## and ##x##

[brotherbobby]

Homework Statement

$\text{Resolve into two or more factors}$ : $\quad\boldsymbol{a(a-1)x^2+(2a^2-1)x+a(a+1)}$

Relevant Equations

Formulae : We have (1) $a^2-b^2=(a-b)(a+b)$ and (2) $ax\pm bx=x(a\pm b)$

Statement : I copy and paste the problem as it appeared in the text.


Attempt : I confess I couldn't go much far at all. Here's my attempt below in $\text{Autodesk Sketchbook}^{\circledR}$. The underlined , wavy underlined and box brackets below are my attempts to see what terms can be grouped.

No solutions exist but the answer does at the back of the text. I don't want to see it right now.

Request : A hint as to how to go about factorising it.

 

[WWGD]

Is a a constant or a variable?

 

[Mark44]

Is a a constant or a variable?

Can't say for sure, but I don't think it matters.

 

[brotherbobby]

Is a a constant or a variable?

Let's assume it's a variable.

 

[FactChecker]

In the variable $x$, it is a second order polynomial. Do you know how to use the quadratic formula? You can use it to factor it into two first degree polynomials in $x$.
Alternatively, if you multiply it out and collect powers of $a$, it is also a second order polynomial in $a$. So you could do the same thing with $a$ as the variable.
I don't know if there is a reason to prefer one option over the other, but I tend to think of $x$ as the main variable and $a$ as a constant parameter.

 

[brotherbobby]

I don't know if there is a reason to prefer one option over the other, but I tend to think of x as the main variable and a as a constant parameter.

Your method works, though I'd still be on the hunt for something cleaner, without having to use something that "lies ahead" as it were; in this case, quadratic equations. Let me start with the problem statement.

Problem statement :


Attempt : Let us have $\small{a(a-1)x^2+(2a^2-1)x+a(a+1)=0\Rightarrow x=\dfrac{1-2a^2\pm\sqrt{(2a^2-1)^2-4a^2(a^2-1)}}{2a(a-1)}}$ upon using the (well-known) quadratic formula.

The term under the square roots comes to be just 1.

Hence, $x=\dfrac{1-2a^2\pm 1}{2a(a-1)}\Rightarrow x=-\dfrac{1+a}{a}\qquad (1)$,
upon taking the + sign and simplifying.

Taking the negative (-) sign and simplifying, $x=-\dfrac{a}{a-1}\qquad (2)$

From (1) and (2) above, $\quad ax+a+1=0\quad\text{and}\quad ax-x+a=0$

Multiplying, we obtain the factors : $\quad\large{\boxed{(ax+a+1)(ax-x+a)=0}}\quad\Large{\color{green}{\checkmark}}$

This agrees with the answer in the text.


Thank you @FactChecker, but I wonder if there's a more straightforward method, reminiscent of factorising expressions from school algebra where one didn't have to put the expression to 0 to solve for $x$.

 

[Hill]

I wonder if there's a more straightforward method, reminiscent of factorising expressions from school algebra

Like this:
$a(a-1)x^2+(2a^2-1)x+a(a+1)$
$=a(a-1)(x^2+\frac {2a^2-1} {a(a-1)}x+\frac {a(a+1)}{a(a-1)})$
$=a(a-1)(x^2+\frac {a^2+a^2-1} {a(a-1)}x+\frac {a(a+1)}{a(a-1)})$
$=a(a-1)(x^2+\frac {a^2} {a(a-1)}x+\frac {a^2-1} {a(a-1)}x+\frac {a(a+1)}{a(a-1)})$
$=a(a-1)(x^2+\frac {a} {a-1}x+\frac {a+1} {a}x+\frac {a(a+1)}{a(a-1)})$
$=a(a-1)((x+\frac {a} {a-1})x+(x+\frac {a} {a-1})\frac {a+1}a)$
$=a(a-1)(x+\frac {a} {a-1})(x+\frac {a+1}a)$
$=((a-1)x+a)(ax+(a+1))$
?

 

[SammyS]

Homework Statement: $\text{Resolve into two or more factors}$ : $\quad\boldsymbol{a(a-1)x^2+(2a^2-1)x+a(a+1)}$
Relevant Equations: Formulae : We have (1) $a^2-b^2=(a-b)(a+b)$ and (2) $ax\pm bx=x(a\pm b)$

View attachment 343118Statement : I copy and paste the problem as it appeared in the text.

Attempt : I confess I couldn't go much far at all. Here's my attempt below in $\text{Autodesk Sketchbook}^{\circledR}$. The underlined , wavy underlined and box brackets below are my attempts to see what terms can be grouped.

Yes, it can be factored by grouping.

Notice that $\displaystyle \ 2a^2-1 = a^2 + (a^2-1) =a^2 + (a-1)(a+1) \ .$

Group as follows:

$\displaystyle \quad a(a-1)x^2+(2a^2-1)x+a(a+1)=a(a-1)x^2+a^2\,x + (a-1)(a+1)x+a(a+1)$

$\displaystyle \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad =((a-1)x+a)\,ax + ((a-1)\,x+a)(a+1)$

etc.

 

[FactChecker]

Your method works, though I'd still be on the hunt for something cleaner, without having to use something that "lies ahead" as it were; in this case, quadratic equations.

People who have put no information in their profile often post questions where it is impossible to know what their background is and what material they have already covered versus "lies ahead". As far as "cleaner", see my later comment.

View attachment 343123This agrees with the answer in the text.

Good.

Thank you @FactChecker, but I wonder if there's a more straightforward method,

I know that you didn't mean to, but that hits a nerve with me.
For "clean" and "straightforward", it is hard to beat this:
1) Notice that it's second order
2) Use the quadratic equation.
3) Do routine math to get the answer.
We can see in two seconds that it can be done and how to do it.

reminiscent of factorising expressions from school algebra where one didn't have to put the expression to 0 to solve for $x$.

There is a very close connection between the factors of an expression and solving for the zeros.
It's possible to learn a million tricks from school algebra and underestimate the basics.

 

[SammyS]

Homework Statement: $\text{Resolve into two or more factors}$ : $\quad a(a-1)x^2+(2a^2-1)x+a(a+1)$

This can also be factored as a quadratic in $a$ .

Using your $\displaystyle \text{Autodesk Sketchbook}^{\circledR}$ image we get

$\quad a(a-1)x^2+(2a^2-1)x+a(a+1)$

$\displaystyle \quad =(x^2+2x+1)a^2-(x^2-1)a-x$

$\displaystyle \quad =(x+1)^2 a^2+(1-x)(x+1)a-x$

$\displaystyle \quad =((x+1)a+1)((x+1)a-x)$

 

[WWGD]

This can also be factored as a quadratic in $a$ .

Using your $\displaystyle \text{Autodesk Sketchbook}^{\circledR}$ image we get

$\quad a(a-1)x^2+(2a^2-1)x+a(a+1)$

$\displaystyle \quad =(x^2+2x+1)a^2-(x^2-1)a-x$

$\displaystyle \quad =(x+1)^2 a^2+(1-x)(x+1)a-x$

$\displaystyle \quad =((x+1)a+1)((x+1)a-x)$

That's why I had asked whether a was a constant. It wouldn't make much sense to solve for it if it was.

 

[brotherbobby]

This can also be factored as a quadratic in $a$ .

Using your $\displaystyle \text{Autodesk Sketchbook}^{\circledR}$ image we get

$\quad a(a-1)x^2+(2a^2-1)x+a(a+1)$

$\displaystyle \quad =(x^2+2x+1)a^2-(x^2-1)a-x$

$\displaystyle \quad =(x+1)^2 a^2+(1-x)(x+1)a-x$

$\displaystyle \quad =((x+1)a+1)((x+1)a-x)$

Can you tell me @SammyS how you factored the second last step to go to the last one?

Let me copy and paste those two steps for you as an image below.

Many thanks.

 

[MatinSAR]

@brotherbobby I’m not sure if anyone has solved it this way before.
We initially divide by $a(a-1)$, and ultimately, we multiply by the same factor, $a(a-1)$.
$$x^2 + \dfrac {a^2+(a-1)(a+1)}{a(a-1)}x+\dfrac {a+1}{a-1}=0$$ $$x^2 + (\dfrac {a}{a-1}+\dfrac {a+1}{a})x+\dfrac {a+1}{a-1}=0$$ $$(x+\dfrac {a}{a-1})(x+\dfrac {a+1}{a})=0 $$ Finally: $$ (xa-x+a)(xa+a+1)=0 $$

 

[Hill]

factored the second last step to go to the last one?

I'd do it so:
$=(x+1)^2 a^2+(1-x)(x+1)a-x$
$=(x+1)^2 a^2+(x+1)a-x(x+1)a-x$
$=((x+1)a+1)(x+1)a-((x+1)a+1)x$
$=((x+1)a+1)((x+1)a-x)$

BTW, here is a nice factorizing puzzle for practice:
Find two 4-digit numbers that multiply to give $4^8 + 6^8 + 9^8$.

 

[MatinSAR]

Can you tell me @SammyS how you factored the second last step to go to the last one?

Another way is to use identity of a common term: $t^2+(a+b)t+ab=(t+a)(t+b)$
$$t \to (x+1)a$$ $$a \to 1 $$ $$ b \to -x$$ $$(x+1)^2 a^2+(1-x)(x+1)a-x= ((x+1)a+1)((x+1)a -x)$$

 

[SammyS]

Can you tell me @SammyS how you factored the second last step to go to the last one?

Let me copy and paste those two steps for you as an image below.

View attachment 343214

Many thanks.

You asked how to go from

$\displaystyle \quad (x+1)^2 a^2+(1-x)(x+1)a-x$

to

$\displaystyle \quad ((x+1)a+1)((x+1)a-x)$

You have mentioned using grouping to factor in the OP. That's what @Hill uses in Post #16.

Adding some detail to doing the grouping:
First you need to split the "middle term", $\displaystyle \quad (1-x)(x+1)a\ \quad$ into two terms. The leading term (the $a^2$ term) has two factors of $(x+1)\,,$ so in the middle term, you need to split $(1-x)$ factor, keeping the $(x+1)$ factor. This gives you the following.

$\displaystyle \quad (x+1)^2 a^2+(1)(x+1)a-x(x+1)a-x$

From this point, continue as @Hill did.

$=((x+1)a+1)(x+1)a-((x+1)a+1)x$

$=((x+1)a+1)((x+1)a-x)$