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Not sure if this belongs here, but I'm slowly trudging through my studies for Math 1 and wondered if y'all could give feedback and/or corrections on the following factorisation question:

$$ \text{Factorise}: f(x) = x^3+4x^2+3x $$

Firstly, the GCD of the above is $x$:

$$x(x^2+4x+3)$$

Now take $x^2+4x+3$ and factorise that:

$$ x^2+4x+3 $$

Using the box method, enter the first term $x^2$ into the upper left corner, and the last term $3$ into the lower right corner.

\begin{array}{|c|c|} \hline x^2 & \\ \hline & 3 \\ \hline \end{array}

Then find HCF of 3:

$$3\\ 1 | 3 $$

Enter the values $1x$ and $3x$ into the other two boxes:

\begin{array}{|c|c|} \hline x^2 & 1x \\ \hline 3x& 3 \\ \hline \end{array}

Now factorise the rows and columns:

$$ x^2 + 1x = x(x+1)\\ x^2 + 3x = x(x+3)\\ 1x + 3=1(x+3)\\ 3x +3=3(x+3) $$

Therefore: $$x^2+4x+3=(x+1)(x+3)$$ It follows that: $$f(x) = x^3+4x^2+3x=x(x+1)(x+3)$$

Any feedback on method and/or corrections are gladly accepted! Be gentle, I'm a struggling student you know...

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    $\begingroup$ If you want to check your work, multiply out $x(x+1)(x+3)$, and you will find it equals $x^3+4x^2+3x$. $\endgroup$ – vadim123 Nov 18 '16 at 18:40
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Had never heard of the box method before I saw you use it!

When you got to the part of factoring $x^2 + 4x + 3$ I would go and find the roots of it, because with the roots one can also factor the polynomial.

Upon finding that $-1$ and $-3$ are the roots, I would know $x^2 + 4x + 3 = (x - (-3))(x - (-1)) = (x + 1)(x + 3)$.

This works for a general polynomial of degree $n$. If a $n$-degree polynomial has roots $\lambda_1, \cdots, \lambda_n$, then the polynomial is equal to $(x - \lambda_1)\cdots(x - \lambda_n)$

Of course we not always have access to all the polynomial's roots at once, but we can partially factorise and work our way through that. Let us go through the polynomial $p = x^4 + 3x^3 - x^2 - 3x$.

First obvious thing is that $\lambda_1 = 0$ and thus $p = x^4 + 3x^3 - x^2 - 3x = x(x^3 + 3x^2 - x - 3)$.

Now comes the tricky part. Finding the roots of $p_1 = x^3 + 3x^2 - x - 3$. What I always start by doing is trying some small numbers. Guessing $\lambda_2 = 1$ turns out to be fine and thus we can factor $p_1$. Now there is some polynomial $p_2$ of degree 2 that multiplied by $(x - \lambda_2) = (x - 1)$ gives $x^3 + 3x^2 - x - 3$. To calculate such polynomial I refer you to Ruffini's rule, which is just a faster way to factorize a polynomial when you know one root. Then we get $p_2 = x^2 + 4x + 3$ which was what you factorized above. Since this is a 2nd degree polynomial, we could keep trying to guess roots, use your box method, or using the quadratic formula to find its roots.

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    $\begingroup$ The box method is really cool, particularly when in particular multiplying trinomials with trinomials. $\endgroup$ – imranfat Nov 18 '16 at 18:44
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    $\begingroup$ First time for me too ! $\endgroup$ – Claude Leibovici Nov 18 '16 at 18:45
  • $\begingroup$ @imranfat had never heard of it. should I google it and learn it? $\endgroup$ – RGS Nov 18 '16 at 18:47
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    $\begingroup$ Bear in mind guys, I'm a total novice! I like systematic approaches to problems, and this one helps me remember the steps required. Video I used is here: youtube.com/watch?v=_Wb_CT-1VN8&app=desktop $\endgroup$ – Dan Nov 18 '16 at 18:49
  • $\begingroup$ @DanB your approach is fine. Also note that the method I talked about always works. If you are able to find the roots, then you are able to factorise it! I will add something to my answer. $\endgroup$ – RGS Nov 18 '16 at 18:53
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I wouldn't know if this really qualifies as an answer, but what you're asking for is essentially an opinion of methodology.

First of all, your factorisation is correct: you can check the result simply by doing the multiplication.

That said, it looks like you could have saved a lot of effort by using a couple of different techniques.

To begin with, the polynomial $x^3+4x^2+3x$ is clearly divisible by $x$ (as you noted), so we factor it out to get $x(x^2+4x+3)$.

Here is where it gets interesting: to factor the degree 2 polynomial, I usually take one of these two roads.

First: observe that $(x+a)(x+b)=x^2+(a+b)x+ab$, so if we can guess two numbers $a$ and $b$ such that, in our case, we have $ab=3$ and $a+b=4$, we are done.

In this case you might have been able to spot that $3\cdot 1=3$ and $3+1=4$, so $(x+1)(x+3)$ is the factorization we're looking for. Trust me, it gets easier with practice and it's a huge time- and effort-saver for simple degree 2 polynomials with integer coefficients.

Else we could try and guess just one root, and then use polynomial long division. This has the advantage of working with any degree of polynomials, as long as you can guess one root. I usually try $1$, $-1$ and maybe $2$, but if those don't work I'm better off using some other technique.

And finally, for degree 2 polynomials you have the general formula to find the roots, which has the advantage of always working (in the sense that it will always find all real roots there are to find), but can get messy and is, in my honest opinion, rather boring.

Now, this is not to say that any of what you did is incorrect or even strictly less efficient: it is ultimately a matter of personal taste, but I feel that having as many tools as possible in your pocket can only do you good.

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    $\begingroup$ Thanks! It was indeed really a question of methodology, although I wasn't completely sure of the answer! The guessing method sounds like something that with practice would indeed save a lot of time. $\endgroup$ – Dan Nov 18 '16 at 19:00
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Everything is more or less correct about the way you approach the problem. You could have also opted for the middle term factorisation method or the method of vanishing method. However you have to correct one thing ...

Now factorise the rows and columns: $$ x^2 + 1x = x(x+1)\\ x^2 + 3x = x(x+3)\\ 1x + 3=1(x+3)\\ 3x +3=1(x+3) $$


The last line ought to be $3x +3=3(x+1)$.

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  • $\begingroup$ Thank you! I've edited my post. Will look up the methods you've suggested, can never hurt to have more tools in the arsenal! $\endgroup$ – Dan Nov 18 '16 at 19:01
  • $\begingroup$ @DanB You're welcome. $\endgroup$ – SchrodingersCat Nov 18 '16 at 19:02
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To factor $x^3+4x^2+3x$, we notice that we can factor $x$ out. Therefore, we get$$x^3+4x^2+3x=x(x^2+4x+3)\tag1$$ Now, we need to see if $x^2+4x+3$ can be factored as a product of two linear terms. An easy way to factor a monic polynomial is to find two numbers $r,s$ that sum to the negated value of $b$ and have a product of $c$ in $x^2+bx+c$.

In other words, we have $$\begin{align*} & r+s=-b\\ & rs=c\end{align*}\tag2$$ for $x^2+bx+c$. In your example, we see that $-b=-4$ and $c=3$. Messing around, we see that when $r=-1,s=-3$, the requirements are met. Thus,$$x^3+4x^2+3x=x(x-1)(x-3)$$

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X(X+1)(X+3)

Therefore (X+1)(X+3) = X^2+4X+3.

Therefore multiplying this by X you get X^3+4X^2+3X.

So yes well done that is correct.

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Well done. And for those of us unaware of the box method, the following would have also worked: $$x^2+4x+3=x^2+4x+4-1=(x+2)^2-1=(x+2-1)(x+2+1)=(x+1)(x+3)$$

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Avoid the 'box method' which will only work for quadratics with nice integer roots and doesn't really tell you anything about what's going on.

Suppose we have a quadratic $p(x)$ that factorizes as:

$$ p(x) = (x + s)(x + t) $$

where we don't know what $s$ and $t$ are yet. If we multiply out the brackets, then we get:

$$ p(x) = x^2 + sx + tx + st = x^2 + (s+t)x + st $$

Now suppose that we are given $p(x)$ in the form

$$ p(x) = x^2 + bx + c $$

In order to factorize $p(x)$, all we need to do is find two numbers $s$ and $t$ such that $$s+t=b$$ and $$st=c$$

In your case, you want to find $s$ and $t$ such that $s+t=4$ and $st=3$. It shouldn't take you very long to realize that $s=3$ and $t=1$ will do. Then you can immediately factorize the polynomial as $(x+3)(x+1)$.

Slogan: To factorize a quadratic of the form $x^2+bx+c$, just find two numbers that add to give $b$ and multiply to give $c$.


Just for interest

This method should work for all the quadratics you see in Math 1, and I hope that it's a bit clearer what's going on compared to the 'Box Method'. You might be interested in how we solve quadratic equations that don't have nice solutions. For example, suppose we were trying to factorize $$ x^2+x-1 $$ We want to find two numbers that add together to give $1$ and multiply together to give $-1$. It turns out that the right two numbers are $$ \frac{1+\sqrt{5}}{2} $$ and $$ \frac{1-\sqrt{5}}{2} $$ How did I work those out? Well, I'll show you a method that was invented by the ancient Babylonians. Suppose we have two numbers $b$ and $c$ and we're trying to find two numbers $s$ and $t$ such that $s+t=b$ and $st=c$.

If we square the first equation, we get $$ b^2 = (s+t)^2 = s^2 + 2st + t^2 $$ Now since $st=c$, we can subtract $4c$ from the left hand side and $4st$ from the right hand side to get: $$ b^2-4c = s^2 - 2st + t^2 $$ But $s^2 - 2st + t^2=(s-t)^2$, so we may write $$ s-t = \sqrt{b^2-4c} $$ (Here, we are free to assume that $s\ge t$, so this is the positive square root.) Now we can recover $s$ and $t$: $$ s = \frac{(s+t) + (s-t)}{2} = \frac{b+\sqrt{b^2-4c}}{2} $$ $$ t = \frac{(s+t) - (s-t)}{2} = \frac{b-\sqrt{b^2-4c}}{2} $$

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