Tips for factoring quadratic expressions

[MarkFL]

The purpose of this tutorial is to provide students of algebra with techniques and tips for factoring quadratic expressions. In my experience as a tutor, I have found this can be one of the more difficult and challenging topics for students.

I invite anyone with any techniques of their own to add to this topic to give our readers as comprehensive a list of tips/tricks as possible.

Note: all coefficients, both in the quadratic forms and in the factored forms are integers.

Typically, a quadratic expression is given in the form:

(1) $ax^2+bx+c$

and the goal is to express this expression as the product of two linear factors:

$ax^2+bx+c=(dx+e)(fx+g)$

First of all, students may wonder why we bother to factor quadratics. Well, this has to do with what's called the zero-factor property. Many times in the application of quadratic equations, we express the equation in standard form $ax^2+bx+c=0$. In short, if we have two factors equal to zero, then we know that the product must be zero when one or the other of the factors is equal to zero. Since finding the root of a linear factor is very straightforward, this gives us a neat and easy way to find the two roots of the quadratic expression.

The way I begin, is (referring to (1)) by looking at the product $ac$ and try to find two factors of this product whose sum is $b$.

Suppose we are given to factor:

$x^2+8x+12$

The product $ac=1\cdot12=12$ so we want two factors of $12$ whose sum is $8$. Thinking about the factor pairs of $12$ which are $(1,12),\,(2,6),\,(3,4)$ we see the pair $(2,6)$ has a sum of $8$. Since $a=1$, we know the coefficient of $x$ in both factors will also be $1$, and so we have:

$x^2+8x+12=(x+6)(x+2)$

If we're unsure, we may check our work by expanding the right side using the FOIL method:

$(x+6)(x+2)=x^2+2x+6x+12=x^2+8x+12$

So, it checks out. Why does this method work? Let's do a bit of investigation:

To keep things simple for now, let's consider the case where $a=1$:

$x^2+bx+c=(x+d)(x+e)=x^2+ex+dx+de=x^2+(d+e)x+de$

Equating coefficients, we find we require:

$d+e=b$

$de=c$

As you can see, this implies we need to find two factors of $c$ whose sum is $b$. This is why we look for two such factors.

Now, let's step things up just a bit and let $a$ be a prime number. Since a prime number only has itself and $1$ as factors, we know the factored form must be:

$ax^2+bc+c=(ax+d)(x+e)=ax^2+(d+ae)x+de$

So, we see now, we require:

$d+ae=b$

$de=c$

If we mutiply the second equation by $a$, we may write:

$d(ae)=ac$

So, these equations imply that we need two factors of $ac$ whose sum is $b$. Let's now apply this to the quadratic:

$3x^2-13x-10$

We want two factors of $3(-10)=-30$ whose sum is $-13$. These are $(2,-15)$. Since $a=3$, we look for the factor that is divisible by $3$ and that is $-15$. So we set:

$ae=3e=-15,\,e=-5$

$d=2$

and so we now have:

$3x^2-13x-10=(3x+2)(x-5)$

Now, let's examine the case when $a$ is composite, which means there is more than 1 possible factor pair of $a$, and so we need to write the factored form as:

$ax^2+bx+c=(dx+e)(fx+g)=dfx^2+(dg+ef)x+eg$

Equating coefficients, we find:

$df=a$

$dg+ef=b$

$eg=c$

If we multiply the first equation by the third, we have:

$(dg)(ef)=ac$

This, along with the second equation above implies we need two factors of $ac$ whose sum is $b$. Suppose we are given to factor:

$6x^2+13x-28$

We want two factors of $6(-28)=-168$ whose sum is $13$. They are $(-8,21)$.

Now, we look at the factor pairs of $6$ which are $(1,6),\,(2,3)$. Neither of the pair $(-8,21)$ is divisible by $6$, but $-8$ is divisible by $2$ and $21$ is divisible by $3$, so the factor pair of $6$ we need is $(2,3)$. And so we find we may set:

$d=2$

$f=3$

$dg=2g=-8\,\therefore\,g=-4$

$ef=3e=21\,\therefore\,e=7$

and so we have:

$6x^2+13x-28=(2x+7)(3x-4)$

Commentary to this tutorial should be posted here:

http://mathhelpboards.com/commentary-threads-53/commentary-factoring-quadratics-4203.html

 

[agentredlum]

Moderator edit: This commentary topic pertains to the following tutorial:

http://mathhelpboards.com/math-notes-49/factoring-quadratics-3396.html

I agree with everything MarkFL has posted so i'll just give a quick 'scheme' leaving out the details.

Another approach is to factor by grouping, consider,

$6x^2 - 17x - 45 $

$(6)(-45) = -270$

Now find 2 integers that multiply to -270 and sum to -17,

-27 with 10 work

rewrite $-17x$ as $-27x + 10x $

$6x^2 - 27x + 10x - 45 $

use grouping, paying mind to the signs,

$ (6x^2 - 27x) + (10x -45)$

factor,

$3x(2x - 9) + 5(2x - 9)$

$(2x - 9)(3x + 5)$

You can multiply this out using FOIL to confirm the factoring is correct.

:)

 

[mathmaniac1]

Re: Factoring Quadratics

$6x^2 - 17x - 45 $

$(6)(-45) = -270$

Now find 2 integers that multiply to -270 and sum to -17,

-27 with 10 work

How to do it:

We have to find p and q such that pq=6 x (-45)= 2 x 3 x 3 x 3 x 5

and p+q=-17

We observe that 17 is a prime.So p and q cannot have common factors (otherwise p+q will be have a common factor).

So either p or q will have all the 3 "3"s.Let it be p.
We have a 2 and 5 remaining which can go into either p or q.Let us first try both of them in q.
So we have p=27 and q=10

One and only one of p and q is negative and also the greater of p and q has to have the sign "-" (because the sum is negative).

So we have p + q = -27 + 10 = -17 and it works.If it hadn't worked then we had to exchange the 2 and 5 a couple of times until we get it right.

If p + q had a common factor,then we had to allow it to be the GCD of p and q and continue this.

 

[agentredlum]

Re: Factoring Quadratics

How to do it:

We have to find p and q such that pq=6 x (-45)= 2 x 3 x 3 x 3 x 5

and p+q=-17

We observe that 17 is a prime.So p and q cannot have common factors (otherwise p+q will be have a common factor).

So either p or q will have all the 3 "3"s.Let it be p.
We have a 2 and 5 remaining which can go into either p or q.Let us first try both of them in q.
So we have p=27 and q=10

One and only one of p and q is negative and also the greater of p and q has to have the sign "-" (because the sum is negative).

So we have p + q = -27 + 10 = -17 and it works.If it hadn't worked then we had to exchange the 2 and 5 a couple of times until we get it right.

If p + q had a common factor,then we had to allow it to be the GCD of p and q and continue this.

These are nice observations, i especially like the part of all 3's must belong to p or q. As the product AC gets large your observations can in fact reduce the work involved trying to find p and q.

:)

 

[Petrus]

Re: Factoring Quadratics

Hello,
I can add a method :)
Let's say we want to factor \(\displaystyle p(x)=-4x^2+24x-32\) we can use this formula \(\displaystyle p(x)=k*(x-a)(x-b)\)
we want to calculate crittical point (without derivate). In Sweden we learned a formula caled "pq-formula"(too calculate critical point in second polynom) as you can see there can't be any number or negative number infront of \(\displaystyle x^2\) so we start to break out -4 then we got \(\displaystyle -4(x^2-6x+8)\)
now we calculate the critical point \(\displaystyle x^2-6x+8=0\) and we get \(\displaystyle x_1=4\) and \(\displaystyle x_2=2\) now we use the formula \(\displaystyle p(x)=k*(x-a)(x-b)\) where k is our constant and a,b our crit point so we got \(\displaystyle p(x)=-4*(x-4)(x-2)\)
here is pq formula: