# Taking Calculus in a few days and I still don't know how to factorize quadratics

Taking Calculus in a few days and I still don't know how to factorize quadratics with a coefficient in front of the 'x' term. I just don't understand any explanation. My teacher gave up and said just use the formula to find the roots or something like that..

Can someone explain to me simply how I would step by step factorize something like $4x^2 + 16x - 19$ ?

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Are you interested in how to find factors where the quadratic has integer coefficients and happens to have nice rational roots like $x=\frac{a}{b}$? Or do you simply want to know how to get the roots of a quadratic. The example you gave does not have a simple factorisation, the roots are awkward $-2\pm\frac{\sqrt{35}}{2}$. –  almagest Sep 2 '14 at 9:33
@almagest I chose a random question, x = a/b is how I want to be able to solve ones like that –  Idin Sep 2 '14 at 9:34
There are two things you can do. One is simply to solve the quadratic and see if the roots are rational or not. In practice you just memorise the formula for the solutions and use it. It is well worth memorising because it comes up repeatedly in all areas of maths. You might also want to learn how to derive the formula ("completing the square"). The other approach is just to look for nice rational solutions. That is what many people use, but it takes a little more practice. –  almagest Sep 2 '14 at 9:38
This question was shown among related questions and it seems it might be useful for you: math.stackexchange.com/questions/39917/… –  Martin Sleziak Sep 2 '14 at 10:41
Being in the situation you´re describing I would highly advice you to spend the next couple of days on khanacademy.org (not affliated with them whatsoever of course, they provide online courses in a lot of things, and their math courses are extraordinary good) –  David Mulder Sep 2 '14 at 14:11

The term $4x^2+16x$ is almost square of $2x+4$, more precisely $$4x^2+16x=(2x+4)^2-4^2.$$ Therefore, we have \begin{align} 4x^2+16x-19&=(2x+4)^2-4^2-19\\ &=(2x+4)^2-35\\ &=(2x+4)^2-\left(\sqrt{35}\right)^2. \end{align} Knowing that $$p^2-q^2=(p-q)(p+q),\tag1$$ then we have \begin{align} 4x^2+16x-19 &=\left(2x+4-\sqrt{35}\ \right)\left(2x+4+\sqrt{35}\ \right).\qquad\blacksquare \end{align}

Here is a general approach to factorize a quadratic equation. Suppose that we want to factorize quadratic equation $$ax^2+bx+c=0.\tag2$$ Now, multiplying $(2)$ by $4a$ yields $$4a^2x^2+4abx+4ac=0.\tag3$$ The term $4a^2x^2+4abx$ is almost square of $2ax+b$, more precisely $$4a^2x^2+4abx=(2ax+b)^2-b^2,\tag4$$ then $(3)$ turns out to be \begin{align} 4a^2x^2+4abx+4ac&=(2ax+b)^2-b^2+4ac\\ &=(2ax+b)^2-(b^2-4ac)\\ &=(2ax+b)^2-\left(\sqrt{b^2-4ac}\right)^2. \end{align} Using $(1)$, we have $$4a^2x^2+4abx+4ac=\left(2ax+b+\sqrt{b^2-4ac}\ \right)\left(2ax+b-\sqrt{b^2-4ac}\ \right).\tag5$$ Final step, dividing $(5)$ by $4a$ yields $$ax^2+bx+c=\color{blue}{\frac{\left(2ax+b+\sqrt{b^2-4ac}\ \right)\left(2ax+b-\sqrt{b^2-4ac}\ \right)}{4a}}.$$ The process might look complicated, but once you understand the logic behind the process, especially for $(4)$, it will not be necessary anymore to memorize every step and your hand will automatically drive you to factorize every quadratic equation.

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Long time since I was doing this, so just for the sake of nostalgia, what if you reach something like: (2x+4)^2+35 instead of (2x+4)^2-35? –  bolov Sep 2 '14 at 19:33
@bolov Then the equation has no real roots, since $(2x+4)^2+35\ge 35$. Hence $\not\exists x$ such that $(2x+4)^2+35=0$. –  user26486 Sep 2 '14 at 19:41
@mathh thank you. And I feel so stupid now. As I was saying long time since I was doing quadratic equations but that was right in my face. –  bolov Sep 2 '14 at 19:46
@bolov: far better to ask basic questions and learn something than not ask and not learn anything. –  G. Bach Sep 2 '14 at 21:14
@bolov Please don't feel stupid because we are all still learning. $\ddot\smile$ –  Tunk-Fey Sep 3 '14 at 4:40

First of all, to find the roots of $4x^2+16x-19$ we have to calculate the discriminant:

If the second degree polynomial is of the form $$ax^2+bx+c=0$$ the discriminant is given from the formula: $$\Delta=b^2-4 \cdot a \cdot c$$

So the discriminant in this case is the following:

$4x^2+16x-19=0 \Rightarrow \Delta=16^2-4 \cdot 4 \cdot (-19)=256+304 \Rightarrow \Delta=560$

Then the solutions are given from the formula: $$x_{1.2}=\frac{-b \pm \sqrt{\Delta}}{2 \cdot a}$$

Therefore we have the following solutions:

$x_{1,2}=\frac{-16 \pm \sqrt{560}}{2\cdot 4}=\frac{-16 \pm \sqrt{560}}{8}$

$x_1=\frac{-16-4 \sqrt{35}}{8}=-2-\frac{\sqrt{35}}{2} \text{ and } x_2=\frac{-16 +4\sqrt{35}}{8}=-2+\frac{\sqrt{35}}{2}$

Knowing that $$ax^2+bx+c=0 \Rightarrow a(x-x_1)(x-x_2)=0$$

we have the following:

$$4x^2+16x-19=4 \left ( x- \left (-2-\frac{\sqrt{35}}{2} \right ) \right ) \left ( x- \left (-2+\frac{\sqrt{35}}{2} \right ) \right )$$

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Although it may be mathematically accurate, I don't think the questioner will be able to follow what you did here, much less extrapolate the general method from it. –  Foo Bar Sep 2 '14 at 16:34
@FooBar If he knows how to solve a quadratic equation, all he needs to know is that $ax^2+bx+c=a(x-x_1)(x-x_2), \forall a,b,c,x\in\mathbb C, a\neq 0$. But yes, the general method does not follow. It is simply $$a_nx^n+a_{n-1}x^{n-1}+\cdots+a_1x+a_0=a_n(x-x_1)(x-x_2)\cdots(x-x_n), \forall a_i,x\in\mathbb C, a_n\neq 0, i\in\{0,1,2,\ldots,n\}$$,though. –  user26486 Sep 2 '14 at 20:03
I wouldn't have written it this way, but the only thing lacking to be "general" for the quadratics is an explanation of what to do if the discriminant is negative (no real roots) or zero (the quadratic is a square). –  hobbs Sep 3 '14 at 5:38
@hobbs Oh, I assumed that Foo Bar meant 'general' in the sense that from this solution you can't extrapolate on how to find the factorization of every existing polynomial (while Tunk-Feys method applies to all polynoamials, since you can 'complete the cubic', 'complete the quartic', etc.), while he actually meant 'general' as in 'you know how to factorize all quadratics just by reading this answer'. But I might've been correct, we can't know what he meant until he tells us. –  user26486 Sep 3 '14 at 18:11
@mathh maybe :) –  hobbs Sep 3 '14 at 20:29

For well behaved problems with integer solutions:

\begin{align} &\text{Quadratic is of form }ax^2+bx+c:&&\ 3x^2-39x+120=0\\ \\ &\text{We need }a=1,\text{ so }\div \text{ both sides by }3:&&\ x^2-13x+40=0\\ \\ &ax^2+bx+c&\text{now }&\ a=1,\ b=-13,\ c=40\\ &a=1,\text{ so we can find the factors:}&&\ \underline{\hspace{0.5cm}}+\underline{\hspace{0.5cm}}=b,\text{ and }\underline{\hspace{0.5cm}}\times\underline{\hspace{0.5cm}}=c\\ \\ &\text{If in doubt, write factors of }c:&&\ \{1,40\},\{2,20\},\{4,10\},\{5,8\}\\ \\ &\{5,8\}\text{ are ones to use:}&&\ (-5)+(-8)=-13\\ &&&\ (-5)\times(-8)=40\\ \\ &&\text{so }&\ (x-5)(x-8)=0&\\ &&&\ x-5=0\text{ or }x-8=0&\\ \\ &\text{Answer:}&&\ x=5\text{ or }x=8\\ \end{align}

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Please add a source (and crop the top and bottom of the image). As it stands now this answer is just plagiarism. –  bdesham Sep 2 '14 at 13:39
Source is me. I made it for previous student, on my phone now so can't add tex. Will update later. Kindly remove downvote please, unless quality of answer is in question. –  martin Sep 2 '14 at 13:49
I voted more than 15 minutes ago so I can't remove the downvote until you edit the post. Let me know when you do and I'll change the vote. –  bdesham Sep 2 '14 at 13:55
Really nice note! +1 –  Tunk-Fey Sep 2 '14 at 16:55
@Tunk-Fey, cheers! :) –  martin Sep 2 '14 at 17:03

So if you just want to search for nice factors, you are given $Ax^2+Bx+C$ and you want to express it as $(sx+r)(vx+u)$, which means that $sv=A$ and $ru=C$. If you take the example you gave $4x^2+16x-19$, then the only factors of 19 are 1 and 19, and the only factors of 4 are 1,2,4. So it would have to be $(4x+19)(x-1)$ or one of several other possibilities. In practice it is fairly quick to check that none of them work.

Suppose you were looking at $4x^2+15x-19$. So it is either $(2x+h)(2x-k)$ or $(4x+h)(x-k)$ or $(4x-h)(x+k)$ (assuming $h,k$ are both positive). The first won't work because we want an odd coefficient for $x$, not an even one. We must have $h,k=1,19$ in some order. And so on.

This is not a particularly systematic procedure, but it is fast after a little practice.

If you want to go the other route, start by dividing through by the coefficient of $x^2$, so you have something like $x^2+Bx+C$, where $B,C$ may be fractions. Now you need to remember that $(x+a)^2=x^2+2ax+a^2$. So you take $(x+\frac{B}{2})^2=x^2+Bx+\frac{B^2}{4}$. You then need to adjust the constant term, so you end up with $(x+\frac{B}{2})^2=\frac{B^2}{4}-C$. Provided the rhs is not negative, you can not take the square root to get the solution (a positive and a negative square root are possible, so two solutions).

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The first 3 paragraphs are a good thing to try, but they only apply to quadratics with rational roots. If that fails, you need to use other methods. –  Foo Bar Sep 2 '14 at 16:47
If you read the comments under the question, that is precisely what was wanted. –  almagest Sep 2 '14 at 16:52
Nice explanation. +1 –  Tunk-Fey Sep 2 '14 at 16:55

By completing the square: $4x^{2} + 16x - 19 \\ = 4[x^{2} + 4x] - 19 \\ = 4[(x + 2)^{2} - 4] -19 \\ = 4(x+2)^{2} - 16-19 \\ = 4(x+2)^{2} - 35 \\ = (2(x+2) - \sqrt{35})(2(x+2) + \sqrt{35}) \\ = 4[(x+2) - \frac {\sqrt{35}}{2}][(x+2) + \frac {\sqrt{35}}{2}]$

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The reason why you are confused is because this thing is really more complicated than it looks. The thing you do is much easier that the reason you do it.

I'll try to go step by step. Do not be discouraged, but the rabbit hole is kinda deep =P

Step 1: Find the roots of the quadratic (as the roots of your equation are ugly, lets use $r_1$ and $r_2$)

Now, if you want to write this polinomial P using only first degree equations (and numbers) and multiplying them, you know that (x-$r_1$) and (x-$r_2$) must appear. (either them or them multplied by a number, really ...)

Why?

If (x-$r_1$) does not appear, then the expression you created will not be zero, when you try to evaluate it with $r_1$

If you try to say P=(x+$r_5$)(x-$r_7$), and you put $r_1$ in, you dont get zero. ($r_1$+$r_5$) is not zero and ($r_1$-$r_7$) is not zero either. Multipling two non-zero numbers, we get something that is not zero (unless you picked $r_5$ as $-r_1$ or $r_7$ as $r_1$ =P)

So we have (x-$r_1$)(x-$r_2$) as an expression that is zero in the right moments. That is nice, but not enough. It still has to match P in all other places.

Step 2: Pick the polinomial P and the polinomial H=(x-$r_1$)(x-$r_2$). Evaluate both on a number that is not a root (say, 1) and compare. If your polinomial H has a result 4 times smaller, that we can say that P=4(x-$r_1$)(x-$r_2$)

• We knew that the polinomial we were trying to factor had (x-$r_1$) and (x-$r_2$) as "factors that involved x". Why only those ? If you add another "factor with x", you are adding another point for the polinomial to be zero. Another root, that it does not have
• That makes it clear why we COULD take that last step: surely there were no further "factors with x". All we could do from there was to multiply by a number
• Other answers will say that the 4 came from the coefficient of $x^2$. That is also true! say you picked H=(x-$r_1$)(x-$r_2$) and did the multiplication: you would get ($x^2$ -x*something + something else). And this should look like the original P, right ? Looking at H, we can tell that, after the multiplication, $x^2$ will appear without a coefficient. So, to get P, we have to multiply H by the coefficient of $x^2$ in P (a.k.a 4)
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The way to factor any quadratic is to use the curious property of all quadratics that with the proper constant term, the quadratic would be a perfect square. The first two terms determine what the necessary constant term would be. So get the necessary constant term by squaring the first two terms. Then add or subtract enough to both sides of the original equation to make the left side a square. Now we can take the square root of both sides. That's it. This is precisely how the formula for roots of a quadratic is generated.

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Take a simple example first :

Say $x^2+5x+6=x^2+3x+2x+6=x(x+3)+2(x+3)=(x+3)(x+2).$

This can also be done by the Sreedhara-Acharya Method of solving :

$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$, so in that case

$x=\frac{-5\pm\sqrt{5^2-(4\times 1\times6)}}{2\times 1}=\frac{-5\pm 1}{2}=-3$ $or -2$.

In the first case, we factorized the quadratic formula, in the second case we computed the numbers satisfying the quadratic equation.

Also note that a quadratic polynomial can be formed by its roots as follows :

If $a$ and $b$ are the roots of a quadratic equation, then the quadratic equation is given by :

$x^2-(a+b)x+(ab)$. Verify for the quadratic polynomial we took,

Its roots were $-3$ and $-2$ So the equation becomes

$x^2-(-3+(-2))x+((-3)\times(-2))=x^2+5x+6$ as required.

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If we multiply two monomials $$(x-a)(x-b)=x^2-(a+b)x+ab$$ Think of this in terms of the arithmetic mean of the roots $\mu=(a+b)/2$ and the geometric mean $\gamma=\sqrt{ab}$: $$x^2-2\mu+\gamma^2$$ Then \begin{align}\mu^2-\gamma^2& =\left(\frac{a+b}{2}\right)^2-ab\\ & =\frac{a^2+2ab+b^2}{4}-\frac{4ab}{4}\\ & =\frac{a^2-2ab+b^2}{4}\\ & =\left(\frac{a-b}{2}\right)^2\\ & =\rho^2 \end{align} Where $\rho=(a-b)/2$ is the "radius" from the mean $\mu$ to the roots $a,b$. We can recover the roots from $$\{a,b\}=\mu\pm\rho=\mu\pm\sqrt{\mu^2-\gamma^2}$$ Taking your example, we see that $4x^2+16x−19=4(x^2+4x-19/4)=4(x^2-2(-2)x-19/4)$ so $\mu=-2$ and $\gamma=-19/4$, giving us the roots \begin{align} \{a,b\}& =-2\pm\sqrt{(-2)^2-(-19/4)}\\ & =-2\pm\sqrt{4+19/4}\\ & =-2\pm\sqrt{35/4}\\ & =-2\pm\sqrt{35}/2 \end{align} If you need to, you can then rewrite the original expression as \begin{align} 4x^2+16x−19& =4(x-(-2+\sqrt{35}/2))(x-(-2-\sqrt{35}/2))\\ \quad& =4(x+2-\sqrt{35}/2)(x+2+\sqrt{35}/2)\end{align}

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Note that $\rho$ is just half of the discriminant. –  Scott Centoni Sep 3 '14 at 20:28

Polynomial factorization is directly related to root search.

Assume your second degree trinomial $C_2x^2+C_1x+C_0$ has two distinct roots, $r_0$ and $r_1$.

Then $$C_2x^2+C_1x+C_0=C_2(x-r_0)(x-r_1).$$

Indeed, the coefficient of the leading term is $C_2$ on both sides, and the two polynomials evaluate to $0$ at $r_0$ and $r_1$: by definition of $r_0$, $$C_2r_0^2+C_1r_0+C_0=0,$$ and obviously $$C_2(r_0-r_0)(r_0-r_1)=0.$$ Similarly at $r_1$.

This is enough to prove that the two polynomials are identical (deducing the leading quadratic term, there is only one linear polynomial through two given points $(r_0,-C_2r_0^2)$, $(r_1,-C_2r_1^2)$).

The reasoning is still valid for a double root, let $r$, that is such that the polynomial and its first derivative vanish. $$C_2x^2+C_1x+C_0=C_2(x-r)^2.$$ By definition of $r$, $$C_2r^2+C_1r+C_0=0,\\2C_2r+C_1=0,$$ and obviously $$C_2(r-r)^2=0,\\2C_2(r-r)=0.$$ This property generalizes to polynomials of any degree.

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