# 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 $4x^2 + 16x – 19$ ?

#### Solutions Collecting From Web of "Taking Calculus in a few days and I still don't know how to factorize quadratics"

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
\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.

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:

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

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&\\
\\
\end{align}

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).

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

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)

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.

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.

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}

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.