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 $4x^2 + 16x - 19$ ?

Solution 1:

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}

Addendum :

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.

Solution 2:

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:

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

Solution 3:

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}

Solution 4:

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).

Solution 5:

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}] $