Factoring Quadratics

If we have a quadratic of the form $y=ax^2 + bx + c$ it can be useful to put it in the form $y=(px+q)(rx+s)$ , as this will give us two equations that will give us both of the solutions of the quadratic, one is $0=(px+q)$ and the other is $0=(rx+q)$ .

So first if we are given the factored form of the quadratic, we can expand it to give our $y=ax^2 + bx + c$ form:

$\begin{split} y & = (px+q)(rx+s) \\ & = px \cdot rx + px \cdot s + q \cdot rx + q \cdot s \\ & = prx^2 +(ps + qr)x + qs \end{split}$

So in this example $a=pr$ , $b=ps+qr$ and $c=qs$

Moving from the $y=ax^2 + bx + c$ form to the factored form requires first finding two numbers that add together to give $b$ and multiply together to get $ac$ , we then write the quadratic in the form $y=ax^2 + px + rx + c$ we can then factor the first and last terms. So an example, given $y=2x^2 + 5x + 3$ , so we can see that we need $ac=6$ and $b=5$ which means that we can choose from $1+4=5$ and $2+3=5$ . In this case we can see that $2 \cdot 3 = 6$ , so we can use $2+3=5$ , we then write the quadratic out using this:

$\begin{split} y & = 2x^2 + 5x + 3 \\ & = 2x^2 + 2x + 3x + 3 \\ \end{split}$

So we can now factor the first two terms and last two terms separately:

$\begin{split} y & = 2x^2 + 2x + 3x + 3 \\ & = 2x(x + 1) + 3(x + 1) \end{split}$

We now have a clearly obvious common factor $(x+1)$ in this case. We can now factor this out to give:

$\begin{split} y & = 2x(x + 1) + 3(x + 1) \\ & = (2x+3)(x + 1) \end{split}$

So we can check this result by expanding it:

$\begin{split} y & = (2x+3)(x + 1) \\ & = 2x \cdot x + 2x \cdot 1 + 3 \cdot x + 3 \cdot 1 \\ & = 2x^2 + 2x + 3x + 3 \\ & = 2x^2 + 5x + 3 \\ \end{split}$

So we can see that $2x^2 + 5x + 3=(2x + 3)(x + 1)$ as required. From this solution we can see that there are zero points at $0=2x+3$ and $0=x+1$ , so in this case there are zeros at $x=-1$ and $x=-\frac{3}{2}$ , which we can see on the graph of $y=(2x + 3)(x + 1)$

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If we now complete the square of this example we will also get the coordinates of the minimum value. We need $h =- \frac{b}{2a}$ and $k = c - \frac{b^2}{4a}$ , so:

$\begin{split} h & = -\frac{b}{2a} \\ & = -\frac{5}{2 \cdot 2} \\ & = - \frac{5}{4} \end{split}$

and

$\begin{split} k & = c -\frac{b^2}{4a} \\ & = 3 - \frac{5^2}{4 \cdot 2} \\ & = 3 - \frac{25}{8} \\ & = \frac{24}{8} - \frac{25}{8} \\ & = - \frac{1}{8} \\ \end{split}$

The completed the square form is $y = a ( x - h ) ^ 2 + k$ , so $y = 2(x + \frac{5}{4})^2 - \frac{1}{8}$ , so we now know that the minimum is at the coordinates are $(h,k)$ and is at $(-\frac{5}{4},-\frac{1}{8})$ . Wrapping up we have 3 forms of the same quadratic: